3
$\begingroup$

Statement Prove that the degree of any irreducible factor of $x^8-x$ over $\mathbb Z_2$ is $1$ or $3$.

The hint in the back of the book makes a certain claim that I wasn't sure about.

The author says that we take an irreducible factor, call it $g(x)$, and let it's degree be $m$. Then the field $\mathbb Z_2[x]/(g) \approx GF(2^m)$.

Then the author claims that this is a subfield of $GF(8)$, which would also be the field of roots for the polynomial $x^8-x$.

I'm having trouble seeing why $GF(2^m)$ has to be a subfield of $GF(8)$. Sure, there is an element $g \in GF(2^m)$ that corresponds to $x + (g(x))$ which is the zero we get when we construct the extension field, but I don't see why the rest of the elements are necessarily in $GF(8)$.

$\endgroup$
  • $\begingroup$ Short answer: if all the fields you work with are assumed to be subfields of the same algebraically closed field, then $GF(8)$ is the set of things $y$ in that field for which $y^8 = y$. Since elements of $Z_2[X]/(g)$ satisfy that property, you have $Z_2[X]/(g) \subseteq GF(8)$. $\endgroup$ – D_S Apr 1 '16 at 2:52
2
$\begingroup$

Generally when you work with field extensions, you want to assume that everything you consider sits inside some big algebraically closed field. For example in number theory, you regard all the finite extensions of $\mathbb{Q}$ as subfields of $\mathbb{C}$.

Let's fix an algebraic closure $K$ of $\mathbb Z_2$. So we can take all the extension fields of $\mathbb Z_2$ that we consider to be subfields of $K$. For each $m \geq 1$, there is a unique subfield $F$ (not even up to isomorphism, I really mean unique) of $K$ for which $[F : \mathbb Z_2] = m$.

Proof: Let $F$ be a field with $[F : \mathbb Z_2] = m$. Then $F$ has $2^m$ elements. And because $F \setminus \{0\}$ is a group with $2^m-1$ elements, you have that every element of $F$ is a root of the polynomial $X^{2^m} - X$. But the roots of this last polynomial in any algebraically closed field containing $\mathbb Z_2$ are distinct, so the roots of $X^{2^m}-X$ are exactly the elements of $F$. So if we assume $F$ is a subfield of $K$, then $F$ literally consists of those elements in $K$ which are roots of the polynomial $X^{2^m}-X$.

So back to your problem. You might as well assume $GF(2^3)$ to be a subfield of some fixed algebraic closure $K$ of $\mathbb Z_2$. If you take any irreducible factor $g$ of $X^{2^3}-X$, and you look at a field $\mathbb Z_2[x]$ where $x$ is some root of $g$, and you consider this to be a subfield of the same algebraically closed field $K$ as the one containing $GF(2^3)$, then you are going to literally have $\mathbb Z_2[x]$ contained in $GF(2^3)$, because $GF(2^3)$ is exactly the set of $y \in K$ satisfying $y^{2^3} = y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.