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.

  • $\begingroup$ 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? $\endgroup$ – Jonathan Dec 14 '11 at 11:54
  • $\begingroup$ Close to being a duplicate question. 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. $\endgroup$ – Jyrki Lahtonen Dec 14 '11 at 17:34
  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Dec 16 '11 at 6:48

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 $(\hat{a}, \hat{b}, \hat{c})$ 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 $$ r^3 = r + 1 \Rightarrow r = r^{-2} + r^{-1} ~~\text{and}~~ r^2 = 1 + r^{-1} $$ and so $$\begin{align*}ar^2 + br + c &= a(r^{-1} + 1) + b(r^{-2} + r^{-1}) + c\\ &= br^{-2} + (a+b)r^{-1} + (a+c)\\ &= \hat{a}r^{-2} + \hat{b}r^{-1} + \hat{c} \end{align*}$$ 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 $a(r+1)^2+b(r+1)+c$.

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    $\begingroup$ Borrowing notation from Gerry Myerson's answer, the roots of $X^3 + X^2 + 1$ are inverses of the roots of $X^3 + X + 1$. $$r^3 + r + 1 = 0 \Rightarrow r^3(1 + r^{-2} + r^{-3}) = 0 \Rightarrow (r^{-1})^3 + (r^{-1})^2 + 1 = 0.$$ $\endgroup$ – Dilip Sarwate Dec 14 '11 at 13:45

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