Vtyjkyuk

Suppose $K$ is a finite field, $K = mathbb F_p^s$. If we take an irreducible polynomial $f$ of degree $d$ over $K$, then the splitting field $L$ of $f$ is $K(alpha)$ where $f$ is the minimal polynomial of $alpha$. But then $L = mathbb F_p^sd $. Since $mathbb F_p^sd$ is unique, we see that this is the splitting field of every irreducible polynomial of degree $d$ over $K$.

Take $K = mathbb F_2$ and let $P(X) = X^3 + X + 1$, $Q(X) = X^3 + X^2 + 1$. Let $L$ be the splitting field of $P$ and $L'$ be the splitting field of $Q$. The above tells us that $L$ and $L'$ are isomorphic. I would like to construct an explicit isomorphism between $L$ and $L'$.

I know that $L cong mathbb F_2[X] /(X^3 +X + 1)$ and $L' cong mathbb F_2[X] / (X^3 + X^2 + 1)$. Intuitively, I want to find an isomorphism $phi : mathbb F_2[X] to mathbb F_2[X]$ such that $phi((X^3 + X + 1)) = (X^3 + X^2 + 1)$. A little playing around gives me $phi(X) = X+1$. It now feels like I'm falling at the last hurdle: how do I finish the construction of an isomorphism between $L$ and $L'$? I don't think $phi$ makes sense as a map from $L$ to $L'$, yet it seems the map I want.

Borrowing notation from Gerry Myerson's answer, the field
$L = mathbb F_2[x]/(x^3 + x + 1)$ is the set of $8$ polynomials
of degree $2$ or less over $mathbb F_2$ with field addition and
multiplication being polynomial addition and multiplication
modulo $x^3 + x + 1$. Equivalently, $L$ is
the set of elements $ar^2 + br + c$ where
$a,b,c in mathbb F_2$ and $r^3 + r + 1 = 0$. It is also the vector
space $mathbb F_2^3$ whose elements are represented as $3$-tuples $(a,b,c)$
with respect to the basis $r^2, r, 1$. Now,
$$r^3 + r + 1 = 0 Rightarrow r^3(1 + r^-2 + r^-3) = 0
Rightarrow (r^-1)^3 + (r^-1)^2 + 1 = 0$$
so that $r^-1 in L$ is a root of $x^3 + x^2 + 1$.
Thus, $L^prime = mathbb F_2[x]/(x^3 + x^2 + 1)$
is the vector space $mathbb F_2^3$ where we are
representing elements as $3$-tuples $(hata, hatb, hatc)$
with respect to the basis $r^-2, r^-1, 1$. In particular,
dividing by $r^3 = r + 1$ by $r^2$ and $r$ respectively gives
$$
r^3 = r + 1 Rightarrow r = r^-2 + r^-1 ~~textand~~ r^2 = 1 + r^-1
$$
and so
$$beginalign*ar^2 + br + c &= a(r^-1 + 1) + b(r^-2 + r^-1) + c\
&= br^-2 + (a+b)r^-1 + (a+c)\
&= hatar^-2 + hatbr^-1 + hatc
endalign*$$
represent the same element.

You can think of $L$ as the set of all things of the form $ar^2+br+c$ where $a,b,c$ come from $bf F_2$ and $r$ satisfies $r^3+r+1=0$. Now $L$ also contains a zero of $X^3+X^2+1$, so you are looking for the values of $a,b,c$ such that if $s=ar^2+br+c$ then $s^3+s^2+1=0$. You can just multiply everything out, use $r^3+r+1=0$ to get it down to a quadratic in $r$, set the coefficients to zero, and solve. This is probably a mess. There may be an easier way to do it, but this will get you an element of $L$ that satisfies $X^3+X^2+1=0$. Once you have that element, you know the isomorphism you're looking for takes $r$ to that element. Since $r$ generates $L$, you get the entire isomorphism.

Now in fact I think you have managed to find that (in my notation) $r+1$ is the element you are looking for. So your map from $L$ to $L'$ takes $r+1$ in $L$ to a generator, call it $t$, in $L'$. Might be easier to see it as a map from $L'$ to $L$, taking $at^2+bt+c$ to $a(r+1)^2+b(r+1)+c$.