Galois theory: Galois extensions

This lecture is part of an online graduate course on Galois theory.

We define Galois extensions in 5 different ways, and show that 4 f these conditions are equivalent. (The 5th equivalence will be proved in a later lecture.) We use this to show that any finite group is the Galois group of some finite extension.

[tim-cca]

Really good introduction to galois extension. It is much better and clearer than my lectures.

[gunhasirac]

For anyone comes later that are confused, the use of lemma at around 22:00 is legit and is addressed in comments below. Perhaps adding |G| ≦ |Gal(M/M^G)| ≦ [M:M^G] will be clearer.

It’s a general fact flipping twice gives a larger field/group and it’s the same only if it’s Galois. Namely:
If K ⊆ L ⊆ M and G = Gal(M/K) and H < G, then
L ⊆ M^Gal(M/L), and H ⊆ Gal(M/M^H).
This result can be written more compactly by the “prime notation”. With K, M, G fixed, we define L’ = Gal(M/L) and H’ = M^H. Then above result becomes L ⊆ L’’ and H < H’’, which is easier to memorize. Incidentally we also have L’ = L’’’ and H’ = H’’’.

[Snyvale]

At 21:20 we just heard that we should observe that the extension M/M^G has |G| automorphisms, but that seems to be what we are proving in this section (2) => (3), namely M^G=K, in which case also Gal(M/M^G) = Gal(M/K).

[hausdorffm]

Let M/K be a separable normal extension of fields, and consider Gal(M/K). In the following, we can show how separablity and normality work.
Galois group relates some extension problem of inclusion map K -> cl(K) into M -> cl(K), where cl(K) denotes the algebraically closed field of K. Suppose M = K(a) with p(a) = 0 for some p(x) in K[x]. For any extended map, say f: K(a) -> cl(K), its possible values f(a) are roots of p(x). (note that any root is in cl(K) ). So each extension of K -> cl(K) corresponds a root of p. If M/K is separable, this correspondence is injective.
Galois group is map M -> M, On the other hand, the extension is a map M -> cl(K) which is not M -> M. But we can decompose it as M -> M -> cl(K), and this means that the extension is actually a map M -> M. To show this, we use the fact that M/K is a normal extension. Now, I do not understand the proof of this, which is explained at 19:23….



15:32 In the of the lemma, field M/K is assumed as an algebraic extension, i.e., M = K(a1, a2, …, a_n), so implicitly, this lemma was proved only in case of [M:K] is finite. But finiteness can be removed because, formally, if [M;K] is infinite, this holds because “|G| <= infinite” is trivial.
If X = M and map K -> X is inclusion, then this is the case of Galois group.


18:56 Typo. In the inductive approach, he wrote K < K(a1) < K(a2) < …. But it may be K < K(a1) < K(a1, a2) < …


15:19 To define an extension of homomorphsim of field f:K -> X to f’:K(a) -> X for some algebraic element a of K, it is sufficient to define only a value f’(a) which inherits some equation from p(a) =0 for some p(x) of K[x]. So, there is some polynomial p’(x) of X[x] which determines all possible f’(a). But if X is not algebraic closure of K, then all roots are not necessarily contained in X. Thus, if f’(a) is not in X, such a map is excluded. Hence, if p’(x) has at least one root which is not in X, then the number of extensions is strictly less than degree of p’(x).


15:19 Let K(a)/K be an algebraic extension of field K with polynomial p(x) of K[x] with p(a) = 0.
Consider the extension problem of a homomorphism of fields from K -> X into K(a) -> X. Let f; K -> X be a homomorphism of fields K, X and suppose f’: K(a) -> X be its extension. Because p(a) = 0, and f’ = f on K, we get 0 = f’(0) = f’(p(a)) = p(f’(a)) and thus f’(a) is also a root of polynomial p(x) of K[x] (here, in the last equality we use f = identity on K, this condition can be removed.). Therefore possible f’ is determined by the root of p(x). Hence the number of possible f’s are at most degree of p(x). Furthermore if we assume that p(x) is separable on X and all roots are in X, then the number of possible f’ is equal to degree of p(x). By replacing the role of the pair of K(a) and K by K(a, b) and K(a), and using the dimension formula [M:L][L:k] = [M:K], I am not sure but we get for K(a, b, …c)/K.
In case of homomorphism of fields f:K->X is not identity on K I am not sure. A root of polynomial p(x) of K[x] is mapped by extension f’; K(a) -> X. This f’(a) is a root of polynomial p’(x) of X[x] and if p is a monic, then p’ is also monic and hence the number of possible f’(a) is at most degree of p. If p is irreducible polynomial and p’ inherit irreducibility, then the number of possible f’(a) is exactly degree of p. I guess … I am not sure.




Open? Problem 11:23 from an inverse problem of Galois theory.
For a given base field Q and a given group H, find an extension of fields M/Q such that Gal(M/Q) = H.

[maxwibert]

Hi Dr. Borcherds, is the correspondence described by definition (5) in this video a functorial one? Say I had an inclusion map from L1 to L2 in the category of fields containing K, would it necessarily correspond to a group morphism from Gal(M/L2) to Gal(M/L1) in some nice commuting diagram or something? If so would such a morphism's kernel or cokernel have any particular significance?

[littlecousin]

Can someone explain what the polynomial in 26:16 is? Is it saying for each element in the (finite) set of generators a, consider p_a, a separable polynomial which we know splits inside M and then define p=\prod_a p_a (where this is well-defined because this is a finite extension)? But how do we guarantee that p is also separable? It doesn’t seem to be separable.

[pupfer]

Historical interlude.
Galois found a normal and separable girl, but her husband was not satisfied that she will become Galois' extension, so he challenged Galois to a duel in which Galois died.

[haojiehong]

Hi Professor, I think you misused the lemma when proving (2)→(3). The lemma says #G≤[M:K], but in your proof of (2)→(3), you got #G≤[M:M^G] by the lemma. This is opposite to a result of E.Artin, which says if G is a finite group, then [M:M^G]≤(G:1).