# Algebraic closure

1. Jul 15, 2007

### barbiemathgurl

given a field F and two algebraic closures of F, are those two the isomorphic?

and why doesnt this show that C and A (algebraic numbers) arent isomorphic?

2. Jul 15, 2007

### StatusX

Yes they are. And C is the algebraic closure of R, while A is the algebraic closure of Q, and in particular doesn't even contain all of R.

3. Jul 15, 2007

### Kummer

Yes.

This is a consequence of the Isomorphism extentening theorem. Let $$F$$ be a field and $$\bar F$$ and $$\bar F'$$ be two algebraic closures of $$F$$. The identity isomorphism $$\iota: F\mapsto F$$ can be extended to the isomorphism $$\phi: \bar F \mapsto \phi\left[ \bar F \right]$$. Now define the inverse map $$\phi^{-1}: \phi \left[ \bar F \right] \mapsto \bar F$$ and notice that $$\phi^{-1}$$ is onto. This map can be extended so that $$\psi: \bar F' \to \psi \left[ \bar F \right]$$ but $$\psi \left[ \bar F \right] = \bar F$$ since $$\psi$$ is onto because a possible smaller map $$\phi^{-1}$$ was onto. Hence $$\psi$$ provides an isomorphism between $$\bar F$$ and $$\bar F'$$.

Because C is not an algebraic closure of Q. It is algebraically closed but it is not algebraic over Q. So by definition it is not an algebraic closure of Q.

4. Jul 17, 2007

### mathwonk

these proofs are in the free notes on my web page.

5. Jul 17, 2007

### Kummer

Nice page, I might read your Field Theory notes when I have time.

6. Jul 17, 2007

### mathwonk

thank you.

i am posting a new set today, briefer, on abelian groups and rings, finitely generated modules over pids, ratl and jordan forms, spectral theorems and duality, groups, fields, and finite galois groups. 100 pages total, as opposed to 400 pages for the previous notes (but omitting tensors). the field theory notes include again artins proof of existence of algebraic closures, and uniqueness.

groups fields and galois theory are only about 31 pages total in this set of notes. groups take about 11 pages and fields and galois theory take the other 20, so it should be less time consuming to peruse the field theory part.

Last edited: Jul 17, 2007
7. Jul 18, 2007

### Kummer

The proof I learned is a nice approach by Axiom of Choice.

Outline.
--------
1)Define $$S=\{ F\leq E | E \mbox{ algebraic over }F\}$$ (and show it is a set).
2)Order $$S$$ by inclusion, i.e. $$\subseteq$$
3)Use Zorn's Lemma.
4)Argue by contradiction that $$\bar F$$ - the maximal element , must be algebraically closed via Kroneckor's Theorem.

8. Jul 19, 2007

### mathwonk

unfortunately that class S does not appear to be a set. hence that approach does not work without some more tedious work restricting the definition of S. Such a tedious approach is taken in Hungerford.

the artin approach finesses the task of dealing only with sets, and also is a direct generalization of the proof that each polynomial has a root in some extension.

recall that proof looks at the field k[X]/(f) where f is irreducible.

sketch: consider all irreducible polynomials f over k, definitely a set, in fact a subset of k[X]. Then choose one variable Xf for each such f and look at the polynomial ring R in all of them.

then use zorn to produce a maxl ideal containing all polynomials of form f(Xf), and mod out R by it.

this is the esential step of producing a field where every irreducible polynomial of k has a root.

Last edited: Jul 19, 2007
9. Jul 19, 2007

### Kummer

It is a set. The condition that "E is algebraic over F" just makes it a set. However, in general "E extension field over F" is no longer a set. So much care is taken to work with set S.

10. Jul 19, 2007

### mathwonk

i disagree. you have to dos oemthing to restrict the fields you consider. i am not expert here, but it seems obvious you could take all sets in the world of the same cardinality as a given extension field, and give them all the structure of that extensions field, and you would get way too many algebraic extensions. you want somehow to consider only one algebraic extension field in each isomorphism class of extensions dont you?

11. Jul 19, 2007

### mathwonk

see Hungerford, page 259, footnote: "As anyone familiar with the paradoxes of set theory might suspect, the class of all algebraic extension fields of K need not be a set, and therefore cannot be used in such an argument."

of course this is just an assertion, not a proof. but I agree with him.

Last edited: Jul 19, 2007
12. Jul 20, 2007

### Kummer

See Fraliegh 7/e page 290. This approach is used. Hence the proof is more set theoretic than algebraic.

13. Jul 20, 2007

### mathwonk

I am surprized fraleigh says this, as it still seems clearly wrong to me. I do not have access to Fraleigh though.

But I better wait and read fraleigh. As presented here however, it seems to me obviously not a set.

for exmple i have given a construction that shows every set of the same cardinality as the given field occurs as an algebraic extension field of the given field. although all these field extensions are isomorphic, they all occur in your set S as distinct elements. this seems like a lot of sets to me.

i.e. every set in the world large enough, has a huge number of subsets of the same cardinaity as our given field. so the number of such field extensions is greater than the number of sets of cardinality at least as large as our field. it is surely impossible for S to be a set.

Any opinions from the resident set theory guys? Hurkyl?

14. Jul 20, 2007

### mathwonk

let me quote from the great algebraist richard brauer's harvard notes on galois theory, 1957-8, rev 1963-4.

p.169. "we first give a FALSE proof. consider the set S of all algebraic extension fields L of K............[then he gives a zorn's lemma argument on S.].....
The snag in the argument is of course that the algebraic extension fields of K do not form a "set". It is shown by russell's paradox that we have to be much more careful in working with the notion SET."

well i am confident of this situation, and surprized that a modern mathematician would make such a mistake. better read fraleigh carefully, to see whether he did not take only one extension in each isomorphisn class.
even that would require proof.

this sort of situation is practically the only time when mathematicians take set theory seriously, i.e. using zorn carefully.

Last edited: Jul 20, 2007
15. Jul 20, 2007

### mathwonk

rotman, dummit-foote, sah, lang, jacobson, all give artin's proof. older books like zariski samuel, and also hungerford make some effort to use a specially restricted class of extension fields and then use zorn; van der waerden disliking to use zorn, restricts to countable fields; but none of them makes the assertion that all algebraic extensions form a set.

we better check that fraleigh reference.

Last edited: Jul 20, 2007
16. Jul 20, 2007

### mathwonk

well i have searched online fraleigh's 6th edition where he does it correctly, p.409, thm 4.3.22? restricting his fields to have elements inside a given set constructed from polynomials of the original field. this approach is the one used in older books like zariski - samuel and van der waerden. but surely he would not have become more naive by the 7th edition.

perhaps the discussion referenced above on p. 290, 7th edition, is a naive discussion that he probably says somewhere does not quite work. it is a famous rule that teachers should beware of giving a false proof, as students remember it, but not that it is false.

17. Jul 20, 2007

### morphism

I got my hands on Fraleigh 7, and he does it the way he did it in the 6th edition, i.e. the way mathwonk describes in his post above this one.

18. Sep 11, 2007

### mathwonk

just so you can find it easier.

19. Sep 11, 2007

### Kummer

Thank you.

20. Sep 11, 2007

### matt grime

mathwonk is indeed correct - the class Kummer describes in his first step is a proper class, and not a set. But it is a skeletally small category so everything is OK if done properly.