08-401/The Fundamental Theorem
The statement appearing here, which is a weak version of the full fundamental theorem of Galois theory, is taken from Gallian's book and is meant to match our discussion in class. The proof is taken from Hungerford's book, except modified to fit our notations and conventions and simplified as per our weakened requirements.
Here and everywhere below our base field will be a field of characteristic 0.
Theorem. Let be a splitting field over . Then there is a bijective correspondence between the set of intermediate field extensions lying between and and the set of subgroups of the Galois group of the original extension :
The bijection is given by mapping every intermediate extension to the subgroup of elements in that preserve ,
and reversely, by mapping every subgroup of to its fixed field :
This correspondence has the following further properties:
- It is inclusion-reversing: if then and if then .
- It is degree/index respecting: and .
- Splitting fields correspond to normal subgroups: If in is the splitting field of a polynomial in then is normal in and .
The four lemmas below belong to earlier chapters but we skipped them in class (the last one was also skipped by Gallian).
Zeros of Irreducible Polynomials
Lemma 1. An irreducible polynomial over a field of characteristic 0 has no multiple roots.
Proof. See the proof of Theorem 20.6 on page 362 of Gallian's book.
Uniqueness of Splitting Fields
Lemma 2. Let be an isomorphism of fields, let be a polynomial and let , and let and be splitting fields for and over and , respectively. Then there is an isomorphism (generally not unique) that extends .
Proof. See the proof of Theorem 20.4 on page 360 of Gallian's book.
The Primitive Element Theorem
The celebrated "Primitive Element Theorem" is just a lemma for us:
Lemma 3. Let and be algebraic elements of some extension of . Then there exists a single element of so that . (And so by induction, every finite extension of is "simple", meaning, is generated by a single element, called "a primitive element" for that extension).
Proof. See the proof of Theorem 21.6 on page 375 of Gallian's book.
Splitting Fields are Good at Splitting
Lemma 4. (Compare with Hungerford's Theorem 10.15 on page 355). If is a splitting field of some polynomial over and some irreducible polynomial has a root in , then splits in .
Proof. Let be a splitting field of over . We need to show that if is a root of in , then (so all the roots of are in and hence splits in ). Consider the two extensions
The "smaller fields" and in these two extensions are isomorphic as they both arise by adding a root of the same irreducible polynomial () to the base field . The "larger fields" and in these two extensions are both the splitting fields of the same polynomial () over the respective "small fields", as is a splitting extension for and we can use the sub-lemma below. Thus by the uniqueness of splitting extensions (lemma 2), the isomorphism between and extends to an isomorphism between and , and in particular these two fields are isomorphic and so . Since all the degrees involved are finite it follows from the last equality and from that and therefore . Therefore .
Sub-lemma. If is a splitting extension of some polynomial and is an element of some larger extension of , then is also a splitting extension of .
Proof. Let be all the roots of in . Then they remain roots of in , and since completely splits already in , these are all the roots of in . So
and is obtained by adding all the roots of to .
Proof of The Fundamental Theorem
Proof of . More precisely, we need to show that if is an intermediate field between and , then . The inclusion is easy, so we turn to prove the other inclusion. Let be an element of which is not in . We need to show that there is some automorphism for which ; if such a exists it follows that and this implies the other inclusion. So let be the minimal polynomial of over . It is not of degree 1; if it was, we'd have that contradicting the choice of . By lemma 4 and using the fact that is a splitting extension, we know that splits in , so contains all the roots of . Over a field of characteristic 0 irreducible polynomials cannot have multiple roots (lemma 1) and hence must have at least one other root; call it . Since and have the same minimal polynomial over , we know that and are isomorphic; furthermore, there is an isomorphism so that yet . But is a splitting field of some polynomial over and hence also over and over . By the uniqueness of splitting fields (lemma 2), the isomorphism can be extended to an isomorphism ; i.e., to an automorphism of . but then so , yet , as required.
Proof of . More precisely we need to show that if is a subgroup of the Galois group of over , then . The inclusion is easy. Note that is finite since we've proven previously that Galois groups of finite extensions are finite and hence is finite. We will prove the following sequence of inequalities:
This sequence and the finiteness of imply that these quantities are all equal and since it follows that as required.
The first inequality above follows immediately from the inclusion .
By the Primitive Element Theorem (Lemma 3) we know that there is some element so that . Let be the minimal polynomial of over . Distinct elements of map to distinct roots of , but has exactly roots. Hence , proving the second inequality above.
Let be an enumeration of all the elements of , let (with as above), and let be the polynomial
Clearly, . Furthermore, if , then left multiplication by permutes the 's (this is always true in groups), and hence the sequence is a permutation of the sequence , hence
and hence . Clearly , so , so , proving the third inequality above.
Property 1. If then and if then .
Proof of Property 1. Easy.
Property 2. and .
Proof of Property 2. If , then as was shown within the proof of . But every is for some , so for every between and . The second equality follows from the first and from the multiplicativity of the degree/order/index in towers of extensions and in towers of groups:
Property 3. If in is the splitting field of a polynomial in then is normal in and .
Proof of Property 3. We will define a surjective (onto) group homomorphism whose kernel is . This shows that is normal in (kernels of homomorphisms are always normal) and then by the first isomorphism theorem for groups, we'll have that .
Let be in and let be an element of . Let be the minimal polynomial of in . Since is a splitting field, lemma 4 implies that splits in , and hence all the other roots of are also in . As is a root of , it follows that and hence . But since is an isomorphism, and hence . Hence the restriction of to is an automorphism of , so we can define .
Clearly, is a group homomorphism. The kernel of is those automorphisms of whose restriction to is the identity. That is, it is . Finally, as is a splitting extension, so is . So every automorphism of extends to an automorphism of by the uniqueness statement for splitting extensions (lemma 2). But this means that is onto.