<s>[INST] If $\phi$ : $F_2[X]/(X^3+X+1)$ $\rightarrow$ $F_2[X]/(X^3+X^2+1)$ is an isomorphism, then prove that $\phi(a)=b+1$, with $a$, $b$ the classes of $X$ Consider two fields:
$K: $ $F_2[X]/(X^3+X+1)$. Let $a$ be the class of $X$ (so $a=X+(X^3+X+1))$
$L: $ $F_2[X]/(X^3+X^2+1)$. Let $b$ be the class of $X$ (so $b=X+(X^3+X^2+1))$
$F_2$ denotes the field with $2$ elements. Field $L$ is ismorphic with $K$. 
If $\phi$ : $F_2[X]/(X^3+X+1)$ $\rightarrow$ $F_2[X]/(X^3+X^2+1)$ is an isomorphism, then prove that $\phi(a)=b+1$.
They give as a hint: $b+1$ is a root of $Y^3+Y+1$.
I came up with the following:
Fill in $a$ in $K$, so you get $a^3+a+1$.
Fill in $b+1$ in $L$, so you get $(b+1)^3+(b+1)^2+1 = b^3+b^2+b+1+b^2+1= b^3+b+1$, which is equal to the $a$ part above.
But I don't know if this is a way to actually prove that $\phi(a)=b+1$. How can I do this?
 [/INST] Define $[g(X)]$ to be the class of $g(X)$.
There are 8 elements in $F_2[X]/(X^3+X+1):$
$[0],[1],[X],[X^2],[X^3],[X^4],[X^5],[X^6]$.
Note that:$[X^3] = [X+1],
 [X^4] = [X^2+X],
 [X^5] = [X^2+X+1],
 [X^6] = [X^2+1],[X^7] = [1]$
$[X],[X^2],[X^4]$ are roots of $y^3+y+1$
and $[X^3],[X^5],[X^6]$ are roots of $y^3+y^2+1$.
So actually there are three isomorphisms:
$φ_1([X]) = [X^3] = [X+1],
φ_2([X]) = [X^5] = [X^2+X+1],
φ_3([X]) = [X^6] = [X^2+1]$
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