# How to prove the field extension is algebraically closed

1. Apr 5, 2012

### lugita15

Suppose that E is a field extension of F, and every polynomial f(x) in F[x] has a root in E. Then E is algebraically closed, i.e. every polynomial f(x) in E[x] has a root in E.

I've been told that this result is really difficult to prove, but it seems really intuitive so I find that surprising. Where can I find a proof of this result?

Any help would be greatly appreciated.

2. Apr 5, 2012

### DonAntonio

No wonder you've been told that: it is false. Let $E:= ℂ(t)$ be the field of rational functions on the (transcendental) variable t. Then E is a field extension of $ℂ$ in which every pol. of $ℂ[x]$ has a root, but E certainly is not alg. closed as, for example, the pol. $x^2-t\in ℂ(t)[x]$ has no root in it...

Now, the claim is true if one adds the condition that E/F is an algebraic extension.

DonAntonio

3. Apr 5, 2012

### lugita15

OK, so adding that condition where can I find the proof of the result?

4. Apr 5, 2012

### DonAntonio

The proof isn't THAT easy but it isn't THAT hard, either.

Enjoy, and for this read the whole paper. The first part is just the well-known and "standard" proof by Artin.

DonAntonio

5. Apr 5, 2012

### mathwonk

my free algebra notes 844.I.1, do this in detail, in the first 12 pages.

http://www.math.uga.edu/%7Eroy/844-1.pdf [Broken]

Last edited by a moderator: May 5, 2017
6. Apr 5, 2012

### DonAntonio

The question wasn't about the existence of alg. closures of fields, which Artin's proof, that also appears in your notes, does, but the fact that after the first construction of an extension E/F s.t. every non-constant pol. in E[x] has a root we can ALREADY stop there since F is alg. closed. I couldn't find this point in your notes, though perhaps I oversaw it.

DonAntonio

Last edited by a moderator: May 5, 2017
7. Apr 7, 2012

### mathwonk

oops, my mistake, thank you.

8. Apr 8, 2012

### lugita15

Is it possible to prove the result without invoking the notion of perfect fields?

9. Apr 8, 2012

### DonAntonio

I doubt it since the trick focuses on extensions K/F which can be put in the form $K=F(\alpha)$ , for some $\alpha\in K$ , and one

achieves this using separability, but also: why? Separability shouldn't, imo, be a problem for someone trying to tackle what you're trying...

DonAntonio