# Splitting fields of polynomials over finite fields

I can't follow a statement in my notes:

"Let $K$ be a finite field, with $f \in K[X]$ an irreducible polynomial of degree $d$. Then any finite extension $L/K$ is normal, and so if $L$ contains one root of $f$ then it contains all the roots of $f$. Therefore, the splitting field $L$ of $f$ is of the form $K(\alpha)$, where $f$ is the minimal polynomial for $\alpha$."

I can see why $L$ must be normal (any finite extension of a finite field is Galois), and so by definition if $L$ contains one root of $f$ then it contains all the roots of $f$. I don't follow the next sentence at all:

i) Why must $L$ be the splitting field of $f$? EDIT: Is this $L$ now a 'new' $L$?

ii) If $L$ is the splitting field of $f$, why must it be of the form $K(\alpha)$?

• For ii), I can see why it would be true if $f$ were separable (primitive element theorem), but not why it's true for inseparable $f$ – Jonathan Dec 13 '11 at 18:15
• Another way of explaining the problem in your comment would be to observe that over a finite field all the irreducible polynomials are separable. An irreducible inseparable polynomial is always a $p$th power of a polynomial with coefficients in $K^{1/p}$. But when $K$ is finite, the Frobenius automorphism is onto, and hence $K^{1/p}=K$. Thus the polynomial would be a $p$th power of a polynomial in $K[x]$, and hence couldn't be irreducible after all. – Jyrki Lahtonen Dec 17 '11 at 8:29

i) Your edit is correct. The statement is telling you that any finite extension of a finite field is normal. Since the splitting field of $f$ is necessarily finite (with degree at most $d!$), then it must also be normal.
ii) $K(\alpha)$ is a finite extension of $K$, and so (by the previous sentence) is a normal extension. It contains a root of $f$, so must therefore contain all the roots of $f$ by normality, and therefore contains the splitting field of $f$. To see that $K(\alpha)$ is precisely the splitting field of $f$, observe that a splitting field of $f$ must contain $K$ and $\alpha$ and so contains $K(\alpha)$.