Are Algebraic Closures of a Field Isomorphic?

[barbiemathgurl]

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

and why doesn't this show that C and A (algebraic numbers) arent isomorphic?

 

[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.

 

[Kummer]

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

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'.

and why doesn't this show that C and A (algebraic numbers) arent isomorphic?

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.

 

[mathwonk]

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

 

[Kummer]

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

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

 

[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.

 

[Kummer]

the field theory notes include again artins proof of existence of algebraic closures, and uniqueness.

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.

 

[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.

 

[Kummer]

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.

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.

 

[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 don't you?

 

[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.

 

[Kummer]

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

 

[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?

 

[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.

 

[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.

 

[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.

 

[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.

 

[mathwonk]

just so you can find it easier.

 

[Kummer]

Thank you.

 

[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.

 

[Kummer]

But it is a skeletally small category so everything is OK if done properly.

What does that mean? I do not know much about axiomatic set theory and category theory but I would like to know the basics of what you mean by that.

(Of course, I can just avoid using Algebraic Closure for much of field theory by working with splitting fields as well. In fact, I have two books on abstract algebra. It is interesting how once repeatedly uses Algebraic Closure and one does not even mention it in the book.)

 

[Hurkyl]

A skeleton for a category is a full subcategory with the properties that:
(1) Any two isomorphic objects in the skeleton are equal
(2) Any object of the category is isomorphic to an object of the skeleton

A category is small iff it has a set of objects and a set of morphisms.

A category is skeletally small if it has a small skeleton.



Intuitively, a skeleton is what you get when you replace everything with its isomorphism class, so being skeletally small means that there is a set of isomorphism classes.

From a technical standpoint, when basing everything on ZFC, the trick to implementing this idea is to find a canonical representative for each isomorphism class (because each isomorphism class is a proper class, so you don't have much freedom in manipulating them), and then showing there is a set of representatives.


Actually, this is far easier if you're allowed to assume a large cardinal. Then, you have a (large) set of all small algebraic extensions of your base field.

Another way to make this easier is to assume a strong form of the axiom of choice that applies to proper classes as well as sets.

 

[mathwonk]

using zorn to produce a maximal element of the set of all proper ideals in a ring, allows artin's proof below where the closure is obtained from the union of a countable number of extension fields: (these are my class notes from fall 2006.)

Proposition 15: If k is any field, there is an extension E of k in which every irreducible polynomial in k[X] has a root.
proof: Let S be a set of variables Xf, one for each irreducible polynomial f in k[X], and let k be the polynomial ring in the variables Xf. For each irreducible polynomial f(X) in k[X], consider the polynomial f(Xf) in k. Let T be the set of all these polynomials in k. We claim T does not generate the unit ideal of k. E.g. if there were a finite linear combination ∑gifi(Xfi) = 1, then by extending k to a field where all of the polynomials fi have roots, and substituting in these roots, we would have 0 = 1.
Hence there is some maximal ideal M containing the set T. Then E = k/M is a field, and in this quotient field every class f([Xf]) = [f(Xf)] = [0]. I.e. [Xf] is a root of f. Thus every irreducible polynomial f in k[X] has a root in E. QED.

Proposition 16: If k is any field, there is an extension E of k in which every polynomial in k[X] splits completely into linear factors.
proof: Let E1 be a field containing k, in which every irreducible polynomial in k[X] has a root. Then let E2 be an extension of E1 where every irreducible polynomial in E1[X] has a root. Etc... Then take the union E of all these fields. If f is a polynomial of degree n in k[X]. Then f splits completely into linear factors in En[X], hence also in E. QED.

Def 17: A field E such that every polynomial in E[X] has a root in E, hence splits completely, is called algebraically closed. An algebraically closed field which is algebraic over a subfield k, is called an algebraic closure of k.

Theorem 18: Any field k has an algebraic closure.
proof: Let F be the field constructed above, in which every polynomial in k[X] splits into linear factors, and let E be the subfield generated by the roots of all polynomials over k. Then E is algebraic over k. Let g be a polynomial in E[X], and let a be a root of g in an extension field of E. Then a is algebraic over E, and E is algebraic over k, so a is algebraic over k. Thus a is a root of some polynomial in k[X] and hence a is already contained in E. QED.