# $[K:F]=p^2$ implies $\exists \alpha, \beta\in K$ such that $K=F(\alpha, \beta)$

Let $K/F$ be a field extension, such that $[K:F]=p^2$, where $p\in\mathbb{N}$ is prime. Show that there must exist elements $\alpha, \beta\in K$ so that $K=F(\alpha,\beta)$.

Proof:

Consider $GF(p^2)$. $\exists m(x)\in K[x]$ such that $\deg(m(x))=2$, where $m(x)$ is the minimal polynomial over some element in $K(x)$. That is, $[GF(p^2):K]=2$. Now, $[GF(p^2):F]=[GF(p^2):K][K:F]=p^2$, which implies that $2\vert p^2$, thus $p$ is even, that is $p=2$. Hence, $\exists \alpha, \beta\in K$ such that $K=F(\alpha,\beta)$.

I'm not very convinced that my proof is correct, so I'd like to discuss with you how to improve it / make it correct / make sense.

[In fact, I didn't know how exactly to approach this problem on an exam, so I wrote my proof something like the above (recalling it from memory now).]

• Is it standard to write $\exists ab \in X$ or is it a typo of $\exists a,b \in X$? – Myridium Apr 15 '17 at 5:14
• @Myridium Sorry, it was a typo. – sequence Apr 15 '17 at 5:18

## 2 Answers

Well you proved $p = 2$, so your proof is wrong. (Any irreducible $9$ degree polynomial will yield an extension of degree $9 = 3^2$).

To prove the statement, first note that since $K$ is a finite extension of $F$, it is algebraic over $F$. So take some $\alpha \in K\setminus F$. If $K = F(\alpha)$, then we are done since we can take $\beta = \alpha$ as well. Otherwise, $p^2 = [K:F] = [K:F(\alpha)][F(\alpha):F] \implies [K:F(\alpha)]=p=[F(\alpha):F]$ since $p$ is a prime.

Then taking any $\beta \in K\setminus F(\alpha)$ shows $[F(\alpha)(\beta):F(\alpha)] > 1$ and $p = [K:F(\alpha)] = [K:F(\alpha)(\beta)][F(\alpha)(\beta):F]$. We conclude $K = F(\alpha)(\beta) = F(\alpha,\beta)$, as desired.

This part:

$$[GF(p^2):F]=[GF(p^2):K][K:F]$$ assumes that $K \subset GF(p^2)$ which is not true in general.

Solution: Pick some $\alpha \in K \backslash F$. Then $F(\alpha) \subset K$ and hence $$[K:F]=[K:F(\alpha)][F(\alpha):F]$$

If $[F(\alpha):F]=p$ repeat the argument. If $[F(\alpha):F]$ than any $\beta$ will work.