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’’’.
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| 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.
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).
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?
Hmm so given an inclusion from L1 to L2 we get an obvious map from Gal(M/L2) to Gal(M/L1) since an automorphism of M fixing L2 must fix L1. This map is injective since any non-identity automorphism of M fixing L2 is a non-identity automorphism of M fixing L1 (so the kernels are always trivial). Then constructing the cokernel is gonna require Gal(M/L2) to be normal in Gal(M/L1), which might have some significance we see later. I believe this is functorial (though not 100%), given a commuting square of fields containing K (contained in M), the corresponding square of Gal's is gonna commute (since essentially nothing is happening to the underlying automorphism of M).
@@austintheyognaut My impression is that the cokernel is trivially well defined because Galois extensions are normal extensions, which almost by definition must correspond to normal subgroups, no?
Depends what you mean by the category of field extensions between K and M. For example an automorphism of L1 doesn't induced an automorphism of Gal(M/L1) in any natural way. If you just think of it as a poset category then it is a duality. I.e. the inclusion of L1 into L2 induces an inclusion of Gal(M/L2) into Gal(M/L1). Further Gal(M/L2) is a normal subgroup of Gal(M/L1) if and only if L2 is a normal extension of L1, and in this case the cokernel is Gal(L2/L1).
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.
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.
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).
Yeah I have Ian Stewart's Galois Theory textbook on my shelf and the way that these theorems are proven is very different. Like there's a whole chapter on the linear independence of monomorphisms that seems absent from these video lectures.
@@haojiehong4876 The fact that G is a subgroup of Gal(M/M^G) holds for G any group of automorphisms of M. Since M^G is by definition the field of all elements of M fixed by all elements of G, Gal(M/M^G) is the group of all automorphisms of M which fix all elements of M that are fixed by all elements of G. So since any element g of G is an automorphism of M that fixes all elements of M that are fixed by all elements of G, g must be in Gal(M/M^G), and so G is a subgroup of Gal(M/M^G). Your intuition about the relative sizes is correct. But both are true. The statement being proved in the video is that they are equal to each other.
Really good introduction to galois extension. It is much better and clearer than my lectures.
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’’’.
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| 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.
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).
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?
Hmm so given an inclusion from L1 to L2 we get an obvious map from Gal(M/L2) to Gal(M/L1) since an automorphism of M fixing L2 must fix L1. This map is injective since any non-identity automorphism of M fixing L2 is a non-identity automorphism of M fixing L1 (so the kernels are always trivial). Then constructing the cokernel is gonna require Gal(M/L2) to be normal in Gal(M/L1), which might have some significance we see later.
I believe this is functorial (though not 100%), given a commuting square of fields containing K (contained in M), the corresponding square of Gal's is gonna commute (since essentially nothing is happening to the underlying automorphism of M).
@@austintheyognaut My impression is that the cokernel is trivially well defined because Galois extensions are normal extensions, which almost by definition must correspond to normal subgroups, no?
Depends what you mean by the category of field extensions between K and M. For example an automorphism of L1 doesn't induced an automorphism of Gal(M/L1) in any natural way. If you just think of it as a poset category then it is a duality. I.e. the inclusion of L1 into L2 induces an inclusion of Gal(M/L2) into Gal(M/L1). Further Gal(M/L2) is a normal subgroup of Gal(M/L1) if and only if L2 is a normal extension of L1, and in this case the cokernel is Gal(L2/L1).
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.
Brilliant, thank you
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.
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).
Yeah I have Ian Stewart's Galois Theory textbook on my shelf and the way that these theorems are proven is very different. Like there's a whole chapter on the linear independence of monomorphisms that seems absent from these video lectures.
I think it might be because Gal(M/K) = G is a subgroup of Gal(M/M^G), which by applying the lemma to M^G contained in M, would give |G|
@@probablyapproximatelyok8146 This doesn't make sense. Since K
@@haojiehong4876 The fact that G is a subgroup of Gal(M/M^G) holds for G any group of automorphisms of M. Since M^G is by definition the field of all elements of M fixed by all elements of G, Gal(M/M^G) is the group of all automorphisms of M which fix all elements of M that are fixed by all elements of G. So since any element g of G is an automorphism of M that fixes all elements of M that are fixed by all elements of G, g must be in Gal(M/M^G), and so G is a subgroup of Gal(M/M^G).
Your intuition about the relative sizes is correct. But both are true. The statement being proved in the video is that they are equal to each other.
Think of it as |G| = |Aut(M/M^G)|
なるほどね。
yeee
もっとゆっくりみたいな。