# 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 $X$ [duplicate]

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?

Define $[g(X)]$ to be the class of $g(X)$.
There are 8 elements in $F_2[X]/(X^3+X+1):$
$,,[X],[X^2],[X^3],[X^4],[X^5],[X^6]$.
Note that:$[X^3] = [X+1], [X^4] = [X^2+X], [X^5] = [X^2+X+1], [X^6] = [X^2+1],[X^7] = $
$[X],[X^2],[X^4]$ are roots of $y^3+y+1$ and $[X^3],[X^5],[X^6]$ are roots of $y^3+y^2+1$.
$φ_1([X]) = [X^3] = [X+1], φ_2([X]) = [X^5] = [X^2+X+1], φ_3([X]) = [X^6] = [X^2+1]$