From: mareg@mimosa.csv.warwick.ac.uk () Subject: Re: should be easy.. (algebra question) Date: Thu, 1 Nov 2001 10:02:45 +0000 (UTC) Newsgroups: sci.math,sci.math.research Summary: Towers of radical field extensions In article <3bde0d1e$0$1917$b45e6eb0@senator-bedfellow.mit.edu>, dpwood@mit.edu (David P Woodruff) writes: >Hi, > >I have a 2-radical extension K of F,i.e., there exist intermediate fields >Li such that F < L1 < L2 < ... < Ln =K, where [Li:L(i-1)] = 2, [L1:F] = 2. >Let x be in K. How do I show there exists intermediate fields Ri such that >F < R1 < R2 < ... < Rm = F(x), where [Ri:R(i-1)] = 2, [R1:F] = 2? This is not hard using Galois theory and a little group theory. Let L be the normal closure of K over F, let G be the Galois group of [L:F] and let Gi be the subgroup fixing Li. Then we have Gn < G{n-1} < ... < G1 < G, where each |Gi:G{i+1}| = 2. In fact G must be a 2-group, because each G{i+1} is normal in Gi, so the 2-residual O^2(G) of G (= smallest normal subgroup of G in which G/O^2(G) is a 2-group) lies in each Gi by induction on i, so it lies in Gn, and then O^2(G)=1 by definition of normal closure. But your field F(x) corresponds to a subgroup H of H that contains Gn, and by standard properties of p-groups, we can find a chain of subgroup H = Hm < H{m-1} < ... < H1 = G with each |Hi:H{i+1}|=2, and then the fixed fields Ri of the Hi are what you want. >Also, if I have a radical field extension K of F, where there exist >intermediate fields Si, such that F < S1 < S2 <.... < Sn = K, where >[Si:S(i-1)] = 2, [S1 : F] = p for some prime p. Let x be in K. I want to show >that the field F(x) is a radical 2-extension of F, or a radical 2-extension >of F(u), where u^p is in F. I think this is wrong. Here is a counterexample. Take F=Q, S1=W where r^3=2, S2=S1 where w^2+w+1=0 (w = cube root of 1), S3=S2, where x1^2 = r+r^2, S4=S3 where x2^2=(wr)+(wr)^2, S5=S4 where x3^2=(w^2r)+(w^2r)^2. Then S5 is a normal extension of Q - note that x1,x2,x3 are conjugates. The Galois group G of [S5:Q] is of order 48, and is an extension of the permutation module M for Sym(3) with Sym(3). (S2 is the fixed field of M.) (Sym(3) = symmetric group of degree 3.) Now let x = x1+x2+x3. The subgroup of G fixing F(x) is, and [Q:Q] = 8. (In fact the minimal polynomial of x over Q is x^8 + 48x^4 -384x^2 +576 (using computer)). Since M is the direct sum of irreducible modules of degrees 1 and 2 for Sym(3), there is no chain of subgroups from S3 to G in which layers have index 2, so Q(x) is not a radical extension of Q. Derek Holt. ============================================================================== From: barr@barrs.org (Michael Barr) Subject: Re: should be easy.. (algebra question) Date: 1 Nov 2001 06:01:59 -0800 Newsgroups: sci.math,sci.math.research barr@barrs.org (Michael Barr) wrote in message news:<19daab49.0110301347.76171077@posting.google.com>... > dpwood@mit.edu (David P Woodruff) wrote in message news:<3bde0d1e$0$1917$b45e6eb0@senator-bedfellow.mit.edu>... > > Hi, > > > >I have a 2-radical extension K of F,i.e., there exist intermediate > >fields Li such that F < L1 < L2 < ... < Ln =K, where [Li:L(i-1)] = 2, > >[L1:F] = 2. Let x be in K. How do I show there exists intermediate > >fields Ri such that F < R1 < R2 < ... < Rm = F(x), where [Ri:R(i-1)] = > >2, [R1:F] = 2? > > >Also, if I have a radical field extension K of F, where there exist > >intermediate fields Si, such that F < S1 < S2 <.... < Sn = K, where > >[Si:S(i-1)] = 2, [S1 : F] = p for some prime p. Let x be in K. I want to show > >that the field F(x) is a radical 2-extension of F, or a radical 2-extension > >of F(u), where u^p is in F. > Forget my original post which was wrong. I made some assumptions about F(x), which were clearly, in retrospect, not correct. I think the following is correct. It unfortunately uses Galois theory, but I see no alternative. It also makes some hypotheses about characteristic and, for the second question, about having a primitive pth root of 1 in F. I see no alternative to those. Say that K is a 2-tower extension of F if there is a tower of extensions of degree 2 between F and K. Assume char(F) \ne 2. (I do not see it in char 2). If K is a 2-tower Galois extension, then every intermediate field is too. For if E is an intermediate field and Gal(E/F) is its Galois group, then there is a composition series of Gal(K/F) that passes through Gal(E/F) and all the composition factors are Z_2 and that does it. Next I claim that if K is a 2-tower/F, then so is its normal closure, which will complete the argument. I do it by induction on degree. Let me first, as I like to do, look at a small case. When [K:F] = 2, then K is already normal, so look at [K:F] = 4. In this case, there are elements a and b in F, so that K = F(sqrt(a+sqrt(b))) and the normal closure is K(sqrt(a-sqrt(b))). But since sqrt(b) in F, the latter has degree 2 (or maybe 1) over K and hence there is a 2-tower. Another way of seeing this is that the normal closure of K is the splitting field of (X^2-a)^2 - b and this polynomial splits into two quadratic factors in F(sqrt(b)). For the general case, assume that you have a tower F < L_1 < ... < L_{n-1} < L_n of extensions of degree 2. Inductively suppose that there is a tower F < K_1 < ... K_m of extensions of degree 2 and that K_m is a normal closure of L_{n-1}. Let G = Gal(L_{n-1}/F). Suppose a in L_{n-1} is such that L_n = L_{n-1}(sqrt(a)). Then adjoin to K_m, one by one the square roots of sigma(a) for all sigma in G. Each is an extension of degree 1 or 2. At the end, you have made a succession of adjunctions of degree 1 or 2 and so the field K at the end is a 2-tower. To see it is Galois over F, let R be its normal closure over F. If tau is any element of the Galois group, tau(K_m) = K_m because K_m is a Galois extension of F. Hence for any sigma in G, there is a rho in G such that tau(sigma(a)) = rho(a) and hence tau(sqrt(sigma a)) is a sqare root root of rho(a) which is also in K. But then K is invariant under all elements of Gal(K/F), which implies it is already normal. As observed at the beginning, this implies every intermediate field is a 2-tower. Your second problem I can do if F contains a primitive pth root of 1 and also char(F) != p. In that case L_1 is Galois and you do the same trick, at each stage, adjoin not merely the square root of an element, but also the square roots of each of its conjugates to get a Galois extension at each stage. When you are finished you have a Galois extension K of degree 2^m.p for some m. Let G = Gal(K/F) and H = Gal(K/F(x)). There are two possibilities. If H is a 2 group, there is a 2-Sylow subgroup Q of G that contains H and there is a tower of extensions of degree 2 from Fix(Q) to Fix(H) = F(x). Moreover, Q is normal. The reason is that K contains L_1 and L_1 is normal so that Gal(K/L_1) is a normal subgroup of G. But Gal(K/L_1) is a 2-group and clearly of the same 2-order as G, so that Gal(K/L_1) is *the* 2-Sylow subgroup of G. Fix(Q) is an extension of F of degree p and, given that F has a primitive pth root of 1, it is known that it is gotten by adjoining a pth root. (So the assumption of a primitive pth root of 1 is used twice, once to see that a pth root extension is normal and once here.) If H is not a 2 group, then by choosing a composition series for G that passes through H, we get that F(x) is already a 2-tower above F and the pth root part is unnecessary. Michael Barr ============================================================================== From: mareg@mimosa.csv.warwick.ac.uk () Subject: Re: should be easy.. (algebra question) Date: Thu, 1 Nov 2001 22:00:50 +0000 (UTC) Newsgroups: sci.math,sci.math.research In article <19daab49.0111010601.1807eb30@posting.google.com>, barr@barrs.org (Michael Barr) writes: >barr@barrs.org (Michael Barr) wrote in message news:<19daab49.0110301347.76171077@posting.google.com>... >> dpwood@mit.edu (David P Woodruff) wrote in message news:<3bde0d1e$0$1917$b45e6eb0@senator-bedfellow.mit.edu>... >> > Hi, >> > >> >I have a 2-radical extension K of F,i.e., there exist intermediate >> >fields Li such that F < L1 < L2 < ... < Ln =K, where [Li:L(i-1)] = 2, >> >[L1:F] = 2. Let x be in K. How do I show there exists intermediate >> >fields Ri such that F < R1 < R2 < ... < Rm = F(x), where [Ri:R(i-1)] = >> >2, [R1:F] = 2? >> >> >Also, if I have a radical field extension K of F, where there exist >> >intermediate fields Si, such that F < S1 < S2 <.... < Sn = K, where >> >[Si:S(i-1)] = 2, [S1 : F] = p for some prime p. Let x be in K. I want to show >> >that the field F(x) is a radical 2-extension of F, or a radical 2-extension >> >of F(u), where u^p is in F. >> > >Forget my original post which was wrong. I made some assumptions >about F(x), which were clearly, in retrospect, not correct. I think >the following is correct. It unfortunately uses Galois theory, but I >see no alternative. It also makes some hypotheses about >characteristic and, for the second question, about having a primitive >pth root of 1 in F. I see no alternative to those. > [Proof for first problem snipped] > >Your second problem I can do if F contains a primitive pth root of 1 and >also char(F) != p. In that case L_1 is Galois and you do the same trick, >at each stage, adjoin not merely the square root of an element, but also >the square roots of each of its conjugates to get a Galois extension at >each stage. When you are finished you have a Galois extension K of >degree 2^m.p for some m. Let G = Gal(K/F) and H = Gal(K/F(x)). There >are two possibilities. If H is a 2 group, there is a 2-Sylow subgroup Q >of G that contains H and there is a tower of extensions of degree 2 from >Fix(Q) to Fix(H) = F(x). Moreover, Q is normal. The reason is that K >contains L_1 and L_1 is normal so that Gal(K/L_1) is a normal subgroup >of G. But Gal(K/L_1) is a 2-group and clearly of the same 2-order as G, >so that Gal(K/L_1) is *the* 2-Sylow subgroup of G. Fix(Q) is an >extension of F of degree p and, given that F has a primitive pth root of >1, it is known that it is gotten by adjoining a pth root. (So the >assumption of a primitive pth root of 1 is used twice, once to see that >a pth root extension is normal and once here.) OK, so fa! >If H is not a 2 group, >then by choosing a composition series for G that passes through H, we >get that F(x) is already a 2-tower above F and the pth root part is >unnecessary. But there may not be a composition series for G passing through H. Such a series exists only when H is a subnormal subgroup of G. For example, if G is A4 and |H|=3, then there is no such series. I gave an explicit example in an earlier post (where G was of order 48 and H of order 6), so I think the claim is false. Derek Holt. > >Michael Barr > ============================================================================== From: barr@barrs.org (Michael Barr) Subject: Re: should be easy.. (algebra question) Date: 7 Nov 2001 03:33:05 -0800 Newsgroups: sci.math,sci.math.research mareg@mimosa.csv.warwick.ac.uk () wrote in message news:<9rsgmi$qcn$1@wisteria.csv.warwick.ac.uk>... > In article <19daab49.0111010601.1807eb30@posting.google.com>, > barr@barrs.org (Michael Barr) writes: > >barr@barrs.org (Michael Barr) wrote in message news:<19daab49.0110301347.76171077@posting.google.com>... > >> dpwood@mit.edu (David P Woodruff) wrote in message news:<3bde0d1e$0$1917$b45e6eb0@senator-bedfellow.mit.edu>... > > OK, so fa! > > >If H is not a 2 group, > >then by choosing a composition series for G that passes through H, we > >get that F(x) is already a 2-tower above F and the pth root part is > >unnecessary. > > But there may not be a composition series for G passing through H. > Such a series exists only when H is a subnormal subgroup of G. > For example, if G is A4 and |H|=3, then there is no such series. > > I gave an explicit example in an earlier post (where G was of order 48 > and H of order 6), so I think the claim is false. > > Derek Holt. > > > > >Michael Barr > > But your counter-example is not a 2-group. I am not sure about p-groups in general, but 2 is special. Suppose G is a 2-group and H < G. I claim there is a composition series going through H. The proof is by a double induction, first on the order of G and second on the index of H in G. It is obvious for #(G) = 2 and suppose #(G) = 2^n and we assume it is true for smaller orders. If #(G/H) = 2, then H is normal and there is certainly a composition series passing through H. Assume now H is any subgroup and it is true for subgroups of smaller index. Let z be an element of order 2 in the center (which is known to exist). Case 1: z in H. In that case H/ is a subgroup of G/ in which case, the first inductin hypothesis says there is a normal series in G/ that passes through H/ which obviously implies there is a normal series in G that passes through H. Case 2: z not in H. In that case H is a normal subgroup of (the subgroup generated by H and z in which H has index 2) and by the second induction hypothesis, there is normal series going from to G. There is obviously one up to H. I am going to ask a group theorist friend about the situation for odd primes. Michael Barr ============================================================================== From: mareg@primrose.csv.warwick.ac.uk () Subject: Re: should be easy.. (algebra question) Date: Wed, 7 Nov 2001 22:10:34 +0000 (UTC) Newsgroups: sci.math,sci.math.research In article <19daab49.0111070333.14c32e2e@posting.google.com>, barr@barrs.org (Michael Barr) writes: >mareg@mimosa.csv.warwick.ac.uk () wrote in message news:<9rsgmi$qcn$1@wisteria.csv.warwick.ac.uk>... >> In article <19daab49.0111010601.1807eb30@posting.google.com>, >> barr@barrs.org (Michael Barr) writes: >> >barr@barrs.org (Michael Barr) wrote in message news:<19daab49.0110301347.76171077@posting.google.com>... >> >> dpwood@mit.edu (David P Woodruff) wrote in message news:<3bde0d1e$0$1917$b45e6eb0@senator-bedfellow.mit.edu>... >> >> OK, so fa! >> >> >If H is not a 2 group, >> >then by choosing a composition series for G that passes through H, we >> >get that F(x) is already a 2-tower above F and the pth root part is >> >unnecessary. >> >> But there may not be a composition series for G passing through H. >> Such a series exists only when H is a subnormal subgroup of G. >> For example, if G is A4 and |H|=3, then there is no such series. >> >> I gave an explicit example in an earlier post (where G was of order 48 >> and H of order 6), so I think the claim is false. >> >> Derek Holt. >> >> > >> >Michael Barr >> > > >But your counter-example is not a 2-group. I am not sure about >p-groups in general, but 2 is special. Suppose G is a 2-group and H < >G. I claim there is a composition series going through H. Yes, this property is true for all finite p-groups. There are various equivalent ways of saying this, like every subgroup of a finite p-group is subnormal, or every proper subgroup of a finite p-group is strictly contained in its normalizer. One way of proving it is first to show that the centre of a nontrivial p-group is nontrivial, which is a well-known argument involving adding up the orders of the conjugacy classes. Then prove H proper subgroup of p-group G => H strictly contained in its normalizer by induction. Either Z(G) (the centre of G) is contained in H, and we can induct on H/Z(G) in G/Z(G), or Z(G) is not contained in H and then H is properly contained in HZ(G). Derek Holt. ============================================================================== From: barr@barrs.org (Michael Barr) Subject: Re: should be easy.. (algebra question) Date: 7 Nov 2001 15:32:34 -0800 Newsgroups: sci.math,sci.math.research barr@barrs.org (Michael Barr) wrote in message news:<19daab49.0111070333.14c32e2e@posting.google.com>... > mareg@mimosa.csv.warwick.ac.uk () wrote in message news:<9rsgmi$qcn$1@wisteria.csv.warwick.ac.uk>... > > In article <19daab49.0111010601.1807eb30@posting.google.com>, > > barr@barrs.org (Michael Barr) writes: > > >barr@barrs.org (Michael Barr) wrote in message news:<19daab49.0110301347.76171077@posting.google.com>... > > >> dpwood@mit.edu (David P Woodruff) wrote in message news:<3bde0d1e$0$1917$b45e6eb0@senator-bedfellow.mit.edu>... > > > > OK, so fa! > > > > >If H is not a 2 group, > > >then by choosing a composition series for G that passes through H, we > > >get that F(x) is already a 2-tower above F and the pth root part is > > >unnecessary. > > > > But there may not be a composition series for G passing through H. > > Such a series exists only when H is a subnormal subgroup of G. > > For example, if G is A4 and |H|=3, then there is no such series. > > > > I gave an explicit example in an earlier post (where G was of order 48 > > and H of order 6), so I think the claim is false. > > > > Derek Holt. > > > > > > > >Michael Barr > > > > > But your counter-example is not a 2-group. I am not sure about > p-groups in general, but 2 is special. Suppose G is a 2-group and H < > G. I claim there is a composition series going through H. The proof > is by a double induction, first on the order of G and second on the > index of H in G. It is obvious for #(G) = 2 and suppose #(G) = 2^n > and we assume it is true for smaller orders. If #(G/H) = 2, then H is > normal and there is certainly a composition series passing through H. > Assume now H is any subgroup and it is true for subgroups of smaller > index. Let z be an element of order 2 in the center (which is known > to exist). Case 1: z in H. In that case H/ is a subgroup of G/ > in which case, the first inductin hypothesis says there is a normal > series in G/ that passes through H/ which obviously implies > there is a normal series in G that passes through H. Case 2: z not in > H. In that case H is a normal subgroup of (the subgroup > generated by H and z in which H has index 2) and by the second > induction hypothesis, there is normal series going from to G. > There is obviously one up to H. I am going to ask a group theorist > friend about the situation for odd primes. > > Michael Barr As I suspected, in a p-group, there is a normal series running through any subgroup. It is sufficient to show that every subgroup of index p is normal. If G is a p-group and H < G has index p, then G acts on G/H by translation. This represents G into S_p. Since p is prime S_p has no subgroup of order p^2, so the image of G in S_p must be of order p and so its kernel has index p. But the kernel is included in H since only H acts as the identity on the identity coset. But then H is the kernel. As a result, the following holds: Theorem: Suppose F contains a primitive pth root of 1 and n is an integer not divisible by p. Suppose F < F_1 < F_2 < ... < F_k is a sequence of field extensions such that F_1/F is a Galois extension of degree n and for i > 1, F_i is a galois extension of F_{i-1} of degree p (hence by a pth root). Then any intermediate field E is also the top of a tower F < E_1 < ... < E_l = E in which each step is a Galois extension, first of degree some m not divisible by p and the remaining ones are by pth roots. The proof is essentially the same.