From: Dave Rusin Date: Mon, 23 Nov 1998 17:59:04 -0600 (CST) To: Bernd.Hilgenberg@rt.bosch.de Subject: Re: noncommutative divisionrings Well! The construction of central simple division algebras over a number field is a pretty thing. I won't bore you with the details, but it is related to the Galois cohomology of the field in the center of the algebras. Here is a construction which makes division algebras of dimension 9 over the center F. Let K/F be a Galois extension of degree 3 -- for example, F = Q (the rational field) and K = Q[X]/(X^3-21*X-7). We make an algebra A out of the three-dimensional vector space over K whose basis is x0, x1, x2 that is, every element of A in uniquely expressible as a linear combination k0 x0 + k1 x1 + k2 x2. Addition is done in the natural way. For multiplication it suffices to use the rules x_i k = kbar x_i where kbar is the image under of k under the Galois mapping sigma^i. In plain speech this means we use the presentation of K as a three-dimensional vector space over Q spanned by alpha, beta, gamma, which are three conjugates in K under the action of the Galois group. Then we have the multiplicative structure within K itself, and x_0 serving as the identity element of the ring; x_1 alpha = beta x_1 and likewise beta is sent to gamma and gamma to alpha; x_2 alpha = gamma x_2 and likewise beta is sent to alpha and gamma to beta; To decribe the rest of the product structure, I need only tell you that x_i x_j = x_{i+j} as long as i+j < 3; otherwise x_i x_j = f x_{i+j} for some fixed element f of the rational field Q. That is, the x_i generate a subring isomorphic to Q[f^(1/3)] in A. (Indeed, the whole algebra _almost_ looks like the extension ring K[f^(1/3)] of K, except that these x_i do not commute with all of K.) Now you see that you can describe A as a subring of the set of 3x3 matrices over K; indeed, we simply map k0 x0 + k1 x1 + k2 x2 to the matrix M whose i,j entry is k_{j-i} if j >=0 and f k_{j-i+3} otherwise. This kind of algebra is called a _cyclic algebra_. You can see that replacing K by the complex field C and using only a two-dimensional vector space A over C in this way, you construct exactly the field of quaternions: x_0=1, x_1 = j ; x_1 k = kbar x_1 where kbar is ordinary complex conjugation. In this example, f = x_1^2 = j^2 = -1. More generally, we eventually find this ring is a division algebra iff f is not in the image of the norm map N : K -> F. For example we may take f=5 since 5 is not the norm of any element in K. Almost everything that I have said generalizes and in fact with sufficient generality you can in this way more or less describe _all_ division algebras with a given center. This is what the Brauer group is. dave >I have read that noncommutative and nonassoziative divisionalgebras >are quadratic. I'm not sure I believe this. Were you perhaps reading about division algebras _over the real field_ ? ============================================================================== Date: Wed, 25 Nov 1998 08:40:43 +0100 From: bhilgen To: Dave Rusin Subject: Re: noncommutative divisionrings Hello, please let me ask some questions about galois-fields, because I have only a few information about this field. All I know is the construction F[x] / irreducible polynom . In your example F=Q and the irreducible polynom is X^3-21*X-7 ? > For multiplication it suffices to use the rules > x_i k = kbar x_i > where kbar is the image under of k under the Galois mapping sigma^i. what is the Galois mapping sigma ? Can you define this mapping specielly for X where X = generator of K=Q[X]/(X^3-21*X-7) = a + b*X +c*X*X with X*X*X= 21*X+7 and a,b,c are in Q ? I guess that sigma( u * v ) = sigma( u )*sigma( v ) for all u,v in K? Is sigma^3=Id in our example ? > In plain speech this means we use the presentation of K as a > three-dimensional vector space over Q spanned by alpha, beta, gamma, > which are three conjugates in K under the action of the Galois group. > Then we have the multiplicative structure within K itself, and > x_0 serving as the identity element of the ring; > x_1 alpha = beta x_1 and likewise beta is sent to gamma and gamma to alpha; > x_2 alpha = gamma x_2 and likewise beta is sent to alpha and gamma to beta; > > To decribe the rest of the product structure, I need only tell you that > x_i x_j = x_{i+j} as long as i+j < 3; otherwise > x_i x_j = f x_{i+j} for some fixed element f of the rational field Q. > That is, the x_i generate a subring isomorphic to Q[f^(1/3)] in A. > (Indeed, the whole algebra _almost_ looks like the extension ring > K[f^(1/3)] of K, except that these x_i do not commute with all of K.) > > Now you see that you can describe A as a subring of the set of 3x3 matrices > over K; indeed, we simply map k0 x0 + k1 x1 + k2 x2 to the matrix M > whose i,j entry is k_{j-i} if j >=0 and f k_{j-i+3} otherwise. > if I understand all this it should be that x1 has the regular representaion as [ 0, 0, f ] [ 1, 0, 0 ] "tensor" sigma [ 0, 1, 0 ] where [ 0, 0, f ] [ 1, 0, 0 ] [ 0, 1, 0 ] is linear a linear map on K^3 = { k0 x0 + k1 x1 + k2 x2 } but the Galois mapping sigma is not a linear map on K^3 , because otherwise K is in the centre ??!!?? Therefore I have some trouble to write the element x1 as a a 3x3 matrix over K (I hope these matrices are readable) > More generally, we eventually find this ring is a division algebra iff > f is not in the image of the norm map N : K -> F. For example we may > take f=5 since 5 is not the norm of any element in K. what is the norm map ? I guess N(u):= u*sigma(u)*sigma^2(u) for u in K=Q[X]/(X^3-21*X-7) ? If sigma( u * v ) = sigma( u )*sigma( v ) and sigma^3=Id than it follows sigma( N(u) )= N(u) ( is in Q ? ) > >I have read that noncommutative and nonassoziative divisionalgebras > >are quadratic. > I'm not sure I believe this. you are right, I forgotten to say that the algebra must be alternative > Were you perhaps reading about division algebras _over the real field_ ? in your text index/products.html you recomment the springer book "Numbers" by Ebbinghaus I buy that book and find at the end of Chapter9 part2 a reference to : Bruck, Kleinfeld "the structure of alternative division rings" Proc. Am. Math. Soc. 2 1951 page 878-890 In Lemma 4.1 Bruck, Kleinfeld say that for an alternative ring R without divisors of zero and 1+1 =/= 0 generated by x, y, z such that u:=(x, y, z):= (x*y)*z - x*(y*z) =/= 0 and v:= (x*x, y, z) and w:= (x, x*y, z) then (u*u)*(x*x) -(u*v)*x -(w*w) =0 where (u*u), (u*v), (w*w) are in the centre C of R C:={ c/ c*a=a*c and (c,a,b)=0 for all a,b in R } From this they conclude Theorem 4.1 that an alternative ring without divisors of zero and and 1+1 =/= 0 which is NOT ASSOCIATIVE is quadratic ( the square x*x of every element x in R can be written as x*x= a*x +b with suitable a,b in centre C). I turns out that such an algebra is like O ( Cayley-algebra ) ? see "Nummbers" by Ebbinghaus Now it is a surprise for me to see, that if I replace in Theorem 4.1 NOT ASSOCIATIVE through ASSOCIATIVE ( which is a somehow stronger condition ) we cannot show that the algebra is quadratic. ( I have tried to show this for several weeks but give up , and now I knew thanks your help that we cannot show this) thanks again for your help and plaese give me a hint about the galois-map sigma Bernd Hilgenberg -- Bernd Hilgenberg Robert BOSCH GmbH Abteilung: K8/EIC1 Telefon: 07121/35-1016 mailto:Bernd.Hilgenberg@rt.bosch.de ============================================================================== From: Dave Rusin Date: Wed, 25 Nov 1998 01:55:38 -0600 (CST) To: Bernd.Hilgenberg@rt.bosch.de Subject: Re: noncommutative divisionrings I will have to respond to your letter tomorrow (it is very late here!) but let me just comment on some of your statements: > I hope these matrices are readable) Yes, it is fine. > I forgotten to say that the algebra must be alternative Ah! That is a very strong restriction. Yes, I believe it is natural to think that such division algebras are quadratic. > Now it is a surprise for me to see, that if I replace in Theorem 4.1 >NOT ASSOCIATIVE through ASSOCIATIVE ( which is a somehow stronger >condition) we cannot show that the algebra is quadratic. Yes, it is a little surprising. I guess I would have expected that somehow the algebras would have an associative part (not necessarily quadratic) and a non-associative part (where there is so little room allowed in the alternative axiom for three non-associating elements, that is is reasonable to think they must be quadratic), and somehow all the elements of the algebra are made up from these two parts. But on further reflection, I guess there is no room for two such "parts" in a division algebra -- one characteristic they share is that they have no nontrivial ideals, and thus no way to "glue together the parts" unless one part or the other is essentially empty. I will have to give this some more thought. dave ============================================================================== From: Dave Rusin Date: Mon, 7 Dec 1998 09:21:11 -0600 (CST) To: Bernd.Hilgenberg@rt.bosch.de Subject: Re: nonquadratic divisionalgebras Sorry not to have responded earlier. >the springer book Rotman,J.: Galois Theroie This is a fine choice, but many texts about algebraic number theory or abstract algebra can give you the same information if this is not available. You will really need to be familiar with this material to accomplish what you want. >You define K:=Q[x]/( x^3-21*x-7 ) >but how is this relatet to galios-fields No, no -- these are two different concepts. You are quite right: a Galois field means a finite field. I was talking about a _Galois extension_ which means that K is a larger field which is formed by adjoining to Q _all_ the roots of this irreducible polynomial. I had to choose that polynomial carefully to have a Galois extension of degree 3. For example, if I had said "K = Q[x]/( x^3 - 2 ) " then K would be an extension field of Q having degree 3 (that is, K is a 3-dimensional vector space over Q) but this K is _not_ a Galois field: the polynomial x^3-2 is irreducible over Q and has _just one_ root in K. If you wanted a Galois extension of Q which contained K you would have to include also a root of x^2+x+1 (that is, a primitive cube root of unity). The connection with Galois fields is that when the base field is finite, _all_ finite extensions are Galois extensions. I know this is confusing, but with some exposure to the ideas it is not so bad. >Can you expain this map [the Galois-sigma map] Every element of K may be written a + b x + c x^2 for some rational a,b,c. We define s(a+bx+cx^2) to be a + b s(x) + c s(x)^2, where s(x) is any root of X^3-21X-7. Here x is one of the roots and, _since this is a Galois extension_, the other two roots are also in K. In fact you can check that the other two roots are (14-x-x^2)/3 and (-14-2x+x^2)/3. You can take either of these for s(x), and so you have two different choices for what this mapping s is. dave ==============================================================================