5
$\begingroup$

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.

  • 0
    I think my main difficulty is coming from how to view the quotient ring. Am I right to be viewing elements of $L$ as $ f(X) + (X^3 + X + 1)$? Or should I be viewing them as $ f(\alpha)$, where $\alpha$ is the image of $X$ under the quotient map?2011-12-14
  • 0
    Close to being a [duplicate question.](http://math.stackexchange.com/q/40326/11619) Mind you, I think that Gerry's answer here together with Dilip's nice observation is a very nice addition to the answers offered there. Not voting to close.2011-12-14
  • 0
    The mapping $\varphi$ doesn't **have to** come from an **isomorphism** from $F_2[x]$ to itself. It suffices that the mapping induced to the quotient rings is. In this case the quotient rings are actually fields. Any homomorphism between fields is injective, and the two fields both have 8 elements, so... The key is that you get a **well-defined** map, i.e. one that sends multiples of $x^3+x+1$ to multiples of $x^3+x^2+1$. The mappings $\varphi_2:x\mapsto x^2+1$ and $\varphi_3:x\mapsto x^2+x$ also work here.2011-12-16

2 Answers 2