Is the splitting field equal to the quotient $k[x]/(f(x))$ for finite fields?

Question

maybe that's an idiot question. Given a finite field $k$ and some irreducible polynomial $f(x) \in k[x]$, then $k_f \cong k[x]/(f(x))$? I know that it's true if $k$ is the prime field and I think that the statement is not true for general finite fields, however I could not find any counter example.

Thanks in advance.

Answer 1

Yes, this is true. Basically it is one of the reformulations of the fact that all extensions of fields $\ell/k$ with both $\ell$ and $k$ finite are Galois.

Another way of making this very concrete is that we know the Galois group to be generated by the Frobenius automorphism $F(x)=x^q$, $q=|k|$. So if $\alpha$ is one of the roots of $f(x)$ in $k_f$, then $\alpha^q$, $\alpha^{q^2}$, $\alpha^{q^3}$, $\ldots$ are the others.

It is easy to see that if $m$ is the smallest positive integer with the property that $F^m(\alpha)=\alpha$, then the polynomial $$ (x-\alpha)(x-F(\alpha))\cdots(x-F^{m-1}(\alpha)) $$ is invariant under $F$. Hence it belongs to $k[x]$, hence it must be equal to $f(x)$.

Answer 2

Show that for $\;f(x)=x^3-2\in\Bbb Q[x]\;$ , the extension

$$\Bbb Q[x]/(f(x))\cong\Bbb Q(\sqrt[3]2)$$

is not normal (hint: what are the roots of $\,f(x)\,$ ?)

Edit for the case of finite fields:

Let $\,k:=\Bbb F_{p^n}\;,\;\;p\,$ a prime and $\,n\in\Bbb N\,$ , be a finite field. Now, as with any other finite field, we know $\,k\,$ is the set of all the roots of $\,q_n(x):=x^{p^n}-x\in\Bbb F_p[x]\,$ in some algebraic closure of the prime field.

Let $\,f(x)\in k[x]\,$ be irreducible of degree $\,m\,$ , so that $\,k[x]/(f(x))\cong k(\alpha)\cong\Bbb F_{p^r}\,$ , for some $\,r\in\Bbb N\,$ and $\,\alpha\,$ a root of $\,f(x)\,$

But then both $\,q_r(x)\,,\,f(x)\in k[x]\,$ and they both have $\,\alpha\,$ as a common root, so that, since $f$ is irreducible,

$$f(x)\,\mid\,q_r(x)\implies q_r(x)=x^{p^r}-x=f(x)g(x)\;\;\text{in}\;\;k[x]\,$$

and this means all the roots of $\,f(x)\,$ are also roots of $\,q_r(x)\,$ , which means $\,f(x)\,$ splits in $\,\Bbb F_{p^r}\,$ and we're done.

Answer 3

As per your comment (showing normality), let $K/\mathbb{F}_q$ be a field extension with $|K/\mathbb{F}_q|=d$, and $q=p^n$. I'll make use of the following theorem.

Theorem. Let $K/F$ be a field extension. Then $K/F$ is Galois if and only if $K$ is the splitting field of some separable polynomial over $F$.

Now $K$ is the splitting field of $x^{q^d}-x$ over $\mathbb{F}_p$ (all finite fields arise as the splitting field of $x^s-x$ over $\mathbb{F}_p$, where $s$ is some power of $p$), and hence it is the splitting field of $x^{q^d}-x$ over $\mathbb{F}_q$. So $K/\mathbb{F}_q$ is Galois (and hence normal).