# On weakly $1$-absorbing prime ideals of commutative rings††thanks: Key
Words: $1$-absorbing prime ideal, Weakly $1$-absorbing prime ideal, Prime
ideal, Weakly prime ideal. ††thanks:  2010 Mathematics Subject Classification:
13A15, 13C05.

M. J. Nikmehra, R. Nikandishb and A. Yassinea
aFaculty of Mathematics, K.N. Toosi University of Technology,
P.O. BOX 16315-1618, Tehran, Iran
<EMAIL_ADDRESS><EMAIL_ADDRESS>
bDepartment of Mathematics, Jundi-Shapur University of Technology,
P.O. BOX 64615-334, Dezful, Iran
<EMAIL_ADDRESS>

###### Abstract

Let $R$ be a commutative ring with identity. In this paper, we introduce the
concept of weakly $1$-absorbing prime ideals which is a generalization of
weakly prime ideals. A proper ideal $I$ of $R$ is called weakly $1$-absorbing
prime if for all nonunit elements $a,b,c\in R$ such that $0\neq abc\in I$,
then either $ab\in I$ or $c\in I$. A number of results concerning weakly
$1$-absorbing prime ideals and examples of weakly $1$-absorbing prime ideals
are given. It is proved that if $I$ is a weakly $1$-absorbing prime ideal of a
ring $R$ and $0\neq I_{1}I_{2}I_{3}\subseteq I$ for some ideals
$I_{1},I_{2},I_{3}$ of $R$ such that $I$ is free triple-zero with respect to
$I_{1}I_{2}I_{3}$, then $I_{1}I_{2}\subseteq I$ or $I_{3}\subseteq I$. Among
other things, it is shown that if $I$ is a weakly $1$-absorbing prime ideal of
$R$ that is not $1$-absorbing prime, then $I^{3}=0$. Moreover, weakly
$1$-absorbing prime ideals of PID’s and Dedekind domains are characterized.
Finally, we investigate commutative rings with the property that all proper
ideals are weakly $1$-absorbing primes.

##  1 Introduction

We assume throughout this paper that all rings are commutative with identity.
Let $R$ be a ring and $I$ be an ideal of $R$. The set of nilpotent elements of
$R$, the set of zero-divisors of $R$, the set of integers, and integers modulo
$n$ are denoted by $\sqrt{0}$, $Z(R)$, $\mathbb{Z}$ and $\mathbb{Z}_{n}$,
respectively. By a proper ideal $I$ of $R$ we mean an ideal with $I\neq R$. A
ring $R$ is called local if it has a unique maximal ideal. A ring $R$ is
called a reduced ring if it has no non-zero nilpotent elements; i.e.,
$\sqrt{0}=0$. For any undefined notation or terminology in commutative ring
theory, we refer the reader to [10].

In recent years, various generalizations of prime ideals have been studied by
several authors. For instance, in 1978, Hedstrom and Houston [7] defined the
strongly prime ideal, that is a proper ideal $P$ of $R$ such that for $a,b\in
K$ with $ab\in P$, either $a\in P$ or $b\in P$ where $K$ is the quotient field
of $R$. In 2003, Anderson and Smith [1] introduced the notion of a weakly
prime ideal, i.e., a proper ideal $P$ of $R$ with the property that for
$a,b\in R$, $0\neq ab\in P$ implies $a\in P$ or $b\in P$. So a prime ideal is
weakly prime. In 2005, Bhatwadekar and Sharma [6] introduced the notion of
almost prime ideal which is also a generalization of prime ideal. A proper
ideal $I$ of an integral domain $R$ is said to be almost prime if for $a,b\in
R$ with $ab\in I\setminus I^{2}$, then either $a\in I$ or $b\in I$, and it is
clear that every weakly prime ideal is an almost prime ideal. Another
generalization of prime ideal is $2$-prime; Indeed, a nonzero proper ideal $I$
of $R$ is called $2$-prime if whenever $a,b\in R$ and $ab\in I$, then
$a^{2}\in I$ or $b^{2}\in I$ (See [5] and [9] for more details). The notion of
$2$-absorbing ideals was introduced and investigated in 2007 by Badawi [2]. A
nonzero proper ideal $I$ of $R$ is called $2$-absorbing if whenever $a,b,c\in
R$ and $abc\in I$, then $ab\in I$ or $ac\in I$ or $bc\in I$. In [3], Badawi
and Darani extended the concept of weakly prime ideal to weakly $2$-absorbing
ideal. A proper ideal $I$ of $R$ is said to be a weakly $2$-absorbing ideal of
$R$ if whenever $a,b,c\in R$ with $0\neq abc\in I$ implies $ab\in I$ or $ac\in
I$ or $bc\in I$. In [4], $1$-absorbing primary ideal was introduced and
studied. A proper ideal $I$ of $R$ is called $1$-absorbing primary if for all
nonunit $a,b,c\in R$ such that $abc\in I$, then either $ab\in I$ or
$c\in\sqrt{I}$. Recall that $1$-absorbing prime ideals, as a generalization of
prime ideals, were introduced and investigated in [11]. In this paper, we
extend the concepts of weakly prime ideal and $1$-absorbing primary ideal to
weakly $1$-absorbing prime ideal.

A proper ideal $I$ of $R$ is called (weakly) $1$-absorbing prime if for all
nonunit elements $a,b,c\in R$ such that $(0\neq)abc\in I$, then either $ab\in
I$ or $c\in I$. Clearly, every weakly prime ideal is a weakly $1$-absorbing
prime ideal. However, the converse is not true.

This paper is organized as follows. In section $2$, first we give the
definition of weakly $1$-absorbing prime ideals (Definition 2.1). For
nontrivial weakly $1$-absorbing prime ideals see Example 2.3. Also, it is
proved (Theorem 2.6) that if $I$ is a weakly $1$-absorbing prime ideal of $R$
that is not $1$-absorbing prime, then $I^{3}=0$. In Theorem 2.10 we give a
condition under which every weakly $1$-absorbing prime ideal of $R$ is
$1$-absorbing prime. Among other things, it is shown (Theorem 2.12) that the
radical of a weakly $1$-absorbing prime ideal of a ring $R$ need not be a
prime ideal of $R$. Finally, it is proved (Theorem 2.23) that if $I$ is a
weakly $1$-absorbing prime ideal of a ring $R$ and $0\neq
I_{1}I_{2}I_{3}\subseteq I$ for some ideals $I_{1},I_{2},I_{3}$ of $R$ such
that $I$ is free triple-zero with respect to $I_{1}I_{2}I_{3}$, then
$I_{1}I_{2}\subseteq I$ or $I_{3}\subseteq I$. In Section $3$, we characterize
rings with the property that all proper ideals are weakly $1$-absorbing prime.

##  2 Weakly $1$-absorbing prime ideals

In this section, the concept of weakly $1$-absorbing prime ideals is
introduced and investigated. We start with the following definitions.

###### Definition 2.1

Let $R$ be a ring. A proper ideal $I$ of $R$ is called weakly $1$-absorbing
prime if for all nonunit elements $a,b,c\in R$ such that $0\neq abc\in I$,
then $ab\in I$ or $c\in I$.

###### Definition 2.2

Suppose that $I$ is a weakly $1$-absorbing prime ideal of a ring $R$ and
$a,b,c\in R$ are nonunit elements.

$(1)$ We say $(a,b,c)$ is a triple-zero of $I$ if $abc=0$, $ab\notin I$ and
$c\notin I$.

$(2)$ Suppose that $I_{1}I_{2}I_{3}\subseteq I$, for some ideals
$I_{1},I_{2},I_{3}$ of $R$. We say $I$ is free triple-zero with respect to
$I_{1}I_{2}I_{3}$, if $(a,b,c)$ is not a triple-zero of $I$, for every $a\in
I_{1},b\in I_{2}$, and $c\in I_{3}$.

It is easy to see that if $I$ is a weakly $1$-absorbing prime ideal that is
not $1$-absorbing prime, then there exists a triple-zero of $I$. Note that
every weakly prime ideal is a weakly $1$-absorbing prime ideal, and every
weakly $1$-absorbing prime ideal is a weakly $2$-absorbing prime ideal. The
following example provides a weakly $1$-absorbing prime ideal that is not
$1$-absorbing prime, a weakly $2$-absorbing ideal that is not weakly
$1$-absorbing prime and a weakly $1$-absorbing prime ideal that is not weakly
prime.

###### Example 2.3

$(1)$ Consider the ideal $J=\\{0,4\\}$ of $\mathbb{Z}_{8}$. Let
$R=\mathbb{Z}_{8}(+)J$ and $I=\\{(0,0),(0,4)\\}$. It is easy to see that $I$
is a weakly $1$-absorbing prime ideal, since $abc\in I$ for some $a,b,c\in
R\setminus I$ if and only if $abc=(0,0)$. But, $(2,0)(2,0)(2,0)\in I$ and both
$(4,0)\notin I$ and $(2,0)\notin I$, thus $I$ is a weakly $1$-absorbing prime
ideal that is not $1$-absorbing prime.

$(2)$ Suppose that $R=R_{1}\times R_{2}\times R_{3}$, where
$R_{1},R_{2},R_{3}$ are fields and $I=R_{1}\times\\{0\\}\times\\{0\\}$. One
can easily see that the ideal $I$ is not weakly $1$-absorbing prime, since
$(0,0,0)\neq(1,0,1)(1,0,1)(1,1,0)=(1,0,0)\in I$ and neither
$(1,0,1)(1,0,1)=(1,0,1)\in I$ nor $(1,1,0)\in I$. But $I$ is a weakly
$2$-absorbing ideal of $R$, by [3, Theorem 3.5]. Hence every weakly
$2$-absorbing ideal need not be weakly $1$-absorbing prime.

$(3)$ Consider the ideal $J=\\{0,4,8\\}$ of the ring $\mathbb{Z}_{12}$.
Clearly, $J$ is a weakly $1$-absorbing prime ideal of $\mathbb{Z}_{12}$ that
is not weakly prime.

###### Proposition 2.4

Let $R$ be a ring, $I$ a weakly $1$-absorbing prime ideal of $R$ and $c$ be a
nonunit element of $R\setminus I$. Then $(I:c)$ is a weakly prime ideal of
$R$.

###### Proof.

Assume that $0\neq ab\in(I:c)$ for some nonunit element $c\in R\setminus I$
such that $a\notin(I:c)$. We may assume that $a,b$ are nonunit elements of
$R$. As $0\neq abc=acb\in I$ and $ac\notin I$ and $I$ is a weakly
$1$-absorbing prime ideal of $R$, we have $b\in I\subseteq(I:c)$. Hence
$(I:c)$ is a weakly prime ideal of $R$.  $\Box$

Let $R$ be a local ring and $I$ be a weakly $1$-absorbing prime ideal of $R$
which is not $1$-absorbing prime. To prove $I^{3}=0$, the following lemma is
needed.

###### Lemma 2.5

Let $I$ be a weakly $1$-absorbing prime ideal of a local ring $R$ and let
$(a,b,c)$ be a triple-zero of $I$ for some nonunit elements $a,b,c\in R$. Then
the following statements hold.

$(1)$ $abI=acI=bcI=0$.

$(2)$ $aI^{2}=bI^{2}=cI^{2}=0$.

###### Proof.

$(1)$ Assume that $abI\neq 0$. Then there exists $x\in I$ such that $abx\neq
0$. Therefore $ab(c+x)\neq 0$. Since $ab\notin I$ and $I$ is a weakly
$1$-absorbing prime ideal, $(c+x)\in I$, and hence $c\in I$, a contradiction.
Thus $abI=0$. Now suppose that $acI\neq 0$. Then there exists $x\in I$ such
that $acx\neq 0$. Therefore $a(b+x)c\neq 0$. Since $c\notin I$ and $I$ is a
weakly $1$-absorbing prime ideal, $a(b+x)\in I$, and so $ab\in I$, a
contradiction. Thus $acI=0$. Similarly, one can easily see that $bcI=0$.

$(2)$ Assume that $axy\neq 0$ for some $x,y\in I$. By part $(1)$,
$a(b+x)(c+y)=axy\neq 0$. Since $R$ is local, the set of nonunit elements of
$R$ is an ideal of $R$. Therefore $(b+x),(c+y)$ are nonunit elements of $R$.
This means that either $a(b+x)\in I$ or $c+y\in I$. Hence either $ab\in I$ or
$c\in I$, a contradiction. Thus $aI^{2}=0$. Similarly, one can easily show
that $bI^{2}=cI^{2}=0$.  $\Box$

###### Theorem 2.6

Let $I$ be a weakly $1$-absorbing prime ideal of a local ring $R$ that is not
$1$-absorbing prime. Then $I^{3}=0$.

###### Proof.

Suppose that $I$ is a weakly $1$-absorbing prime ideal of a ring $R$ that is
not $1$-absorbing prime. Then there exists a triple-zero $(a,b,c)$ of $I$ for
some nonunit elements $a,b,c\in R$. Suppose that $I^{3}\neq 0$. Hence $xyz\neq
0$ for some $x,y,z\in I$. By Lemma 2.5, $(a+x)(b+y)(c+z)=xyz\neq 0$. Since $R$
is local, the set of nonunit elements of $R$ is an ideal of $R$. Therefore
$(a+x),(b+y)$ and $(c+z)$ are nonunit elements. Hence either $(a+x)(b+y)\in I$
or $(c+z)\in I$, and so either $ab\in I$ or $c\in I$, a contradiction. Thus
$I^{3}=0$.  $\Box$

It is worth mentioning that if $I$ is an ideal of a ring $R$ such that
$I^{3}=0$, then $I$ need not be a weakly $1$-absorbing prime ideal. For
example, let $R=\mathbb{Z}/16\mathbb{Z}$ and $I=8\mathbb{Z}/16\mathbb{Z}$.
Then $I^{3}=0$, but
$\overline{0}\neq\overline{2}\cdot\overline{2}\cdot\overline{2}=\overline{8}\in
I$ and $\overline{2},\overline{4}\notin I$.

Now, we state the following corollary.

###### Corollary 2.7

$(1)$ Let $I$ be a weakly $1$-absorbing prime ideal of a local ring $R$ that
is not $1$-absorbing prime. Then $\sqrt{I}=\sqrt{0}$.

$(2)$ If $R$ is a reduced local ring and $I\neq 0$ is a proper ideal of $R$,
then $I$ is a weakly $1$-absorbing prime ideal of $R$ if and only if $I$ is a
$1$-absorbing prime ideal of $R$.

###### Theorem 2.8

Let $I$ be a weakly $1$-absorbing prime ideal of a local ring $R$ that is not
$1$-absorbing prime. Then the following statements hold.

$(1)$ If $w\in\sqrt{0}$, then either $w^{2}\in I$ or $w^{2}I=wI^{2}=\\{0\\}$.

$(2)$ $\sqrt{0}^{2}I^{2}=\\{0\\}$.

###### Proof.

$(1)$ Suppose that $w\in\sqrt{0}$ such that $w^{2}I\neq\\{0\\}$. Suppose also
that $n$ is the least positive integer such that $w^{n}=0$. Then $n\geq 3$ and
$w^{2}(x+w^{n-2})=w^{2}x\neq 0$ for some $x\in I$. Therefore, either $w^{2}\in
I$ or $(x+w^{n-2})\in I$, since $R$ is local (which means that $(x+w^{n-2})$
is nonunit) and $I$ is weakly $1$-absorbing prime. If $(x+w^{n-2})\in I$, then
$w^{n-2}\in I$, and so $w^{2}\in I$, as $w^{n-4}\notin I$. Hence $w^{2}\in I$.
Thus either $w^{2}\in I$ or $w^{2}I=\\{0\\}$, for every $w\in\sqrt{0}$. Next,
we show that $wI^{2}=\\{0\\}$. Suppose that $wI^{2}\neq\\{0\\}$ and
$w^{2}\notin I$. Then $w^{2}I=\\{0\\}$ and $wxy\neq 0$ for some $x,y\in I$.
Let $m$ be the least positive integer such that $w^{m}=0$. This means that
$m\geq 3$ and $w^{2}I=0$. Therefore $w(w+x)(w^{m-2}+y)=wxy\neq 0$. But $R$ is
local (which means that $(y+w^{n-2})$ is non unit) and $I$ is weakly
$1$-absorbing prime, hence either $w^{2}\in I$ or $w^{m-2}\in I$, a
contradiction. Thus $wI^{2}=\\{0\\}$.

$(2)$ Suppose that $a,b\in\sqrt{0}$. It follows from part $(1)$ that if either
$a^{2}\notin I$ or $b^{2}\notin I$, then $abI^{2}=\\{0\\}$. Hence assume that
$a^{2}\in I$ and $b^{2}\in I$, which means that $ab(a+b)\in I$. We distinguish
two cases, according as $(a,b,a+b)$ is a triple-zero of $I$ or not. If
$(a,b,a+b)$ is a triple-zero of $I$, then $abI=\\{0\\}$, by Lemma 2.5 (1), and
thus $\sqrt{0}^{2}I^{2}=\\{0\\}$. If $(a,b,a+b)$ is not a triple-zero of $I$,
then it is easily seen that $ab\in I$, and so $\sqrt{0}^{2}I^{2}=\\{0\\}$, by
Theorem 2.6.  $\Box$

Corollary 2.7 and Theorem 2.8 lead to the following corollary.

###### Corollary 2.9

Let $I,J,K$ be weakly $1$-absorbing prime ideals of a local ring $R$ such that
none of them is $1$-absorbing prime. Then
$I^{2}JK=IJ^{2}K=IJK^{2}=I^{2}J^{2}=I^{2}K^{2}=J^{2}K^{2}=\\{0\\}$.

If $I$ is a proper ideal of $R$, it is well known that $\sqrt{I}$ is a prime
ideal of $R$ if and only if $\sqrt{I}$ is a primary ideal of $R$. In the
following result, we give a condition under which every weakly $1$-absorbing
prime ideal of $R$ is $1$-absorbing prime.

###### Theorem 2.10

Suppose that $I$ is a proper ideal of $R$ such that $\sqrt{0}\subseteq I$ and
$\sqrt{0}$ is a prime (primary) ideal of $R$. Then $I$ is a weakly
$1$-absorbing prime ideal of $R$ if and only if $I$ is a $1$-absorbing prime
ideal of $R$.

###### Proof.

Suppose that $I$ is a weakly $1$-absorbing prime ideal of the ring $R$ and
$xyz\in I$ for some nonunit elements $x,y,z\in R$. If $xyz\neq 0$, then either
$xy\in I$ or $z\in I$. Therefore assume that $xyz=0$ and $z\notin I$. Since
$xyz=0\in\sqrt{0}$ which is a prime ideal of $R$, we conclude that
$xy\in\sqrt{0}\subseteq I$. Thus $I$ is a weakly $1$-absorbing prime ideal of
$R$ if and only if $I$ is a $1$-absorbing prime ideal of $R$. $\Box$

In the following result, we give a condition under which a weakly
$1$-absorbing prime ideal of $R$ is not $1$-absorbing prime.

###### Theorem 2.11

Suppose that $I$ is a weakly $1$-absorbing prime ideal of a local ring $R$ and
$\\{0\\}$ has a triple-zero $(a,b,c)$ for some nonunit elements $a,b,c\in R$
such that $ab\notin\sqrt{0}$. Then $I$ is not a $1$-absorbing prime ideal of
$R$ if and only if $I\subseteq\sqrt{0}$.

###### Proof.

Suppose that $I$ is not a $1$-absorbing prime ideal of $R$. Hence, by part
$(1)$ of Corollary 2.7, $I\subseteq\sqrt{0}$. Conversely, suppose that
$I\subseteq\sqrt{0}$ and $\\{0\\}$ has a triple-zero $(a,b,c)$ for some
nonunit elements $a,b,c\in R$ such that $ab\notin\sqrt{0}$. Then $ab\notin I$
and $c\notin I$. Hence $(a,b,c)$ is a triple-zero of $I$, and thus $I$ is not
a $1$-absorbing prime ideal of $R$. $\Box$

One can easily see that if $I$ is a $1$-absorbing prime ideal of a ring $R$,
then there exists just one prime ideal of $R$ that is minimal over $I$. In the
following result, we show that for every positive integer $n\geq 2$, there
exist a ring $R$ and a nonzero weakly $1$-absorbing prime ideal $I$ of $R$
such that there are exactly $n$ prime ideals of $R$ that are minimal over $I$.

###### Theorem 2.12

Suppose that $n\geq 2$ is a positive integer. Then there exist a ring $R$ and
a nonzero weakly $1$-absorbing prime ideal $I$ of $R$ such that $I$ has
exactly $n$ minimal prime ideals of $R$.

###### Proof.

Suppose that $n\geq 2$ is a positive integer and
$D=\mathbb{Z}_{8}\times\cdots\times\mathbb{Z}_{8}$ ($n$ times). Clearly,
$M=\\{\overline{0},\overline{4}\\}$ is an ideal of $\mathbb{Z}_{8}$. Define
the $D$-module $M$ such that $xM=a_{1}M$ for every $x=(a_{1},\ldots,a_{n})\in
D$ and consider the idealization ring $R=D(+)M$ and the ideal
$I=\\{(0,\ldots,0)\\}(+)M$ of $R$. It follows from [8, Theorem 25.1 (3)] that
every prime ideal of $R$ is of the form $P(+)M$ for some prime ideal $P$ of
$D$ and since for every $a,b,c\in R\setminus I$ and $abc\in I$, we deduce that
$abc=((0,\ldots,0),0)$. Hence $I$ is a nonzero weakly $1$-absorbing prime
ideal of $R$, and thus there are exactly $n$ prime ideals of $R$ that are
minimal over $I$.  $\Box$

In the next result, one can see that in a reduced ring, the radical of a
nonzero weakly $1$-absorbing prime ideal a prime ideal.

###### Theorem 2.13

Let $R$ be a reduced ring and $I$ a nonzero weakly $1$-absorbing prime ideal
of $R$. Then $\sqrt{I}$ is a prime ideal of $R$.

###### Proof.

Let $0\neq ab\in\sqrt{I}$ for some $a,b\in R$. Without loss of generality, we
may assume that $a,b$ are nonunit. Then $(ab)^{n}\in I$ for some positive
integer $n$, and so $(ab)^{n}\neq 0$, because $R$ is reduced. Therefore $0\neq
a^{m}a^{(n-m)}b^{n}\in I$ for some positive integer $m$, and hence either
$a^{m}a^{(n-m)}=a^{n}\in I$ or $b^{n}\in I$. Thus $\sqrt{I}$ is a weakly prime
ideal of $R.$ But $R$ is reduced and $I\neq\\{0\\}$, hence $\sqrt{I}$ is a
prime ideal of $R$, by [1, Corollary 2].  $\Box$

It was shown in [11, Theorem 2.4] that if $R$ is a non-local ring, then every
$1$-absorbing prime ideal is prime. In the following theorem, we show that if
$R$ is a non-local ring and $I$ is a proper ideal of $R$ having the property
that $ann(x)$ is not a maximal ideal of $R$ for every element $x\in I$, then
$I$ is a weakly $1$-absorbing prime ideal if and only if $I$ is a weakly prime
ideal.

###### Theorem 2.14

Let $R$ be a non-local ring and $I$ a proper ideal of $R$ having the property
that $ann(x)$ is not a maximal ideal of $R$ for every element $x\in I$. Then
$I$ is a weakly $1$-absorbing prime ideal if and only if $I$ is a weakly prime
ideal.

###### Proof.

It is easy to see that every weakly prime ideal of $R$ is a weakly
$1$-absorbing prime ideal of $R$. Then assume that $I$ is a weakly
$1$-absorbing prime ideal of $R$ and let $0\neq ab\in I$ for some $a,b\in R$.
Without loss of generality, we may assume that $a,b$ are nonunit. Since
$ab\neq 0$, $ann(ab)$ is a proper ideal of $R$, and so $ann(ab)\subset L$, for
some maximal ideal $L$ of $R$. But $R$ is a non-local ring, hence there exists
a maximal ideal $M$ of $R$ such that $M\neq L$. Suppose that $m\in M\setminus
L$. Then $m\notin ann(ab)$, which means that $0\neq mab\in I$. Since $I$ is a
weakly $1$-absorbing prime ideal of $R$, either $ma\in I$ or $b\in I$. If
$b\in I$, then the proof is complete. So suppose that $b\notin I$. Therefore
$ma\in I$. But $m\notin L$ and $L$ is a maximal ideal of $R$, hence $m\notin$
J$(R)$. This means that there exists an $r\in R$ such that $1+rm$ is a nonunit
element of $R$. We take the following two cases: Case one: if $1+rm\notin
ann(ab)$, then $0\neq(1+rm)ab\notin I$. Since $I$ is a weakly $1$-absorbing
prime ideal of $R$ and $b\notin I$, we conclude that $(1+rm)a=a+rma\in I$. But
$rma\in I$, so $a\in I$ and $I$ is a weakly prime ideal of $R$. Case two:
Suppose that $1+rm\in ann(ab)$. Since $ann(ab)$ is not a maximal ideal of $R$
and $ann(ab)\subset L$, there exists an element $w\in L\setminus ann(ab)$.
Therefore $0\neq wab\in I$. But $I$ is a weakly $1$-absorbing prime ideal of
$R$ and $b\notin I$, thus $wa\in I$. We have $1+rm+w$ is a nonzero nonunit
element of $L$, because $1+rm\in ann(ab)\subset L$ and $w\in L\setminus
ann(ab)$. Thus $0\neq(1+rm+w)ab\in I$. Now, since $I$ is a weakly
$1$-absorbing prime ideal of $R$ and $b\notin I$, we have
$(1+rm+w)a=a+rma+wa\in I$. Hence $a\in I$, and thus $I$ is a weakly prime
ideal of $R$.  $\Box$

Let $R$ be an integral domain with the quotient field $K$. Recall that a
proper ideal $I$ of $R$ is called invertible if $II^{-1}=R$, where
$I^{-1}=\\{r\in K:rI\subseteq R\\}$. An integral domain is called a Dedekind
domain if every nonzero proper ideal of $R$ is invertible. In the following
results, weakly $1$-absorbing prime ideals of Dedekind domains and principal
ideal domains are completely described.

###### Theorem 2.15

Let $R$ be a Noetherian integral domain that is not a field and $I$ an ideal
of $R$. Then $(1)\Rightarrow(2)\Rightarrow(3)$.

$(1)$ $R$ is a Dedekind domain;

$(2)$ If $I$ is a weakly $1$-absorbing prime ideal of $R$, then $I=M$ or
$I=M^{2}$ where $M$ is a maximal ideal of $R$;

$(3)$ If $I$ is a weakly $1$-absorbing prime ideal of $R$, then $I=P$ or
$I=P^{2}$ where $P=\sqrt{I}$ is a prime ideal of $R$.

###### Proof.

$(1)\Rightarrow(2)$ Suppose that $R$ is a Noetherian integral domain that is
not a field and $I$ is a weakly $1$-absorbing prime ideal of $R$ such that
$P=\sqrt{I}$. Since $R$ is a Dedekind domain, we conclude that every nonzero
prime ideal of $R$ is a maximal ideal of $R$. Therefore $P$ is a maximal ideal
of $R$, by Theorem 2.13. This means that $I$ is a primary ideal of $R$ such
that $P^{2}\subseteq I$. Hence, by [11, Theorem 2.10], $I=M$ or $I=M^{2}$
where $M$ is a maximal ideal of $R$.

$(2)\Rightarrow(3)$ This is obvious. $\Box$

In view of Theorem 2.15, we have the following result.

###### Corollary 2.16

Let $R$ be a principal ideal domain that is not a field and $I$ be a nonzero
proper ideal of $R$. If $I$ is a weakly $1$-absorbing prime ideal of $R$, then
$I=pR$ or $I=p^{2}R$ for some nonzero prime element $p$ of $R$.

In the following theorems we show that weakly $1$-absorbing prime ideals are
really of interest in indecomposable rings.

###### Theorem 2.17

Suppose that $R=R_{1}\times R_{2}$ is a decomposable ring, where $R_{1}$ and
$R_{2}$ are rings and $I$ is a proper ideal of $R_{1}$. Then the following
statements are equivalent.

$(1)$ $I\times R_{2}$ is a weakly $1$-absorbing prime ideal of $R$.

$(2)$ $I\times R_{2}$ is a $1$-absorbing prime ideal of $R$.

$(3)$ $I$ is a $1$-absorbing prime ideal of $R_{1}$.

###### Proof.

$(3)\Rightarrow(2)$ and $(2)\Rightarrow(1)$ These are clear.

$(1)\Rightarrow(3)$ Suppose that $I\times R_{2}$ is a weakly $1$-absorbing
prime ideal of $R$ and $abc\in I$ for some nonunit elements $a,b,c\in R_{1}$.
Suppose also that $0\neq x\in R_{2}$. Then
$(0,0)\neq(a,1)(b,1)(c,x)=(abc,x)\in I\times R_{2}$, and so either
$(a,1)(b,1)=(ab,1)\in I\times R_{2}$ or $(c,x)\in I\times R_{2}$. Hence,
either $ab\in I$ or $c\in I$. Thus $I$ is a $1$-absorbing prime ideal of
$R_{1}$.  $\Box$

###### Theorem 2.18

Suppose that $R=R_{1}\times R_{2}$ is a decomposable ring, where $R_{1}$ and
$R_{2}$ are rings and $I_{1},I_{2}$ are nonzero ideals of $R_{1}$ and $R_{2}$,
respectively. Then the following statements are equivalent.

$(1)$ $I_{1}\times I_{2}$ is a weakly $1$-absorbing prime ideal of $R$.

$(2)$ $I_{1}=R_{1}$ and $I_{2}$ is a $1$-absorbing prime ideal of $R_{1}$ or
$I_{2}=R_{2}$ and $I_{1}$ is a $1$-absorbing prime ideal of $R_{1}$ or
$I_{1},I_{2}$ are prime ideals of $R_{1},R_{2}$, respectively.

$(3)$ $I_{1}\times I_{2}$ is a $1$-absorbing prime ideal of $R$.

$(4)$ $I_{1}\times I_{2}$ is a prime ideal of $R$.

###### Proof.

$(1)\Rightarrow(2)$ Let $I_{1}\times I_{2}$ be a weakly $1$-absorbing prime
ideal of $R$. If $I_{1}=R_{1}$ ($I_{2}=R_{2}$), then by Theorem 2.17, $I_{2}$
($I_{1}$) is a $1$-absorbing prime ideal of $R_{2}$ ($R_{1}$). Therefore
assume that $I_{1}$ and $I_{2}$ are proper ideals. Suppose that $a,b\in R_{2}$
such that $ab\in I_{2}$ and $0\neq x\in I_{1}$. Then
$(0,0)\neq(x,1)(1,a)(1,b)=(x,ab)\in I_{1}\times I_{2}$. But $I_{1}$ is proper,
hence $(1,a)\notin I_{1}\times I_{2}$ and $(1,b)\notin I_{1}\times I_{2}$.
Without loss of generality, we may assume that $x,a,b$ are nonunit. Since
$I_{1}\times I_{2}$ is a weakly $1$-absorbing prime ideal of $R$, we conclude
that $(x,1)(1,a)=(x,a)\in I_{1}\times I_{2}$. Thus $I_{2}$ is a prime ideal of
$R_{2}$. Similarly, it can be easily shown that $I_{1}$ is a prime ideal of
$R_{1}$.

$(2)\Rightarrow(3)$ If $I_{1}=R_{1}$ and $I_{2}$ is a $1$-absorbing prime
ideal of $R_{2}$ or $I_{2}=R_{2}$ and $I_{1}$ is a $1$-absorbing prime ideal
of $R_{1}$, then the result follows from Theorem 2.17. Assume that
$I_{1},I_{2}$ are prime ideals of $R_{1},R_{2}$, respectively. Since
$I_{1},I_{2}$ are nonzero ideals of $R_{1}$ and $R_{2}$, assume that
$(0,0)\neq(a,b)\in I_{1}\times I_{2}$. Then
$(0,0)\neq(a,1)(a,1)(1,b)=(a^{2},b)\in I_{1}\times I_{2}$. This means that
either $(a,1)(a,1)\in I_{1}\times I_{2}$ or $(1,b)\in I_{1}\times I_{2}$,
since $I_{1}\times I_{2}$ is a weakly $1$-absorbing prime ideal of $R$. Hence
either $(a,1)\in I_{1}\times I_{2}$ or $(1,b)\in I_{1}\times I_{2}$, as
$I_{1}$ is a prime ideals of $R_{1}$, a contradiction. Thus $I_{1}\times
I_{2}=R_{1}\times I_{2}$ or $I_{1}\times I_{2}=I_{1}\times R_{2}$ and in both
cases $I_{1}\times I_{2}$ is a $1$-absorbing prime ideal of $R$.

$(3)\Rightarrow(4)$ Since $R$ is not a local ring, the proof follows from [11,
Theorem 2.4].

$(4)\Rightarrow(1)$ This is straightforward. $\Box$

###### Theorem 2.19

Suppose that $R=R_{1}\times R_{2}$ is a decomposable ring, where $R_{1}$ and
$R_{2}$ are rings and $I_{1}$ is a nonzero proper ideal of $R_{1}$ and $I_{2}$
is a proper ideal of $R_{2}$. Then $(1)\Rightarrow(2)$. In fact,
$(2)\Rightarrow(1)$ need not be true.

$(1)$ $I_{1}\times I_{2}$ is a weakly $1$-absorbing prime ideal of $R$ that is
not a $1$-absorbing prime ideal.

$(2)$ $I_{1}$ is a weakly prime ideal of $R_{1}$ that is not a prime ideal and
$I_{2}=\\{0\\}$ is a prime ideal of $R_{2}$.

###### Proof.

$(1)\Rightarrow(2)$ Suppose that $I_{1}\times I_{2}$ is a weakly $1$-absorbing
prime ideal of $R$ that is not a $1$-absorbing prime ideal and
$I_{2}\neq\\{0\\}$. Hence, by Theorem 2.18, $I_{1}\times I_{2}$ is a
$1$-absorbing prime ideal of $R$, a contradiction, and thus $I_{2}=\\{0\\}$.
Now, we show that $I_{2}$ is a prime ideal of $R_{2}$. Suppose that $a,b\in
R_{2}$ such that $ab\in I_{2}$ and $0\neq x\in I_{1}$. Then
$(0,0)\neq(x,1)(1,a)(1,b)=(x,ab)\in I_{1}\times I_{2}$. But $I_{1}$ is proper,
hence $(1,a)\notin I_{1}\times I_{2}$ and $(1,b)\notin I_{1}\times I_{2}$.
Without loss of generality, we may assume that $x,a,b$ are nonunit. Since
$I_{1}\times I_{2}$ is a weakly $1$-absorbing prime ideal of $R$,
$(x,1)(1,a)=(x,a)\in I_{1}\times I_{2}$. Thus $I_{2}=\\{0\\}$ is a prime ideal
of $R_{2}$. We show that $I_{1}$ is a weakly prime ideal of $R_{1}$. Let
$a,b\in R_{1}$ such that $0\neq ab\in I_{1}$. Without loss of generality, we
may assume that $a,b$ are nonunit. Since
$(0,0)\neq(b,1)(1,0)(ab,1)=(ab^{2},0)\in I_{1}\times I_{2}$ and $(ab,1)\notin
I_{1}\times\\{0\\}$ and $I_{1}\times I_{2}$ is a weakly $1$-absorbing prime
ideal of $R$, we conclude that $(b,0)\in I_{1}\times\\{0\\}$, and hence $b\in
I_{1}$. Thus $I_{1}$ is a weakly prime ideal of $R_{1}$. Assume that $I_{1}$
is a prime ideal of $R_{1}$. Since $I_{1}$ is a nonzero ideal of $R_{1}$,
assume that $0\neq a\in I_{1}$. Then $(0,0)\neq(a,1)(a,1)(1,0)=(a^{2},0)\in
I_{1}\times I_{2}$. This means that either $(a,1)(a,1)\in I_{1}\times I_{2}$
or $(1,0)\in I_{1}\times I_{2}$, as $I_{1}\times I_{2}$ is a weakly
$1$-absorbing prime ideal of $R$. Hence either $(a,1)\in I_{1}\times I_{2}$ or
$(1,0)\in I_{1}\times I_{2}$, as $I_{1}$ is a prime ideal of $R_{1}$, a
contradiction. Thus $I_{1}$ is a weakly prime ideal of $R_{1}$ that is not a
prime ideal and $I_{2}=\\{0\\}$ is a prime ideal of $R_{2}$.

$(2)\Rightarrow(1)$ Assume that $I_{1}$ is a weakly prime ideal of $R_{1}$
that is not a prime ideal and $I_{2}=\\{0\\}$ is a prime ideal of $R_{2}$. We
show that $I_{1}\times I_{2}$ need not be a weakly $1$-absorbing prime ideal
of $R$. Since $I_{1}\neq 0$, there exists $0\neq x\in I_{1}$, and so
$(0,0)\neq(a,1)(a,1)(1,0)=(a^{2},0)\in I_{1}\times\\{0\\}$. Since neither
$(a,1)(a,1)=(a^{2},1)\in I_{1}\times\\{0\\}$ nor $(1,0)\notin
I_{1}\times\\{0\\}$ and $I_{1}$ is proper, we conclude that $I_{1}\times
I_{2}$ is not a weakly $1$-absorbing prime ideal of $R$.  $\Box$

In the following, we show that in a decomposable ring $R=R_{1}\times
R_{2}\times R_{3}$ every nonzero weakly $1$-absorbing prime ideal of $R$ is
$1$-absorbing prime.

###### Theorem 2.20

Suppose that $R=R_{1}\times R_{2}\times R_{3}$ is a decomposable ring, where
$R_{1}$, $R_{2}$ and $R_{3}$ are rings and $I=I_{1}\times I_{2}\times I_{3}$
is a nonzero proper ideal of $R$. Then $I$ is a weakly $1$-absorbing prime
ideal if and only if $I$ is a $1$-absorbing prime ideal.

###### Proof.

Let $I=I_{1}\times I_{2}\times I_{3}$ be a nonzero weakly $1$-absorbing prime
ideal of $R$. Then there exists an element $(0,0,0)\neq(a,b,c)\in I$. Since
$(a,1,1)(1,b,1)(1,1,c)=(a,b,c)$, either $(a,b,1)\in I$ or $(1,1,c)\in I$.
Hence either $I_{3}=R_{3}$ or $I_{1}=R_{1}$ and $I_{2}=R_{2}$, and so
$I=I_{1}\times I_{2}\times R_{3}$ or $I=R_{1}\times R_{2}\times I_{3}$. Thus,
by Theorem 2.17, $I$ is a $1$-absorbing prime ideal of $R$. The converse is
clear.  $\Box$

###### Corollary 2.21

Suppose that $R=R_{1}\times R_{2}\times R_{3}$ is a decomposable ring, where
$R_{1}$, $R_{2}$ and $R_{3}$ are rings and $I=I_{1}\times I_{2}\times I_{3}$
is a nonzero ideal of $R$. Then the following statements are equivalent.

$(1)$ $I=I_{1}\times I_{2}\times I_{3}$ is a weakly $1$-absorbing prime ideal
of $R$.

$(2)$ $I=I_{1}\times I_{2}\times I_{3}$ is a $1$-absorbing prime ideal of $R$.

$(3)$ Either $I=I_{1}\times I_{2}\times I_{3}$ such that for some
$k\in\\{1,2,3\\}$, $I_{k}$ is a $1$-absorbing prime ideal of $R_{k}$, and
$I_{j}=R_{j}$ for every $j\in\\{1,2,3\\}-\\{k\\}$, or $I=I_{1}\times
I_{2}\times I_{3}$ such that for some $k,m\in\\{1,2,3\\}$, $I_{k}$ is a prime
ideal of $R_{k}$, $I_{m}$ is a prime ideal of $R_{m}$, and $I_{j}=R_{j}$ for
every $j\in\\{1,2,3\\}-\\{k,m\\}$.

$(4)$ $I=I_{1}\times I_{2}\times I_{3}$ is a prime ideal of $R$.

###### Proof.

$(1)\Leftrightarrow(2)$ It follows from Theorem 2.20.

$(2)\Rightarrow(3)$ Since $I=I_{1}\times I_{2}\times I_{3}$ is a $1$-absorbing
prime ideal of $R$, we have either $I_{3}\neq R_{3}$ or $I_{2}\neq R_{2}$ or
$I_{1}\neq R_{1}$. If $I_{3}\neq R_{3}$, then by the proof of Theorem 2.20,
$I_{2}=R_{2}$ and $I_{1}=R_{1}$, and so we are done. If $I_{2}\neq R_{2}$,
then by the proof of Theorem 2.20, either $I_{3}=R_{3}$ and $I_{1}=R_{1}$ or
$I_{3}=R_{3}$ and $I_{1}\neq R_{1}$. The first case is clear, so assume that
$I_{3}=R_{3}$ and $I_{1}\neq R_{1}$. We show that $I_{1}$ is a prime ideal of
$R_{1}$ and $I_{2}$ is a prime of $R_{2}$. Suppose that $a,b\in R_{1}$ such
that $ab\in I_{1}$, and $c,d\in R_{2}$ such that $cd\in I_{2}$. Then
$(0,0,0)\neq(a,1,1)(1,c,1)(b,d,1)=(ab,cd,1)\in I$. Hence either $a\in I_{1}$
or $b\in I_{1}$, and thus $I_{1}$ is a prime ideal of $R_{1}$. Similarly,
since $(0,0,0)\neq(a,1,1)(b,c,1)(1,d,1)=(ab,cd,1)\in I$, we conclude that
either $c\in I_{2}$ or $d\in I_{2}$. Hence $I_{2}$ is a prime ideal of
$R_{2}$. Finally, assume that $I_{1}\neq R_{1}$ and $I_{3}=R_{3}$. By an
argument similar to that we applied above, we conclude that $I_{1}$ is a prime
ideal of $R_{1}$ and $I_{2}$ is a prime ideal of $R_{2}$.

$(3)\Rightarrow(2)$ This is clear.

$(2)\Leftrightarrow(4)$ It follows from [11, Theorem 2.4].  $\Box$

The next theorem states that if $I$ is a weakly $1$-absorbing prime ideal of a
ring $R$ and $0\neq I_{1}I_{2}I_{3}\subseteq I$ for some ideals
$I_{1},I_{2},I_{3}$ of $R$ such that $I$ is free triple-zero with respect to
$I_{1}I_{2}I_{3}$, then $I_{1}I_{2}\subseteq I$ or $I_{3}\subseteq I$. First,
we need the following lemma.

###### Lemma 2.22

Let $I$ be a weakly $1$-absorbing prime ideal of a ring $R$. If $abJ\subseteq
I$ for some nonunit elements $a,b\in R$ and a proper ideal $J$ of $R$ such
that $(a,b,c)$ is not a triple-zero of $I$ for every $c\in J$, then $ab\in I$
or $J\subseteq I$.

###### Proof.

Suppose that $abJ\subseteq I$, but $ab\notin I$ and $J\nsubseteq I$. Then
there exists an element $j\in J\setminus I$. But $(a,b,j)$ is not a triple-
zero of $I$ and $abj\in I$ and $ab\notin I$ and $j\notin I$, a contradiction.
$\Box$

###### Theorem 2.23

Suppose that $I$ is a proper ideal of a ring $R$. Then the following
statements are equivalent.

$(1)$ $I$ is a weakly $1$-absorbing prime ideal of $R$.

$(2)$ For any proper ideals $I_{1},I_{2},I_{3}$ of $R$ such that $0\neq
I_{1}I_{2}I_{3}\subseteq I$ and $I$ is free triple-zero with respect to
$I_{1}I_{2}I_{3}$, we have either $I_{1}I_{2}\subseteq I$ or $I_{3}\subseteq
I$.

###### Proof.

$(1)\Rightarrow(2)$ Suppose that $I$ is a weakly $1$-absorbing prime ideal of
$R$ and $0\neq I_{1}I_{2}I_{3}\subseteq I$ for some proper ideals
$I_{1},I_{2},I_{3}$ of $R$ such that $I_{1}I_{2}\nsubseteq I$ and $I$ is free
triple-zero with respect to $I_{1}I_{2}I_{3}$. Then there are nonunit elements
$a\in I_{1}$ and $b\in I_{2}$ such that $ab\notin I$. Since $abI_{3}\subseteq
I$, $ab\notin I$ and $(a,b,c)$ is not a triple-zero of $I$ for every $c\in J$,
it follows from Lemma 2.22 that $J\subseteq I$.

$(2)\Rightarrow(1)$ Suppose that $0\neq abc\in I$ for some nonunit elements
$a,b,c\in R$ and $ab\notin I$. Suppose also that $I_{1}=aR,I_{2}=bR$, and
$I_{3}=cR$. Then $0\neq I_{1}I_{2}I_{3}\subseteq I$ and $I_{1}I_{2}\nsubseteq
I$. Hence $I_{3}=cR\subseteq I$, and thus $c\in I$.  $\Box$

Suppose that $I$ is an ideal of a ring $R$. It was shown in [11, Theorem 2.7],
if $I$ is $1$-absorbing prime and $I_{1}I_{2}I_{3}\subseteq I$ for some proper
ideals $I_{1},I_{2},I_{3}$ of $R$, then $I_{1}I_{2}\subseteq I$ or
$I_{3}\subseteq I$. We end this section with the following question: if $I$ is
weakly $1$-absorbing prime and $0\neq I_{1}I_{2}I_{3}\subseteq I$ for some
proper ideals $I_{1},I_{2},I_{3}$ of $R$, does it imply that
$I_{1}I_{2}\subseteq I$ or $I_{3}\subseteq I$?

##  3 Rings in which every proper ideal is weakly $1$-absorbing prime

In this section we study rings in which every proper ideal is weakly
$1$-absorbing prime. To prove Theorem 3.2, the following lemma is needed.

###### Lemma 3.1

Let $R$ be a ring. Then for every $a,b,c\in J(R)$, the ideal $I=Rabc$ is
weakly $1$-absorbing prime if and only if $abc=0$.

###### Proof.

Suppose that $R$ is a ring, $a,b,c\in J(R)$ and $I=Rabc$ is an ideal of $R$.
One can easily see that $I$ is a weakly $1$-absorbing prime ideal of $R$, if
$abc=0$. So assume that $I$ is a weakly $1$-absorbing prime ideal of $R$ and
$abc\neq 0$. Hence $a,b,c$ are nonunit, and so either $ab\in I$ or $c\in I$.
If $ab\in I$, then $ab=abcx$ for some $x\in R$, and thus $ab(1-cx)=0$. This
follows that $ab=0$, since $cx\in J(R)$ which means that $1-cx$ is unit.
Therefore $abc=0$, a contradiction. If $c\in I$, by a similar argument, we can
see that $c(1-abk)=0$ for some $x\in R$. This implies that $c=0$, a
contradiction. Thus $abc=0$.  $\Box$

###### Theorem 3.2

Let $R$ be a local ring with a unique maximal ideal $M$. Then every proper
ideal of $R$ is weakly $1$-absorbing prime if and only if $M^{3}=\\{0\\}$.

###### Proof.

Suppose that $R$ is local with a maximal ideal $M$ and every proper ideal of
$R$ is weakly $1$-absorbing prime. It follows from Lemma 3.1 that $abc=0$ for
every $a,b,c\in M$, since $abcR$ is a weakly $1$-absorbing prime ideal.
Therefore $M^{3}=\\{0\\}$. Conversely, suppose that $M^{3}=\\{0\\}$ and $I$ is
a nonzero proper ideal of $R$. Suppose also that $0\neq abc\in I$ for some
nonunit elements $a,b,c$ of $R$. Then $a,b,c\in M$. But $M^{3}=\\{0\\}$ and
$0\neq abc$, hence either $a$ is a unit of $R$ or $b$ is a unit of $R$ or $c$
is a unit of $R$, a contradiction. Thus $I$ is a weakly $1$-absorbing prime
ideal.  $\Box$

###### Corollary 3.3

Suppose that $R$ is a local ring with a unique maximal ideal $M$ such that
$M^{2}=\\{0\\}$. Then every proper ideal of $R$ is a $1$-absorbing prime ideal
of $R$.

###### Proof.

Suppose that $R$ is local with a maximal ideal $M$ and $I$ is a proper ideal
of $R$. It follows from Theorem 3.2 that $I$ is weakly $1$-absorbing prime,
because $M^{3}=\\{0\\}$. Therefore assume that $0=abc\in I$ for some nonunit
elements $a,b,c$ of $R$. This means that $a,b,c\in M$. Since $M^{2}=\\{0\\}$,
$abc=0$ and $M^{2}$ is primary, we conclude that either $c\in M^{2}\subseteq
I$ or $ab\in M$. In the second case, it easy to see that $ab=0\in I$. Thus $I$
is a $1$-absorbing prime ideal of $R$.  $\Box$

###### Theorem 3.4

Suppose that $R_{1}$ and $R_{2}$ are rings and $R=R_{1}\times R_{2}$. If
$R_{1}$ and $R_{2}$ are local with maximal ideals $M_{1}$ and $M_{2}$,
respectively, and every proper ideal of $R$ is a weakly $1$-absorbing prime
ideal of $R$, then $M_{1}^{2},M_{2}^{2}$ are zero ideals and either $R_{1}$ or
$R_{2}$ is a field.

###### Proof.

Suppose that $R_{1}$ and $R_{2}$ are local rings with maximal ideals $M_{1}$
and $M_{2}$, respectively, and $R=R_{1}\times R_{2}$. First, suppose that
every proper ideal of $R$ is a weakly $1$-absorbing prime ideal of $R$ and
$a,b\in M_{1}$ such that $ab\neq 0$. Then the ideal $I=abR_{1}\times\\{0\\}$
of $R$ is a weakly $1$-absorbing prime ideal of $R$. Without loss of
generality, we may assume that $a,b$ are nonunit. But
$(0,0)\neq(a,1)(b,1)(1,0)=(ab,0)\in I$ and $I$ is a weakly $1$-absorbing prime
ideal of $R$, hence $(1,0)\in I$, since $(a,1)(b,1)\notin I$, and so $1=abx$
for some $x\in R_{1}$. This means that $1-abx=0$, a contradiction, as $1-abx$
is a unit element of $R_{1}$. Therefore $M_{1}^{2}=\\{0\\}$. By a similar
argument as above and taking the ideal $I=\\{0\\}\times abR_{1}$, one can see
that $M_{2}^{2}=\\{0\\}$. Now, assume that $R_{1}$ and $R_{2}$ are not fields
and look for a contradiction. It follows that $M_{1}\neq\\{0\\}$, and so the
ideal $J=M_{1}\times\\{0\\}$ is a weakly $1$-absorbing prime ideal of $R$. But
$M_{2}^{2}=\\{0\\}$ and $R_{2}$ is not a field, hence there exists $c\in
M_{2}$ such that $c\neq 0$. Suppose that $0\neq m_{1}\in M_{1}$. Then
$(0,0)\neq(m_{1},1)(1,c)(1,c)=(m_{1},c^{2})=(m_{1},0)\in J=M_{1}\times\\{0\\}$
which is a contradiction, since both $(m_{1},1)(1,c)=(m_{1},c)\notin J$ and
$(1,c)\notin J$. Thus either $R_{1}$ or $R_{2}$ is a field.  $\Box$

Finally, we characterize rings in which every ideal is weakly $1$-absorbing
prime. To this end, the following lemma is needed.

###### Lemma 3.5

$(1)$ Suppose that $R=R_{1}\times R_{2}\times R_{3}$, where $R_{1}$, $R_{2}$
and $R_{3}$ are rings. If every proper ideal of $R$ is weakly $1$-absorbing
prime, then $R_{1},R_{2},R_{3}$ are fields.

$(2)$ Let $R$ be a ring and every proper ideal of $R$ be weakly $1$-absorbing
prime. Then $R$ has at most three maximal ideals.

###### Proof.

$(1)$ Let $R=R_{1}\times R_{2}\times R_{3}$ be a decomposable ring, where
$R_{1}$, $R_{2}$ and $R_{3}$ are rings such that every proper ideal of $R$ is
weakly $1$-absorbing prime. Without loss of generality, assume that $R_{1}$ is
not a field. Let $J$ be a non-zero proper ideal of $R_{1}$ and $m$ be a non-
zero element of $J$. Suppose that $I=J\times\\{0\\}\times\\{0\\}$. Then $I$ is
a weakly $1$-absorbing prime ideal of $R$. Since
$(0,0,0)\neq(m,1,1)(1,0,1)(1,1,0)=(m,0,0)\in I$ and both
$(m,1,1)(1,0,1)=(m,0,1)\notin I$ and $(1,1,0)\notin I$, we have a
contradiction. Hence $R_{1},R_{2},R_{3}$ are fields.

$(2)$ Let $M_{1},M_{2},M_{3},M_{4}$ be distinct maximal ideals of $R$ and
$I=M_{1}\cap M_{2}\cap M_{3}$. Since every weakly $1$-absorbing prime is
weakly $2$-absorbing, and by [2, Theorem 2.5], $I$ is not a $2$-absorbing
ideal of $R$, we conclude that $I$ is a weakly $2$-absorbing ideal of $R$ that
is not a $2$-absorbing ideal of $R$. Therefore, by [3, Theorem 2.4],
$I^{3}=\\{0\\}$, a contradiction. Thus $R$ has at most three distinct maximal
ideals.  $\Box$

We end this paper with the following result.

###### Theorem 3.6

Let $R$ be a ring. If $(1)$ every proper ideal of $R$ is weakly $1$-absorbing
prime, then either $(2)$ $R$ is a local ring with a unique maximal ideal $M$
such that $M^{3}=0$ or $(3)$ $R=R_{1}\times R_{2}$, where $R_{1}$ is a local
ring with a unique maximal ideal $M_{1}$ such that $M_{1}^{2}=0$ and $R_{2}$
is a field or $(4)$ $R=R_{1}\times R_{2}\times R_{3}$, where
$R_{1},R_{2},R_{3}$ are fields. Furthermore, $(2)\Rightarrow(1)$ but
$(3)\Rightarrow(1)$ and $(4)\Rightarrow(1)$ need not necessarily be true.

###### Proof.

If $R$ satisfies condition $(2)$, then the condition $(1)$ follows from
Theorem 3.2. Suppose $R$ satisfies condition $(3)$, i.e., $R=R_{1}\times
R_{2}$, where $R_{1}$ is local with a unique maximal ideal $M_{1}$ such that
$M_{1}^{2}=0$ and $R_{2}$ is a field. Since $R_{2}$ is a field and every
proper ideal of $R_{1}$ is a $1$-absorbing prime, we conclude that the ideals
$\\{0\\}\times R_{2}$ and $R_{1}\times\\{0\\}$ are weakly $1$-absorbing prime
ideals of $R$. Suppose that $J$ is a non-zero proper ideal of $R_{1}$. It
follows from Theorem 2.17 that $J\times R_{2}$ is a weakly $1$-absorbing prime
ideal, since $J$ is a $1$-absorbing prime ideal of $R_{1}$, by Corollary 3.3.
Hence, it is enough to prove that $J\times\\{0\\}$ is a weakly $1$-absorbing
prime ideal of $R$. Let $0\neq a\in J$. Then
$(0,0,0)\neq(a,1)(a,1)(1,0)=(a^{2},0,0)\in J\times\\{0\\}$ and both
$(a^{2},1)\notin J\times\\{0\\}$ and $(1,0)\notin J\times\\{0\\}$. Thus the
condition $(3)\Rightarrow(1)$ need not necessarily be true. Suppose that $R$
satisfies condition $(4)$, i.e., $R=R_{1}\times R_{2}\times R_{3}$, where
$R_{1},R_{2},R_{3}$ are fields. By taking $I=R_{1}\times\\{0\\}\times\\{0\\}$,
where $R_{1}$ is a field, one can easily see that the ideal $I$ is not weakly
$1$-absorbing prime, because $(0,0,0)\neq(1,0,1)(1,0,1)(1,1,0)=(1,0,0)\in I$
and neither $(1,0,1)(1,0,1)=(1,0,1)\in I$ nor $(1,1,0)\in I$, showing that the
condition $(4)\Rightarrow(1)$ need not necessarily be true. To complete the
proof, assume that every proper ideal of $R$ is weakly $1$-absorbing prime. It
follows from Lemma 3.5 part $(2)$ that $R$ has at most three maximal ideals,
and so if $R$ is local with a unique maximal $M$, then by Theorem 3.2,
$M^{3}=\\{0\\}$. Thus $R$ satisfies $(2)$. If $R$ has two maximal ideals
$M_{1}$ and $M_{2}$, then $J(R)=M_{1}\cap M_{2}$ is a weakly $1$-absorbing
prime ideal of $R$. Since every proper ideal of $R$ is weakly $1$-absorbing
prime, $Rabc$ is weakly $1$-absorbing prime, for every $a,b,c\in J(R)$. Hence
$abc=0$, by Lemma 3.1, and this means that $J(R)^{3}=M_{1}^{3}\cap
M_{2}^{3}=\\{0\\}$. Thus, by the Chinese Remainder Theorem $R\cong
R/M_{1}^{3}\times R/M_{2}^{3}$. But $R/M_{1}^{3}$ and $R/M_{2}^{3}$ are local
rings and every proper ideal of $R$ is weakly $1$-absorbing prime. Hence, by
Theorem 3.4, either $R/M_{1}^{3}$ or $R/M_{2}^{3}$ is a field and
$N^{2}=J^{2}=0$, where $N$ and $J$ are the maximal ideals of $R/M_{1}^{3}$ and
$R/M_{2}^{3}$, respectively. Thus $R$ satisfies $(3)$. If $R$ has three
maximal ideals $M_{1},M_{2}$ and $M_{3}$, then $J(R)=M_{1}\cap M_{2}\cap
M_{3}$ is a weakly $1$-absorbing prime ideal of $R$ (weakly $2$-absorbing
ideal). But $J(R)$ is not a $2$-absorbing, hence, by [3, Theorem 2.4],
$J(R)^{3}=M_{1}^{3}\cap M_{2}^{3}\cap M_{3}^{3}=\\{0\\}$. Thus, by the Chinese
Remainder Theorem, $R\cong R/M_{1}^{3}\times R/M_{2}^{3}\times R/M_{3}^{3}$.
Since every proper ideal of $R$ is weakly $1$-absorbing prime, Lemma 3.5
implies that $R/M_{1}^{3}$, $R/M_{2}^{3}$ and $R/M_{3}^{3}$ are fields. Thus
$R$ satisfies $(4)$.  $\Box$

## References

  * [1] D. D. Anderson, E. Smith, Weakly prime ideals, Houston J. Math. 29 (2003) 831–840.
  * [2] A. Badawi, On $2$-absorbing ideals of commutative rings, Bull. Austral. Math. Soc. 75 (2007) 417–429.
  * [3] A. Badawi, A. Y. Darani, On weakly $2$-absorbing ideals of commutative rings, Houston J. Math. 39 (2013) 441–452.
  * [4] A. Badawi, E. Yetkin, On $1$-absorbing primary ideals of commutative rings, J. Algebra Appl. (2020), 2050111.
  * [5] C. Beddani, W. Messirdi, $2$-prime ideals and their applications, J. Algebra Appl. 15 (2016) 1650051.
  * [6] S. M. Bhatwadekar, P. K. Sharma, Unique factorization and birth of almost primes, Comm. Algebra 33 (2005) 43–49.
  * [7] J. R. Hedstrom, E.G. Houston, Pseudo-valuation domains, Pacific J. Math. 75 (1978) 137–147.
  * [8] J. A. Huckaba, Commutative Rings with Zero-Divisors, Marcel Dekker, New York/Basil, 1988.
  * [9] R. Nikandish, M. J. Nikmehr, A. Yassine, More on the $2$-prime ideals of commutative rings, Bull. Korean Math. Soc. 57 (2020) 117–126.
  * [10] R. Y. Sharp, Steps in Commutative Algebra, 2nd ed. London Mathematical Society Student Texts, 51. Cambridge University Press, Cambridge, 2000.
  * [11] A. Yassine, M. J. Nikmehr, R. Nikandish, On $1$-absorbing prime ideals of commutative rings, to appear in J. Algebra Appl.


# An Observability-Constrained Magnetic-Field-Aided Inertial Navigation System
— Extended Version

Chuan Huang  Dept. of Electrical Engineering
Linköping University
Linköping, Sweden
<EMAIL_ADDRESS>Gustaf Hendeby  Dept. of Electrical Engineering
Linköping University
Linköping, Sweden
<EMAIL_ADDRESS>Isaac Skog  Div. of Communication Systems
KTH Royal Institute of Technology
Stockholm, Sweden
<EMAIL_ADDRESS>
###### Abstract

A method to construct an observability-constrained magnetic-field-aided
inertial navigation system is proposed. The proposed method builds upon the
previously proposed observability-constrained extended Kalman filter and
extends it to work with a magnetic-field-based odometry-aided inertial
navigation system. The proposed method is evaluated using simulation and real-
world data, showing that (i) the system observability properties are
preserved, (ii) the estimation accuracy increases, and (iii) the perceived
uncertainty calculated by the EKF is more consistent with the true uncertainty
of the filter estimates.

##  I Introduction

In an odometry-aided inertial navigation system (INS), such as a magnetic-
field odometry-aided INS [1, 2, 3, 4, 5, 6], the uncertainty about the
position and yaw of the navigation platform can, if the errors in the initial
state are uncorrelated, never become smaller than the initial position and yaw
uncertainty. This is because odometry and inertial measurements only provide
relative motion information. Hence, for any algorithm used to estimate the
navigation state $x_{k}$ in an odometry-aided INS it should hold that

$P^{\text{\footnotesize p}}_{k}\succeq P^{\text{\footnotesize
p}}_{0}\quad\text{and}\quad P^{\phi}_{k}\succeq P^{\phi}_{0},$ (1)

where $P^{\text{\footnotesize p}}_{k}$ and $P^{\phi}_{k}$ denote the
covariance of the, at time $k$, estimated position and yaw states,
respectively.

In [1], an extended Kalman filter (EKF) algorithm was used to realize a
magnetic-field-aided INS (MAINS). In Fig. 1, the square roots of
$P^{\phi}_{0}$ and $P^{\phi}_{k}$ as calculated by the EKF in the MAINS are
shown. As seen from the figure the square root of $P^{\phi}_{k}$ falls below
that of $P^{\phi}_{0}$. Hence the inequality in (1) does not hold and the
covariance estimate of the EKF is inconsistent. The inconsistency is because
the EKF linearizes the system model around the estimated state, which causes
the yaw to be perceived as observable even though it is not. This type of
inconsistency effect has been observed in multiple EKF-based implementations
of odometry-aided INS, see e.g., [7, 8, 9], and is troublesome if the yaw
information should be fused with information from other systems or be used in
a control or decision system. Next, we will present a way to fix this
inconsistency by modifying the EKF algorithm so that it preserves the
observability properties of the underlying nonlinear system. The presented
method is an extension of the method published in [8]. Our contributions are:

  * •

An extension of how the basis vectors spanning the subspace of the
unobservable state space (hereafter referred to as the unobservable subspace)
can be chosen, simplifying the preservation of the observability properties.

  * •

The derivation of the unobservable subspace for the model used in [1].

  * •

An experimental evaluation that verifies the effectiveness of the proposed
method when applied to the MAINS.

The code and data to produce the presented results can be found at:
https://github.com/Huang-Chuan/OC-MAINS-code.

Figure 1: Example of the perceived yaw uncertainty calculated by the EKF used
to realize magnetic-field based odometry aided INS in [1]. Also shown is the
perceived yaw uncertainty of the proposed observability-constrained EKF
algorithm.

##  II Preserve Observability Properties

The behavior of odometry-aided INS is commonly described by a nonlinear state-
space model of the following form

$\displaystyle x_{k+1}$ $\displaystyle=f(x_{k},u_{k},w_{k}),$
$\displaystyle\quad x\in\mathbb{R}^{n_{x}},$ (2) $\displaystyle y_{k}$
$\displaystyle=h(x_{k})+e_{k},$ $\displaystyle\quad y\in\mathbb{R}^{n_{y}}.$

Here $f$ and $h$ are the nonlinear functions, and $u_{k}$ denotes the control
input. Further, $w_{k}$ and $e_{k}$ denote the process and measurement noise,
respectively. They are assumed to be white noise with covariance $Q_{k}$ and
$R_{k}$, respectively.

Several approaches have been proposed to preserve observability properties of
the model (2) in EKF filtering, see [8, 10, 11, 7, 12, 13], one of which is
called the observability-constrained EKF [8]. The basic idea of the method in
[8] is to modify the Jacobians used in the EKF, such that the basis of the
unobservable subspace, evaluated at state estimates, lies in the nullspace of
the observability matrix evaluated at state estimates of EKF. The unobservable
subspace is a subspace of the nullspace of the local observability matrix [14]
associated with (2), i.e. the observability matrix constructed as

$\mathcal{\bar{O}}_{k}\triangleq\begin{bmatrix}\bar{H}_{k}\\\
\bar{H}_{k+1}\bar{\Phi}(k+1,k)\\\ \vdots\\\
\bar{H}_{k+n_{x}-1}\bar{\Phi}(k+n_{x}-1,k)\end{bmatrix}.$ (3a) Here
$\bar{x}_{k:k+n_{x}-1}$ denotes the nominal trajectory, which is the solution
of (2) with the control sequence ${\bar{u}_{k:k+n_{x}-1}}$ and the process
noise turned off. Further,
$\bar{\Phi}(k+i,k)=\bar{F}_{k+i-1}\bar{F}_{k+i-2}\cdots\bar{F}_{k}$ (3b) where
$\bar{H}_{k}=\frac{\partial h}{\partial
x_{k}}\bigg{|}_{x_{k}=\bar{x}_{k}}\quad\text{and}\quad\bar{F}_{k}=\frac{\partial
f}{\partial
x_{k}}\bigg{|}_{\vbox{\Let@\restore@math@cr\default@tag\halign{\hfil$\m@th\scriptstyle#$&$\m@th\scriptstyle{}#$\hfil\cr
x_{k}&=\bar{x}_{k}\\\ u_{k}&=\bar{u}_{k}\\\ w_k&=0\crcr}}}.$ (3c)

The unobservable subspace $\mathcal{N}_{k}$ can be represented as

$\begin{split}\mathcal{N}_{k}&=\text{span}\\{N(\bar{x}_{k})\\},\end{split}$
(4a)

where $N:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n\times p}$ is a matrix valued
function such that $\bar{\mathcal{O}}_{k}N(\bar{x}_{k})=0$, and $p$ denotes
the dimension of the unobservable subspace. That is, the columns of
$N(\bar{x}_{k})$ are basis vectors that span the unobservable subspace. The
basis of the unobservable subspace is not unique. Hence, right multiplying
$N(\bar{x}_{k})$ with any full-rank matrix
$\mathcal{E}_{k}\in\mathbb{R}^{p\times p}$ will not change the span of the
basis. That is,

$\text{span}\\{N^{\dagger}(\bar{x}_{k})\\}=\text{span}\\{N(\bar{x}_{k})\\},$
(5a) where $N^{\dagger}:\quad\mathbb{R}^{n}\rightarrow\mathbb{R}^{n\times
p}\quad\text{and}\quad N^{\dagger}(x_{k})=N(x_{k})\mathcal{E}_{k}.$ (5b)

This property, which was not used in [8], will be important when modifying the
Jacobians, as it reduces the changes made to the Jacobians to preserve the
observability properties.

Once the basis vectors of the unobservable subspace are determined, they are,
as will be described next, used to modify the Jacobians used in the EKF. Let

$\hat{F}_{k}=\frac{\partial f}{\partial
x_{k}}\bigg{|}_{\vbox{\Let@\restore@math@cr\default@tag\halign{\hfil$\m@th\scriptstyle#$&$\m@th\scriptstyle{}#$\hfil\cr
x_{k}&=\hat{x}_{k|k}\\\ u_{k}&=\hat{u}_k\\\
w_k&=0\crcr}}}\quad\text{and}\quad\hat{H}_{k}=\frac{\partial h}{\partial
x_{k}}\bigg{|}_{x_{k}=\hat{x}_{k|k-1}}$ (6)

be the Jacobians used in the unmodified EKF. Here $\hat{x}_{k|k}$ and
$\hat{x}_{k|k-1}$ denote the posterior and prior estimate of the state, i.e.,
the estimate of $x_{k}$ given measurements up to time $k$ and $k-1$,
respectively. These Jacobians are modified by making the smallest possible (in
terms of the Frobenius norm) changes to their entries while still preserving
the observability properties. This is done by solving the following
optimization problems.

$\displaystyle\quad\quad\quad\ \
\tilde{F}_{k}^{\ast}=\operatorname*{arg\,min}_{\tilde{F}_{k}}\|\tilde{F}_{k}-\hat{F}_{k}\|_{\mathcal{F}}^{2}$
(7a) s.t.
$\displaystyle\text{span}\\{N(\hat{x}_{k+1|k})\\}=\text{span}\\{\tilde{F}_{k}N(\hat{x}_{k|k-1})\\},$
and $\displaystyle\tilde{H}_{k}^{\ast}=$
$\displaystyle\operatorname*{arg\,min}_{\tilde{H}_{k}}\|\tilde{H}_{k}-\hat{H}_{k}\|_{\mathcal{F}}^{2}$
(7b) $\displaystyle\text{s.t.}\quad\tilde{H}_{k}N(\hat{x}_{k|k-1})=0.$

Here $\|\cdot\|_{\mathcal{F}}$ denotes the Frobenius norm. The constraints in
(7a) and (7b) guarantee that the unobservable subspace is preserved and that
the unobservable directions cannot be observed. The observability-constrained
EKF algorithm is shown in Alg. 1.

Algorithm 1 Observability-constrained EKF

0: $\\{\bar{u}_{k},y_{k}\\}_{k=1}^{L}$

0: $\\{\hat{x}_{k|k},P_{k|k}\\}_{k=1}^{L}$ Initialisation : estimated state
$\hat{x}_{1|0}$, covariance matrix $P_{1|0}$

For $k=1$ to $L$ do

Measurement update:

$\text{Calculate}\;\hat{H}_{k}$ using (6) and find $\tilde{H}_{k}^{\ast}$ by
solving (7b)

$S_{k}=\tilde{H}_{k}^{\ast}P_{k|k-1}(\tilde{H}_{k}^{\ast})^{\top}+R_{k}$

$K_{k}=P_{k|k-1}(\tilde{H}_{k}^{\ast})^{\top}S_{k}^{-1}$

$\hat{x}_{k|k}=\hat{x}_{k|k-1}+K_{k}(y_{k}-h(\hat{x}_{k}))$

$P_{k|k}=P_{k|k-1}-K_{k}\tilde{H}_{k}^{\ast}P_{k|k-1}$

Time update:

$\hat{x}_{k+1|k}=f(\hat{x}_{k|k},\bar{u}_{k},0)$

$\text{Calculate}\;\hat{F}_{k}$ using (6) and find $\tilde{F}_{k}^{\ast}$ by
solving (7a)

$G_{k}=\frac{\partial f}{\partial
w_{k}}\big{|}_{x_{k}=\hat{x}_{k|k},u_{k}=\bar{u}_{k},w_{k}=0}$

$P_{k+1|k}=\tilde{F}_{k}^{\ast}P_{k|k}(\tilde{F}_{k}^{\ast})^{\top}+G_{k}Q_{k}G^{\top}$

end for

##  III Application to the MAINS

The MAINS proposed in [1] utilizes an IMU and an array of magnetometers, see
Fig. 2, to perform odometry. The method presented in Section II will next be
used to modify the EKF algorithm used in the MAINS to address the
inconsistency illustrated in Fig. 1.

###  III-A State-space Model

Consider a simplified MAINS system where IMU biases are not included in the
state vector. The IMU biases are excluded as they only complicate the
forthcoming observability analysis and biases never improve the observability.
Later, when evaluating the proposed method the biases are included.

Let the state vector $x_{k}$ be defined as

$x_{k}\triangleq\begin{bmatrix}(p_{k}^{\text{n}})^{\top}\;(v_{k}^{\text{n}})^{\top}\;(q_{k})^{\top}\;(\theta_{k})^{\top}\end{bmatrix}^{\top}.$
(8)

Here the superscript n denotes the navigation frame where the physical
quantity is resolved. Further,
$p_{k}^{\text{n}}\in\mathbb{R}^{3},v^{\text{n}}_{k}\in\mathbb{R}^{3}$ denote
the position and velocity, respectively. Moreover, $q_{k}\in\mathbb{H}$
denotes the unit quaternion that encodes the orientation of the body frame
w.r.t. the navigation frame, and $\theta_{k}\in\mathbb{R}^{\kappa}$ denotes
the coefficients of the polynomial model used to describe the magnetic field;
see [1] for details.

Figure 2: The sensor board used in the MAINS. It has 30 PNI RM3100
magnetometers and an Osmium MIMU 4844 IMU.

Recalling the notation used in [1], the state dynamics is given by

$x_{k+1}=f(x_{k},u_{k},w_{k}^{\theta}),$ (9a) where $\displaystyle
f(x_{k},u_{k},w_{k})$
$\displaystyle=\begin{bmatrix}p_{k}^{\text{n}}+v_{k}^{\text{n}}t_{\text{s}}+(R_{k}s_{k}+\text{g})\frac{t_{\text{s}}^{2}}{2}\\\
v_{k}^{\text{n}}+(R_{k}s_{k}+\text{g})t_{\text{s}}\\\
q_{k}\otimes\text{exp}_{q}\left(\omega_{k}t_{\text{s}}\right)\\\
A^{\dagger}B(\psi_{k})\theta_{k}+w_{k}^{\theta}\end{bmatrix}.$ (9b)

Here $u_{k}\triangleq[s_{k}^{\top}\;\omega_{k}^{\top}]^{\top}$ is the system
input, where $s_{k}\in\mathbb{R}^{3}$ and $\omega_{k}\in\mathbb{R}^{3}$ denote
the specific force and angular velocity in body frame, respectively.
$w_{k}^{\theta}\in\mathbb{R}^{\kappa}$ denotes the process noise affecting the
coefficients of the magnetic field model. Furthermore, $t_{\text{s}}$ denotes
the sampling period, $R_{k}\in SO(3)$ denotes the rotation matrix
corresponding to $q_{k}$, $\otimes$ denotes quaternion multiplication, and
$\text{exp}_{q}(\cdot)$ maps an axis-angle rotation to a quaternion. Moreover,
the vector $\text{g}\in\mathbb{R}^{3}$ denotes the local gravity. The matrices
$A^{\dagger}$ and $B(\cdot)$ are used in [1] to propagate coefficients of the
magnetic field model; see [15] and [1] for details. Lastly,
$\psi_{k}\in\mathbb{R}^{6}$ denotes the pose change between time $k$ and
$k+1$, i.e.,

$\psi_{k}=\begin{bmatrix}R_{k}^{\top}t_{\text{s}}(v^{n}_{k}+\text{g}t_{\text{s}}/2)+s_{k}t_{\text{s}}^{2}/2\\\
\omega_{k}t_{\text{s}}\end{bmatrix},$ (10)

where the first and second elements encode the translation and orientation
change, respectively.

The measurements are from the magnetometer array, whose equation is given by

$y_{k}=Hx_{k}+e_{k},$ (11a) where $H=\begin{bmatrix}0_{3\times
10}&H^{\theta}(r_{1})\\\ \vdots&\vdots\\\ 0_{3\times
10}&H^{\theta}(r_{m})\end{bmatrix}.$ (11b)

Here $H^{\theta}(r_{i})\in\mathbb{R}^{3\times\kappa}$ denotes the measurement
matrix for magnetometer sensor location $r_{i}$, whose expression is given in
[16]. Further, $m$ denotes the number of magnetometer sensors.

###  III-B Linearized Error State Model

Since the state vector contains a quaternion component with a unit norm, the
state space is an embedded submanifold [17], which is not a vector space.
Therefore, the standard derivative typically applied to vector spaces cannot
be applied to linearize the model as in (9b). Instead, the error state
formulation presented in [18] will be used, which uses a small angle
approximation to perform the linearization indirectly.

Let $\delta x_{k}$ denote the error state vector

$\delta x_{k}\triangleq\begin{bmatrix}(\delta
p_{k}^{\text{n}})^{\top}\;(\delta
v_{k}^{\text{n}})^{\top}\;(\epsilon_{k})^{\top}\;(\delta\theta_{k})^{\top}\end{bmatrix}^{\top}.$
(12a) Further, let $\bar{x}_{k}$ denote the nominal state vector
$\bar{x}_{k}\triangleq\begin{bmatrix}(\bar{p}_{k}^{\text{n}})^{\top}\;(\bar{v}_{k}^{\text{n}})^{\top}\;(\bar{q}_{k})^{\top}\;(\bar{\theta}_{k})^{\top}\end{bmatrix}^{\top}.$
(12b)

Here the standard additive error definition is used, except for the
orientation error. For example, $\delta
v_{k}^{\text{n}}=v^{\text{n}}_{k}-\bar{v}_{k}^{\text{n}}$). The orientation
error $\epsilon_{k}\in\mathbb{R}^{3}$ satisfies the equation
$R_{k}\approx\bar{R}_{k}(I_{3}+[\epsilon_{k}]_{\times})$, where
$[\cdot]_{\times}$ is an operator that maps a vector in $\mathbb{R}^{3}$ to a
skew-symmetric matrix such that $\left[\xi\right]_{\times}b=\xi\times b$.

Given the measurement input
$\tilde{u}_{k}=[\tilde{s}_{k}^{\top}\;\tilde{\omega}_{k}^{\top}]^{\top}$, the
linearized error state dynamics is given by [1]

$\delta x_{k+1}=\bar{F}_{k}\delta x_{k}+\bar{G}_{k}w_{k},$ (13a) where
$\bar{F}_{k}\\!=\\!\begin{bmatrix}I_{3}\\!\\!&\\!\\!I_{3}t_{\text{s}}\\!\\!&\\!\\!0\\!\\!&\\!\\!0\\\
0\\!\\!&\\!\\!I_{3}\\!\\!&\\!\\!-\bar{R}_{k}[\tilde{s}_{k}]_{\times}t_{\text{s}}\\!\\!&\\!\\!0\\\
0\\!\\!&\\!\\!0\\!\\!&\\!\\!\text{exp}([\tilde{\omega}_{k}t_{\text{s}}]_{\times})^{\top}\\!\\!&\\!\\!0\\\
0\\!\\!&\\!\\!A^{\dagger}\bar{J}_{k}\bar{R}_{k}^{\top}t_{\text{s}}\\!\\!&\\!\\!A^{\dagger}\bar{J}_{k}[\eta(\bar{R}_{k},\bar{v}^{\text{n}}_{k})]_{\times}\\!\\!&\\!\\!A^{\dagger}B(\bar{\psi}_{k})\end{bmatrix},$
(13b) $\displaystyle\bar{\psi}_{k}$
$\displaystyle=\begin{bmatrix}\bar{R}_{k}^{\top}t_{\text{s}}(\bar{v}^{\text{n}}_{k}+\text{g}t_{\text{s}}/2)\\\
\tilde{\omega}_{k}t_{\text{s}}\end{bmatrix},$ (13c)
$\displaystyle\eta(\bar{R}_{k},\bar{v}^{\text{n}}_{k})$
$\displaystyle=\bar{R}_{k}^{\top}t_{\text{s}}(\bar{v}^{\text{n}}_{k}+\text{g}t_{\text{s}}/2),$
(13d) $\displaystyle\bar{J}_{k}$ $\displaystyle=\left.\frac{\partial
B(\psi_{k})\theta_{k}}{\Delta
p_{k}}\right|_{\psi_{k}=\bar{\psi}_{k},\theta_{k}=\bar{\theta}_{k}}.$ (13e)

Here ${\tilde{s}_{k}\in\mathbb{R}^{3}}$ and
${\tilde{\omega}_{k}\in\mathbb{R}^{3}}$ denote the measured acceleration and
angular velocity, respectively. Further,
$w_{k}\triangleq[(w_{k}^{s})^{\top}\;(w_{k}^{\omega})^{\top}\;(w_{k}^{\theta})^{\top}]^{\top}$
denote the process noise, where $w_{k}^{s}\in\mathbb{R}^{3}$ and
$w_{k}^{\omega}\in\mathbb{R}^{3}$ denote the acceleration and angular velocity
measurement noise, respectively. The explicit form of $\bar{G}_{k}$ is not
given as it is relevant in the observability analysis, interested readers can
find it in [1].

The corresponding measurement model is given by

$\delta y_{k}=H_{\delta x}\delta x_{k}+e_{k},$ (14a) where $H_{\delta
x}=\begin{bmatrix}0_{3\times 9}&H^{\theta}(r_{1})\\\ \vdots&\vdots\\\
0_{3\times 9}&H^{\theta}(r_{m})\end{bmatrix}$ (14b)

and $\delta y_{k}\triangleq y_{k}-H\bar{x}_{k}$.

###  III-C Unobservable subspace and Interpretations

Let

$N(x_{k})\triangleq\begin{bmatrix}I_{3}&0_{3\times 1}\\\ 0_{3\times
3}&-[v^{\text{n}}_{k}]_{\times}\text{g}\\\ 0_{3\times
3}&R_{k}^{\top}\text{g}\\\ 0_{\kappa\times 3}&0_{\kappa\times
1}\end{bmatrix},$ (15)

then the basis of the unobservable subspace associated with the linearized
error state model is given by the column vectors in $N(\bar{x}_{k})$. See
appendix for a proof. The first three columns of $N(\bar{x}_{k})$ correspond
to a body frame translation and the last column with the first three
corresponds to a navigation frame rotation around the gravity vector. Note
that the unobservable velocity component is caused by the fact that the yaw
angle cannot be determined. Thus the direction of the velocity in the
navigation frame is ambiguous.

###  III-D Suggested Modifications of the Jacobians

Since the measurement model (14) is linear and $H_{\delta_{x}}$ fulfills the
constraint in (7b), it is left unmodified, i.e.,
$\tilde{H}_{k}^{\ast}=H_{\delta_{x}}$. Only the Jacobian involved in the state
transition, i.e.,

$\hat{F}_{k}\\!=\\!\begin{bmatrix}I_{3}\\!\\!\\!&\\!\\!\\!I_{3}t_{\text{s}}\\!\\!&\\!\\!0\\!\\!&\\!\\!0\\\
0&I_{3}\\!\\!&\\!\\!-\hat{R}_{k}[\tilde{s}_{k}]_{\times}t_{\text{s}}\\!\\!&\\!\\!0\\\
0\\!\\!&\\!\\!0\\!\\!&\\!\\!\text{exp}([\tilde{\omega}_{k}t_{\text{s}}]_{\times})^{\top}\\!\\!&\\!\\!0\\\
0\\!\\!&\\!\\!A^{\dagger}\\!\hat{J}_{k}\hat{R}_{k}^{\top}t_{\text{s}}\\!\\!&\\!\\!A^{\dagger}\\!\hat{J}_{k}\hat{R}_{k}^{\top}t_{\text{s}}[\hat{v}^{\text{n}}_{k}+\text{g}\frac{t_{\text{s}}}{2}]_{\times}\hat{R}_{k}\\!\\!&\\!\\!\hat{T}_{k}^{k+1}\end{bmatrix}$
(16)

needs to be modified. Here the accent $\hat{\cdot}$ denotes the posterior
estimate of the quantity and $\hat{T}_{k}^{k+1}\triangleq
A^{\dagger}B(\hat{\psi}_{k})$. Furthermore, the equallity
$[R\xi]_{\times}=R[\xi]_{\times}R^{\top},R\in SO(3),\xi\in\mathbb{R}^{3}$ is
used.

Solving the optimization problem in (7a) is typically difficult, However, a
suboptimal solution can be obtained by narrowing the search space for
$\tilde{F}_{k}$ and transforming the constraint by selecting a specific set of
transformations $\\{\mathcal{E}_{k},\mathcal{E}_{k+1}\\}$ such that the basis
vectors in the constraint are equal, i.e.,

$N(\hat{x}_{k+1|k})\mathcal{E}_{k+1}=\tilde{F}_{k}N(\hat{x}_{k|k-1})\mathcal{E}_{k}.$
(17)

Since the goal is to make minimal changes to the original Jacobian
$\hat{F}_{k}$ but still fulfill the constraints in (7a), the sub-blocks of
matrices $\hat{F}_{k}$ that are independent of the linearization points are
kept unchanged. Furthermore, the sub-block $\hat{T}_{k}^{k+1}$ will not affect
the constraint since the last $\kappa$ rows of $N(\hat{x}_{k|k-1})$ are all
zeros. Hence, it is kept as it is. The remaining subblocks of $\hat{F}_{k}$
must be modified to meet the constraint in (7a). Therefore, the proposed
modified Jacobian $\tilde{F}_{k}$ has the structure

$\tilde{F}_{k}=\begin{bmatrix}I_{3}\\!&\\!I_{3}t_{\text{s}}&0&0\\\
0\\!&\\!I_{3}\\!&\\!\tilde{F}_{k}^{(1,2)}\\!&\\!0\\\
0\\!&\\!0\\!&\\!\tilde{F}_{k}^{(2,2)}\\!&\\!0\\\
0\\!&\\!A^{\dagger}\hat{J}_{k}\hat{R}_{k}^{\top}t_{\text{s}}\\!&\\!A^{\dagger}\hat{J}_{k}\hat{R}_{k}^{\top}t_{\text{s}}\tilde{F}_{k}^{(3,2)}\\!&\\!\hat{T}_{k}^{k+1}\end{bmatrix},$
(18)

where $\tilde{F}_{k}^{(1,2)}$, $\tilde{F}_{k}^{(2,2)}$, and
$\tilde{F}_{k}^{(3,2)}$ are the block matrices to be determined. Further,
transformations $\\{\mathcal{E}_{k},\mathcal{E}_{k+1}\\}$ are chosen as

$\mathcal{E}_{k}=\begin{bmatrix}I_{3}&a_{k}\\\
0&1\end{bmatrix},\quad\mathcal{E}_{k+1}=\begin{bmatrix}I_{3}&a_{k+1}\\\
0&1\end{bmatrix}.\quad$ (19)

Here $a_{k},a_{k+1}\in\mathbb{R}^{3}$ are column vectors. Then the
optimization problem (7a) can be written as

$\displaystyle\tilde{F}_{k}^{\ast}$
$\displaystyle=\operatorname*{arg\,min}_{\tilde{F}_{k}}\|\tilde{F}_{k}-\hat{F}_{k}\|_{\mathcal{F}}^{2}$
(20a) s.t. $\displaystyle
a_{k+1}=a_{k}-[\hat{v}^{\text{n}}_{k|k-1}]_{\times}\text{g}t_{\text{s}},$
(20b)
$\displaystyle[\hat{v}^{\text{n}}_{k+1|k}]_{\times}\text{g}=[\hat{v}^{\text{n}}_{k|k-1}]_{\times}\text{g}-\tilde{F}_{k}^{(1,2)}\hat{R}_{k|k-1}^{\top}\text{g},$
(20c)
$\displaystyle\hat{R}_{k+1|k}^{\top}\text{g}=\tilde{F}_{k}^{(2,2)}\hat{R}_{k|k-1}^{\top}\text{g}$
(20d) $\displaystyle
A^{\dagger}\hat{J}_{k}\hat{R}_{k}^{\top}t_{\text{s}}(-[\hat{v}^{\text{n}}_{k|k-1}]_{\times}\text{g}+\tilde{F}_{k}^{(3,2)}\hat{R}_{k|k-1}^{\top}\text{g})=0$
(20e)

These constraints are obtained by matching the entries on both sides of (17).

The constraint (20b) does not concern the modification of the Jacobian, it is
the result of keeping the first row of $\tilde{F}_{k}$ the same as that of
$\hat{F}_{k}$.

Since the constraint in (20) only contains one unique subblock of
$\tilde{F}_{k}$ the optimization problem in (20) can be split into three
separate optimization problems of the form

$\begin{split}\tilde{F}^{\ast^{(i,2)}}=&\operatorname*{arg\,min}_{\tilde{F}^{(i,2)}}\|\tilde{F}^{(i,2)}-\hat{F}^{(i,2)}\|_{\mathcal{F}}^{2},\\\
\text{s.t.}\quad&\tilde{F}^{(i,2)}u=w.\end{split}$ (21)

Here $\hat{F}^{(i,2)}$ denotes the matrix to be modified and
$u,w\in\mathbb{R}^{3}$. For example, the constraint (20c) corresponds to
$u=\hat{R}_{k|k-1}^{\top}\text{g}$,
$w=[\hat{v}^{\text{n}}_{k|k-1}-\hat{v}^{\text{n}}_{k+1|k}]_{\times}\text{g}$.
The optimization problem has a closed-form solution [8]

$\tilde{F}^{\ast^{(i,2)}}=\hat{F}^{(i,2)}-(\hat{F}^{(i,2)}u-w)(u^{\top}u)^{-1}u^{\top}.$
(22)

Note that in [8], the optimization problem set for $\tilde{F}^{(2,2)}$ is
different, where $\tilde{F}^{(2,2)}$ is constrained to be a rotation matrix
and the object is to minimize the squared norm of the difference of the
quaternions corresponding to the rotation matrices. In this paper, we also
adopted this approach. Furthermore, when dealing with (20e), we consider the
term in the parenthesis to be 0, although in the general case, it can be any
vector in the nullspace of
$A^{\dagger}\hat{J}_{k}\hat{R}_{k}^{\top}t_{\text{s}}$.

##  IV Experimental Evaluation

The proposed observability-constrained MAINS (OC-MAINS) algorithm is compared
with the original MAINS algorithm in [1]. Both algorithms are evaluated using
Monte Carlo simulations and real-world data. As a performance measure, the
root mean squared error (RMSE), defined as

$\text{RMSE}=\left(\frac{1}{M}\sum_{i=1}^{M}\|\hat{\rho}_{k,i}-\rho_{k,i}\|^{2}\right)^{1/2},$
(23)

is used. Here $\rho$ denotes the quantity for which RMSE is computed, $M$ is
the number of simulations, and $i$ denotes the the simulation index.
Furthermore, the perceived uncertainty of the estimate at each timestep is
calculated as

$\text{perceived
uncertainty}=\left(\frac{1}{M}\sum_{i=1}^{M}P^{\rho}_{k,i}\right)^{1/2}.$ (24)

Here $P^{\rho}_{k,i}$ denotes the covariance of the posterior estimate of the
quantity $\rho$ calculated by the EKF during the $i$:th simulation. The
covariance of the yaw is calculated as

$P^{\phi}_{k,i}={\nabla_{\text{yaw}}(\hat{q}_{k,i})}^{\top}P^{\epsilon}_{k,i}\nabla_{\text{yaw}}(\hat{q}_{k,i}).$
(25)

Here $P^{\epsilon}_{k,i}\in\mathbb{R}^{3\times 3}$ denote the block covariance
matrix corresponding to the orientation error. Furthermore,
$\nabla_{\text{yaw}}(\hat{q}_{k,i})\in\mathbb{R}^{3\times 1}$ denotes the
gradient w.r.t the Euler angles of the function that converts $\hat{q}_{k,i}$
to the yaw angle.

###  IV-A Simulation Setup and Result

In the simulation, a sensor board, similar to that shown in Fig. 2, moves in
squares of size 2 $\times$ 2 meters; see Fig. 3(a). The trajectory’s duration
was 8 seconds and the data was sampled at 100 Hz. The IMU measurements were
generated as the true value corrupted by additive white noise and biases. The
magnetometer measurements were generated as the magnetic field from a multi-
dipole model corrupted by additive white noise. The magnitude of the simulated
magnetic field is in Fig. 3(a). In total, 50 simulations with independent
noise, bias, and initial state realizations, were used in the Monte Carlo
simulation evaluation. The same filter parameters were used in both
algorithms.

The results are shown in Fig. 3(b) and Fig. 3(c). The figures show that the
proposed OC-MAINS algorithm generally has a smaller position and yaw RMSE.
Further, the perceived uncertainty of the yaw estimate is more consistent with
the true uncertainty (See also Fig. 1.). Concerning the position estimates,
even though the perceived uncertainty of the OC-MAINS is somewhat more
consistent with the true uncertainty of the position estimates, significant
inconsistency still exists. However, it is worth noting that the perceived
uncertainty of OC MIANS is always higher than the initial uncertainty, thus
satisfying the condition in (1). The original MAINS algorithm does not meet
this condition.

###  IV-B Experimental Setup and Result

In the experiment, a person held the sensor board in Fig. 2 parallel to the
ground, and walked in squares for a few laps. The true trajectory was measured
using a camera-based motion-tracking system. The same filter parameters were
used in both algorithms.

The results are shown in Fig. 3(e) and Fig. 3(f). From Fig. 3(f) it can be
seen that the yaw error of OC-MAINS is significantly lower than that of the
original MAINS. Also, it can be seen that the perceived and true yaw
uncertainty agree when using the OC-MAINS algorithm. From Fig. 3(e) it can be
seen that the position error of the original MAINS and OC-MAINS algorithms are
about the same, but the OC-MAINS algorithm has slightly better performance in
the $y$-axis direction. For the OC-MAINS algorithm, the perceived and true
uncertainty agrees in the $y$\- and $z$-axis directions, whereas in the
$x$-axis there is a significant inconsistency. This could result from
imperfect IMU calibration. Similar to in the simulations, the perceived
uncertainty of OC MIANS is always higher than the initial uncertainty, thus
satisfying the condition in (1).

(a) Estimated and true trajectory, and the magnetic field magnitude.

(b) Position RMSE (solid) and perceived uncertainty (dashed).

(c) Yaw RMSE (solid) and perceived uncertainty (dashed).

Figure 3: Monte Carlo simulation results. The RMSE results are the average
value calculated from 50 independent simulations.

(d) Estimated and true trajectory, and the magnetic field magnitude

(e) Position RMSE (solid) and perceived uncertainty (dashed).

(f) Yaw RMSE (solid) and perceived uncertainty (dashed).

Figure 4: Results from real-world experiment. The RMSE results are the average
value calculated from using 12 randomly selected initialization states.

##  V Summary & Conclusions

A method to construct an observability-constrained magnetic-field-aided INS
has been proposed. The proposed method modifies the state transition matrix in
the EKF to ensure that the observability properties of the nonlinear system
are preserved in the filtering process. The method extends the work in [8] by
introducing a transformation of the basis that spans the unobservable
subspace. This transformation reduces the changes that must be done to the
state transition matrix. The proposed method was evaluated on simulation and
real-world datasets, showing that (i) the observability properties are
preserved, (ii) the estimation accuracy increases, and (iii) the perceived
uncertainty calculated by the EKF is more consistent with the true uncertainty
of the filter estimates.

## Acknowledgment

This work has been funded by Security Link and the Swedish Research Council
(Vetenskapsrådet) project 2020-04253 ”Tensor-field based localization”.

##  Appendix A The Unobservable Subspace Basis

To derive the unobservable subspace basis for the system defined by (13) and
(14), $\bar{F}_{k}$ is rewritten to its equivalent form. Propagating the
nominal state $\bar{x}_{k}$ with $\tilde{u}_{k}$ using (9b) leads to

$\displaystyle\bar{v}^{\text{n}}_{k+1}$
$\displaystyle=\bar{v}^{\text{n}}_{k}+\bar{R}_{k}\tilde{s}_{k}t_{\text{s}}+\text{g}t_{\text{s}},$
(26a) $\displaystyle\bar{q}_{k+1}$
$\displaystyle=\bar{q}_{k}\otimes\text{exp}_{q}\left(\tilde{\omega}_{k}t_{\text{s}}\right).$
(26b)

By rearranging the terms and using the equality
$[R\xi]_{\times}=R[\xi]_{\times}R^{\top}$, where $R\in SO(3)$ and
$\xi\in\mathbb{R}^{3}$, the following holds

$\displaystyle\bar{R}_{k}[\tilde{s}_{k}]_{\times}t_{\text{s}}$
$\displaystyle=[\bar{v}^{\text{n}}_{k+1}-\bar{v}^{\text{n}}_{k}-\text{g}t_{\text{s}}]_{\times}\bar{R}_{k},$
(27a)
$\displaystyle\text{exp}([\tilde{\omega}_{k}t_{\text{s}}]_{\times})^{\top}$
$\displaystyle=\bar{R}_{k+1}^{\top}\bar{R}_{k},$ (27b)

Let $\bar{T}_{k}^{k+1}\triangleq A^{\dagger}B(\bar{\psi}_{k})$ and
$\Delta\bar{p}_{k}^{G}=\bar{R}_{k}\Delta\bar{p}_{k}$, then it holds that

$A^{\dagger}\bar{J}_{k}\bar{R}_{k}^{\top}t_{\text{s}}=\frac{\partial
A^{\dagger}B(\bar{\psi}_{k})\bar{\theta}_{k}}{\partial\Delta\bar{p}_{k}}\bar{R}_{k}^{\top}t_{\text{s}}=\frac{\partial\bar{T}_{k}^{k+1}\bar{\theta}_{k}}{\partial\Delta\bar{p}_{k}^{G}}t_{\text{s}},$
(28a) and
$A^{\dagger}\bar{J}_{k}[\eta(\bar{R}_{k},\bar{v}^{\text{n}}_{k})]_{\times}=\frac{\partial\bar{T}_{k}^{k+1}\bar{\theta}_{k}}{\partial\Delta\bar{p}_{k}^{G}}t_{\text{s}}[\bar{v}^{\text{n}}_{k}+\text{g}t_{\text{s}}/2]_{\times}\bar{R}_{k}.$
(28b)

Next, combining (27) and (28), then $\bar{F}_{k}$ can be rewritten as

$\displaystyle\bar{F}_{k}$
$\displaystyle=\begin{bmatrix}I_{3}\\!&\\!I_{3}t_{\text{s}}\\!&\\!0\\!&\\!0\\\
0\\!&\\!I_{3}\\!&\\!-[\bar{v}^{\text{n}}_{k+1}-\bar{v}^{\text{n}}_{k}-\text{g}t_{\text{s}}]_{\times}\bar{R}_{k}\\!&\\!0\\\
0\\!&\\!0\\!&\\!\bar{R}_{k+1}^{\top}\bar{R}_{k}\\!&\\!0\\\
0\\!&\\!\frac{\partial\bar{T}_{k}^{k+1}\bar{\theta}_{k}}{\partial\Delta\bar{p}_{k}^{G}}t_{\text{s}}\\!&\\!\frac{\partial\bar{T}_{k}^{k+1}\bar{\theta}_{k}}{\partial\Delta\bar{p}_{k}^{G}}t_{\text{s}}[\bar{v}^{\text{n}}_{k}+\text{g}t_{\text{s}}/2]_{\times}\bar{R}_{k}\\!&\\!\bar{T}_{k}^{k+1}\end{bmatrix},$

The matrix $\bar{\Phi}(l,k)$ can then, for $l>k$, be written as

$\displaystyle\bar{\Phi}(l,k)$
$\displaystyle=\begin{bmatrix}I_{3}\\!&\\!\bullet\\!&\\!\bullet\\!&\\!0\\\
0\\!&\\!I_{3}\\!&\\!-[\bar{v}^{\text{n}}_{l}-\bar{v}^{\text{n}}_{k}-\text{g}(l-k)t_{\text{s}}]_{\times}\bar{R}_{k}\\!&\\!0\\\
0\\!&\\!0\\!&\\!\bar{R}_{l}^{\top}\bar{R}_{k}\\!&\\!0\\\
0\\!&\\!\sum\limits_{j=k}^{l-1}D_{j}\\!&\\!\sum\limits_{j=k}^{l-1}D_{j}[\bar{v}^{\text{n}}_{k}+\frac{2(j-k)+1}{2}t_{\text{s}}\text{g}]_{\times}\bar{R}_{k}\\!&\\!\bar{T}^{l}_{k}\end{bmatrix},$
where
$D_{j}=\frac{\partial\bar{T}^{l}_{k}\bar{\theta}_{k}}{\partial\Delta\bar{p}_{j}^{G}}t_{\text{s}}\quad\text{and}\quad\bar{T}^{l}_{k}=\bar{T}^{l}_{l-1}\bar{T}^{l-1}_{l-2}\cdots\bar{T}^{k+1}_{k}.$
(30b)

Here $\bullet$ denotes subblocks not required for the analysis.

Now, it can be verified that the columns of $N(\bar{x}_{k})$ in (15) are the
basis vectors of the unobservable subspace by multiplying every block row in
$\bar{\mathcal{O}}_{k}$ with $N(\bar{x}_{k})$. Let
$\bar{\mathcal{O}}_{k}^{(s,l)}\triangleq H_{\delta x}^{(s)}\bar{\Phi}(l,k)$,
where $H_{\delta x}^{(s)}$ denotes the block row in $H_{\delta x}$
corresponding to the $s$-th magnetometer. For $l=k$, $\bar{\Phi}(k,k)=I$, then

$\displaystyle\bar{\mathcal{O}}_{k}^{(s,k)}=\begin{bmatrix}0_{3\times
3}&0_{3\times 3}&0_{3\times 3}&H^{\theta}(r_{s})\end{bmatrix},$ (31)

and $\bar{\mathcal{O}}_{k}^{(s,k)}N(\bar{x}_{k})=\left[0_{3\times 3}\quad
0_{3\times 1}\right]$, which can be verified by simple calculation.

For $l>k$,

$\bar{\mathcal{O}}_{k}^{(s,l)}=H^{\theta}(r_{s})\\!\begin{bmatrix}0_{3\times
3}\\\ (\sum\limits_{j=k}^{l-1}D_{j})^{\top}\\\
(\sum\limits_{j=k}^{l-1}D_{j}[\bar{v}^{\text{n}}_{k}+\frac{2(j-k)+1}{2}t_{\text{s}}\text{g}]_{\times}\bar{R}_{k})^{\top}\\\
(\bar{T}^{l}_{k})^{\top}\end{bmatrix}^{\top}$ (32)

and

$\displaystyle\bar{\mathcal{O}}_{k}^{(s,l)}N(\bar{x}_{k})$ $\displaystyle=$
$\displaystyle
H^{\theta}(r_{m_{s}})\left[0\quad\sum\limits_{j=k}^{l-1}D_{j}(-[\bar{v}^{\text{n}}_{k}]_{\times}\text{g}\right.$
(33)
$\displaystyle+\>\left.[\bar{v}^{\text{n}}_{k}+\frac{2(j-k)+1}{2}t_{\text{s}}\text{g}]_{\times}\bar{R}_{k}\bar{R}_{k}^{\top}\text{g})\right]$
$\displaystyle=$ $\displaystyle
H^{\theta}(r_{m_{s}})\left[0\quad\sum\limits_{j=k}^{l-1}D_{j}(-[\bar{v}^{\text{n}}_{k}]_{\times}\text{g}+[\bar{v}^{\text{n}}_{k}]_{\times}\text{g}\right.$
$\displaystyle+\left.\>[\frac{2(j-k)+1}{2}t_{\text{s}}\text{g}]_{\times})\text{g}\right]$
$\displaystyle=\left[0_{3\times 3}\quad 0_{3\times 1}\right],$

Here the equality $[\text{g}]_{\times}\text{g}=0$ is used in the last step.

Since $\mathcal{\bar{O}}_{k}^{(s,l)}N(\bar{x}_{k})=0$ $\forall s,l\geq k$ then
$\mathcal{\bar{O}}_{k}N(\bar{x}_{k})=0$ and the column vectors of
$N(\bar{x}_{k})$ are the basis vectors of the unobservable subspace.

##  Appendix B Interpretation of the Unobservable Subspace

The error state vector can be seen as perturbations to the state vector. The
proof proceeds by applying a perturbation to the state and then calculating
the resulting error state vector. If the error state vector lies in the space
spanned by some columns in $N(\bar{x}_{k})$ then those unobservable directions
correspond to the perturbation applied.

Suppose the initial true state
$x=[p^{\text{n}^{\top}}\;v^{\text{n}^{\top}}\;q^{\top}\;\theta^{\top}]^{\top}$,
where the time index will be dropped for brevity. Next, suppose that a
translation $\Delta\in\mathbb{R}^{3}$ is applied to the body frame. That will
change the state to
$x^{\prime}=[(p^{\text{n}}\\!+\\!\Delta)^{\top}\;v^{\text{n}^{\top}}\;q^{\top}\;\theta^{\top}]^{\top}$.
The error state $\delta x=[-\Delta^{\top}\quad 0^{\top}\quad 0^{\top}\quad
0^{\top}]^{\top}$ lies in the space spanned by the first three columns in
$N(\bar{x}_{k})$. Thus, the perturbation implied by the first three columns is
a body frame translation.

Next, suppose the navigation frame is rotated around the gravity vector by an
angle $c\in[0,2\pi)$ and denote the new navigation frame as
$\text{n}^{\prime}$-frame, then the changed state $x^{\prime}$ becomes

$x^{\prime}=\begin{bmatrix}{}_{\text{n}}^{\text{n}^{\prime}}R\cdot
p^{\text{n}}\\\ {}_{\text{n}}^{\text{n}^{\prime}}R\cdot v^{\text{n}}\\\
q\\{{}_{\text{n}}^{\text{n}^{\prime}}R\\}\otimes q\\\ \theta\\\
\end{bmatrix},\quad{}_{\text{n}}^{\text{n}^{\prime}}R=R^{\top}\\{c\,\text{g}\\},$
(34)

where ${}_{\text{n}}^{\text{n}^{\prime}}R$ denotes the rotation matrix that
transforms the coordinate in n-frame to $\text{n}^{\prime}$-frame,
$R\\{c\,\text{g}\\}$ denotes the rotation matrix corresponding to the axis-
angle representation $c\,\text{g}$, and
$q\\{{}_{\text{n}}^{\text{n}^{\prime}}R\\}$ denotes the quaternion
corresponding to the rotation matrix ${}_{\text{n}}^{\text{n}^{\prime}}R$.

Let $\epsilon\in\mathbb{R}^{3}$ denote the orientation error, i.e. the
orientation difference in $x$ and $x^{\prime}$, and $c$ be infinitesimally
small. Recalling the definition of orientation error and using
$R^{\top}\\{c\,\text{g}\\}\approx I_{3}-[c\,\text{g}]_{\times}$, it holds that

$\begin{split}R&\approx({}_{\text{n}}^{\text{n}^{\prime}}R\cdot
R)\cdot(I_{3}+[\epsilon]_{\times})=(R^{\top}\\{c\,\text{g}\\}\cdot
R)\cdot(I_{3}+[\epsilon]_{\times})\\\
&\approx(I_{3}-[c\,\text{g}]_{\times})R(I_{3}+[\epsilon]_{\times})\end{split}$
(35)

where $R$ is the rotation matrix corresponding to the quaternion $q$.
Neglecting the second-order term, then it holds that

$R\approx R+R[\epsilon]_{\times}-[c\,\text{g}]_{\times}R,$ (36)

which leads to

$[c\,\text{g}]_{\times}\approx
R[\epsilon]_{\times}R^{\top}=[R\epsilon]_{\times}.$ (37)

Therefore,

$\epsilon=cR^{\top}\text{g}.$ (38)

The error state for the position is calculated as

$\begin{split}p^{\text{n}}-{}_{\text{n}}^{\text{n}^{\prime}}R\cdot
p^{\text{n}}&=p^{\text{n}}-R^{\top}\\{c\,\text{g}\\}\cdot p^{\text{n}}\\\
&\approx(I_{3}-(I_{3}-[c\,\text{g}]_{\times}))\cdot p^{\text{n}},\\\
&=-c[p^{\text{n}}]_{\times}\text{g}\end{split}$ (39)

where $[\text{g}]_{\times}p^{\text{n}}=-[p^{\text{n}}]_{\times}\text{g}$ is
used. Similarly, the error state for the velocity is
$-c[v^{\text{n}}]_{\times}\text{g}$. Therefore, the error state is

$\delta x=c\begin{bmatrix}-[p^{\text{n}}]_{\times}\text{g}\\\
-[v^{\text{n}}]_{\times}\text{g}\\\ R^{\top}\text{g}^{\top}\\\
0\end{bmatrix}=c\left(-\begin{bmatrix}I_{3}\\\ 0_{3\times 3}\\\ 0_{3\times
3}\\\ 0_{\kappa\times
3}\end{bmatrix}[p^{\text{n}}]_{\times}\text{g}+\begin{bmatrix}0\\\
-[v^{\text{n}}]_{\times}\text{g}\\\ R^{\top}\text{g}^{\top}\\\
0\end{bmatrix}\right)$ (40)

lies in the space spanned by all columns, so they correspond to a navigation
frame rotation around the gravity vector.

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