Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 6 down vote accepted

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 $K$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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