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We were discussing about the possibility of having an algebra over a field which is associative but has not the unity. Does it exist? It has been proposed as a counterexample the set of even numbers. We believe it is wrong because it can't be a vector space. If an associative but not unital algebra exists, how can we write its associative property? We usually write it as a commutative diagram or in terms of maps composition. In these terms the identity seams unavoidable. i.e. $$ m\circ\left(m\otimes id\right) = m\circ\left(id\otimes m\right)\qquad associativity$$ with $m : A\otimes A\rightarrow A$

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Try the following: take your field to be R (the real numbers) and take $A$ to be the $R$-algebra consisting of all $R$-valued sequences $(a_n)_{n=1}^\infty$ which satisfy $\lim_n |a_n| =0$. –  Yemon Choi Apr 12 '10 at 18:48
Does this help? en.wikipedia.org/wiki/Associative_algebra Also, you should check spelling/grammar before posting. Especially for a new user, people are likely to discount your question without even reading it if it looks like not much effort was put into writing it. –  Steven Gubkin Apr 12 '10 at 18:49
Also you can have an identity MAP without having an identity ELEMENT. –  Steven Gubkin Apr 12 '10 at 18:50
To simplify Yemon's example a bit: just consider any vector space $V$ over your field, and define the product to be zero, for all pairs of elements of $V$ . This is an associative algebra over the field which does not have an identity –  Mariano Suárez-Alvarez Apr 12 '10 at 19:08
@Mariano - I did consider a drastic example of the type you describe. The one I gave was motivated by the fact that functional analysis has quite a lot of natural examples which do not have identity elements, but which "share many of the nice properties of algebras with identity". (Specifically, $c_0$ has a bounded approximate identity, from which much niceness follows.) –  Yemon Choi Apr 12 '10 at 19:09

1 Answer 1

The set of even numbers is a non-unital ring, in particular it has an associative multiplication, but you are right that it isn't an algebra over a field. For an example of a non-unital algebra, consider the continuous functions $\mathbb{R} \to \mathbb{R}$ which vanish at a particular point, under pointwise product. More generally, any proper two-sided ideal of an algebra is a non-unital algebra, just as any proper two-sided ideal of a ring is a non-unital ring.

You may think that these examples are "unnatural," so here is a "natural" one: the algebra of compactly supported continuous functions $\mathbb{R} \to \mathbb{R}$ under convolution.

For an example of a non-associative algebra, take, for example, the octonions. Your confusion arises because of the following issue: given a vector space $V$ and a bilinear operation $V \times V \to V$ we can associate to any $a \in V$ the linear operator $L_a$ which is left multiplication by $a$, and these linear operators form an associative algebra. However, composition of the operators $L_a$ need not be the same as $\times$: associativity is equivalent to the statement that $L_a L_b = L_{a \times b}$, which need not be the case in general. I think this is the source of your confusion.

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A small technicality: Proper 2-sided ideals sure can be unital as algebras (or rings) in their own right. They are not unital subalgebras (or subrings), i.e., they can't have the same unit, but that isn't a problem for the question at hand. –  Jonas Meyer Apr 13 '10 at 0:59
An example of Jonas' point: take functions on a disconnected space, and consider the ideal of functions vanishing on one component. –  Tom Church Apr 13 '10 at 1:36
Toy example: 6 is the unit for $(2) \subset Z/_{10}$. –  Sammy Black Apr 13 '10 at 6:50

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