# Splitting field implies Galois extension

I hope this isn't too elementary of a question, but I'm not sure I understand Artin's proof that if $K/F$ is a finite extension, then $K/F$ Galois is equivalent to $K$ being a splitting field over $F$ (this is Theorem 16.6.4 in the second edition). We're working in characteristic zero here. I understand one direction, that Galois implies splitting field: if we let $\gamma_{1}$ generate $K/F$ and $f$ be its minimal polynomial of degree $n$, each automorphism of $K/F$ comes from sending $\gamma_{1}$ to another root of $f$, and in order for $\operatorname{Gal}(K/F)$ to have order $n$ we need all of the roots of $f$ to be in $K$.

However, I still am not seeing the other direction - after stating the above Artin seems to finish by saying that if we define $\gamma_{1},f$ as before, "$K$ is a splitting field over $F$ iff $f$ splits completely in $K$." But isn't it possible that $f$ doesn't split, but $K$ is the splitting field of some other polynomial?

• If $K$ is a splitting field of some $g$, then it is Galois; if it is Galois, then any irreducible polynomial over $F$ will either split over $K$ or be irreducible over $K$, so in fact $f$ will necessarily split completely if $K$ is the splitting field of any polynomial, and hence $K$ will be the splitting field of $f$ as well. – Arturo Magidin Dec 30 '11 at 7:08
• The theorem as you state it is incorrect. It should say "If $K/F$ is a finite extension, then $K/F$ Galois is equivalent to $K$ being a splitting field of a separable polynomial $f$ over $F$." It is important that $f$ is separable. – mareoraft Jun 6 '17 at 20:51
• @mareoraft Characteristic zero implies separable – CapitalPi Jul 1 at 15:51

## 1 Answer

Let's view everything as living inside of an algebraic closure $\bar F$ of $F$. Suppose that $K \subset \bar F$ is a splitting field for some $g \in F[x]$. Then any embedding $\sigma\colon K \to \bar F$ over $F$ lands inside of $K$, because if $\alpha$ is a root of $g$ then so is $\sigma(\alpha)$. If $\beta$ is a root of $f$ in $\bar F$, then as $f$ is irreducible in $F[x]$ there is an $F$-isomorphism $\tau\colon K \to F(\beta)$ such that $\tau(\gamma_1) = \beta$. Hence $\beta \in K$, and it follows that $f$ splits completely in $K$.

• If any of this is unclear, then let me know. – Dylan Moreland Dec 30 '11 at 7:25
• Also: This seems like a bit of work, so I would be surprised if Artin hasn't made this argument before. Unfortunately, I can't find a copy of the second edition online. – Dylan Moreland Dec 30 '11 at 7:30
• Thanks! Your answer makes sense; Artin also seems to take care of this in proving in a previous section that if $K$ is (any) splitting field of $F$, then a polynomial $f$ with a root in $K$ splits completely in $K$ (however, he doesn't reference this result in the proof I mentioned). – LCL Dec 30 '11 at 17:06