## Splitting Fields and Algebraic Closures (Pt. II)

**Point of Post: **This is a continuation of this post.

**Theorem(Isomorphism Extension Theorem): ***Let be two fields and a ring isomorphism. Then, if is a set of polynomials we can consider the set where is the polynomial obtained by applying to the coefficients of . Let be a splitting field for and be a splitting field for . Then, there is an isomorphism which is an extension of . Furthermore, if is some distinguished element we can take to be such that where is any root of .*

**Proof: **Basically we can create monomorphisms where is some “small” subextension of (namely we can just specify where we want the generators to go), the problem is in extending all the way to . Of course, phrased this way it’s clear that Zorn’s lemma is going to be our lemma of choice for this proof [relevant]. Ok, so let’s formally define

We define a partial ordering structure on by demanding that if and only if and is an extension of . Of course, is not empty since . Now, let’s show that every chain in has a maximal element. Indeed, let be a chain. Define then and by defining for any with (by the requirements on the and the fact that we have a chain this is well-defined). Clearly then and for al . Thus, we see that every chain in has an upper bound and so our best friend Zorn hands us some maximal element of . Our job now is to prove (we hope) that must actually be and that [there is, of course, no need to prove that is injective since it’s a ring map out of a field]. To prove that we merely note that if this wasn’t true then necessarily there is some such that does not split in (since otherwise contains all the roots of all the and so must be equal to ). Thus, we can find some root of not conatined in . Then, by the lemma we can extend to which contradicts the maximality of . Thus, we see that . Now, to see that we merely note that is a splitting field for and thus must necessarily be equal to .

Ok, cool so we have proven two things up to this point: splitting fields exist for a single polynomial, and splitting fields for an arbitrary set of polynomials are unique up to isomorphism. Well, there seems to be somewhat of a discrepancy for these two statements. Namely, we have a theorem telling us a fact about the splitting field for arbitrarily polynomials yet, a priori, we only know that the splitting field of finitely polynomials exists. Strange, right? This begs the questions as to whether or not the splitting field of arbitrarily polynomials even exists. Well, for a finite set of polynomials we actually have nothing to wonder about since, as one can easily deduce, a splitting field for is a splitting field for . But, what about infinite polynomials? We surely can’t just take their product. Indeed, for the infinitary case we are going to need slightly more sophisticated machinery.

To create splitting fields for arbitrarily many polynomials in it suffices to find a field containing all the roots for *all* the polynomials in . Indeed, suppose that we have constructed such a field, let’s call it , then to find a the splitting field for we merely let be the union of all the sets of roots in of all the elements in . We can easily see then that is a splitting field for .

To create such a field, let’s consider the field and for each nonconstant let be an indeterminate. Consider then the polynomial ring . Let then be the ideal generated by the for each nonconstant . What we’d like to do is put inside some maximal ideal (using Krull’s theorem), but for this we need to know that . To prove that this is impossible we note that if it’s true we could find and such that

Ok, but now here’s the cool part, let’s rename the variables for and let be the other variables that occur in the so that we can rewrite as

That said, we can find some extension such that contains roots for . Plugging this into gives which is evidently ridiculous. Thus, we see that and so we can put where is maximal. Consider then . Clearly then is an extension of which contains. We may repeat this process for to get another field , and so on. Define then . Clearly then is a field containing , and since any polynomial in has coefficients in some which has a root in . Thus, every polynomial in has a root in . But, this clearly implies that every polynomial has all its roots in since once we have a root, say , we can write for some . But, then has a root in , so we can write , etc. Thus, we know, in particular, every polynomial has a root in , and considering our previous discussion we can finally conclude that :

**Theorem: ***Let be a field and . Then, has a splitting field.*

We call a splitting field for (all the polynomials) an *algebraic closure *of and denote it –note that while this notation may seem ambiguous, it really isn’t since (as we have proven) all algebraic closures for must necessarily be isomorphic.

**References:**

[1] Morandi, Patrick. *Field and Galois Theory*. New York: Springer, 1996. Print.

[2] Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

[3] Lang, Serge. *Algebra*. New York: Springer, 2002. Print.

[4] Conrad, Keith. *Collected Notes on Field and Galois Theory*. Web. <http://www.math.uconn.edu/~kconrad/blurbs/>.

[5] Clark, Pete. *Field Theory*. Web. <http://math.uga.edu/~pete/FieldTheory.print

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