From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: quaternion equations Date: 3 Nov 1995 20:24:58 GMT In article <[identifier deleted]> [Permission pending] wrote: > >There is a classical result (due to Kronecker I think) saying >that any polynomial has a splitting field. Thus, any polynomial >equation can be solved within a sufficiently large field. Is >there any similar result for polynomials over non-commutative >rings? Can, for example, a polynomial equation with quaternion >coefficients always be solved in some larger ring? You have to decide what you mean by "larger ring" and indeed, what you mean by "polynomial"! The classical result you mentioned results from the following simpler result (which you may have confused with the existence of splitting fields): If f is any irreducible polynomial in F[X] then there is an extension field in which f has a root. Indeed the proof is easy: the quotient ring K=F[X]/(f) is a field, contains F, and contains a root of f (namely, X !). In some ways, the same idea can be used for non-commutative rings. If R is a ring let me write R for the ring of all non-commuting polynomials in X with 'coefficients' in R. This is the collection of all finite sums of terms of the form r1*X*r2*X*...*X*rn with each r_i in R. Then you are given an element f in this ring. If you want a ring in which f(x)=0 has a solution, you need only take the quotient ring R_f = R/(f) where (f) is the _two-sided_ ideal generated by f. This is the "universal" solution of your problem; if S is any other, err, R-algebra in which f(x)=0 has a solution, then S contains a subring S' in which there is also a solution and such that S' is a homomorphic image of R_f. There are a number of issues here which make the problem less nice than in the commutative case. First of all, it is not clear in this generality that this R_f contains R. R does, but it's not so easy to see if R meets (f). If it does, then R_f only contains a homomorphic image of R. (That's why I called R_f an "R-algebra" rather than "an extension ring of R", above.) Next, just as in the commutative case, R_f may be "too big" (e.g. if f=f1*f2 then R_f1 and R_f2 are proper homomorphic images of R_f -- they are "smaller" -- and in each one, f has a root). What you want is to look at maximal ideals containing (f) (but not meeting R, I guess); in complete generality, this is not the same as saying f is irreducible. Finally, I observe that R_f need not be a finitely generated R-module, unlike the commutative case. As an example, I consider the linear (!) quaternionic equation f(X)=i*X-X*i-1. If there is an R-algebra in which this equation has a solution x, then we have 0=i*0+0*i=i*f(x)+f(x)*i=i*(iX-Xi-1)+(iX-Xi-1)*i =(-X-iXi-i)+(iXi+X-i)=(-2i). This contradiction shows that R cannot embed in the new ring. In fact (R not having any proper ideals) the homomorphism R --> R_f is zero in this case! If your interest is specifically in quaternions "H", you can proceed as follows. Suppose f is a polynomial you wish to solve in an extension ring S of H. Suppose S is Noetherian. Then there exist maximal 2-sided ideals not meeting H; call one of them I. Then S/I is also a ring in which f has a root, and it still contains H, but now it has no nonzero two-sided ideals not meeting H. If we assume, as is usually the case, that the quaternion 1 is also an identity element for multiplication in S (e.g., don't try something like S=M_2(H) with the quaternions identified with the multiples of [[1,0],[0,0]]!) the the ring S is simple -- it has no two-sided ideals at all. Next you probably want to assume that S is a finite-dimensional vector space over the reals (e.g., don't take S to be the set of endomorphisms of some Hilbert space!). Well, the finite-dimensional simple real algebras are precisely the matrix rings M_n(R), M_n(C), and M_n(H). At your preference you may embed each of these three types of rings in either of the other two types. In each case, the subring H may be identified with a certain algebra of k x k blocks of scalar matrices, with k= 4, 2, or 1 respectively. Then the original equation f=0 to solve may be written as a certain real/complex/quaternion matrix question. For example, the question, "does iX-Xi=1 have a solution in an extension of H?" now reads, if we choose a complex embedding, "does there exist a (2n)x(2n) complex matrix X such that JX-XJ=I, where I is the identity matrix and J is the matrix [[iI,0], [0, iI]] (a matrix of four n x n blocks)?" This question is easy to answer in the negative, since the traces of JX-XJ and I can't be equal. Finally, let me remark that if your interest is not only to have the domain R over which f is defined be the quaternions but to have the larger domain S be a division ring as well, you are basically stuck. There are no division rings containing H which are finite-dimensional H-modules. I have a dim memory that every equation of the simpler form r_n X^n + r_(n-1) X^(n-1) +...+ r_1 X + r_0 = 0 (with each r_i in H) is solvable, and indeed that many of the usual statements of Galois theory apply to division algebras when restricted to polynomials of this special form, but I cannot remember the source of this information, so perhaps I ought not go out on a limb. You know how dim memories are. dave ============================================================================== Date: Tue, 7 Nov 1995 00:38:06 +0100 To: rusin@washington.math.niu.edu (Dave Rusin) From: [Permission pending] Subject: Re: quaternion equations [headers of my post deleted 3/20/96 -- djr] >>There is a classical result (due to Kronecker I think) saying >>that any polynomial has a splitting field. Thus, any polynomial >>equation can be solved within a sufficiently large field. Is >>there any similar result for polynomials over non-commutative >>rings? Can, for example, a polynomial equation with quaternion >>coefficients always be solved in some larger ring? *************************************************************** Thank you very much for your long and comprehensive answer to my question on quaternion equations etc to sci.math. I have some comments and further questions which I hope you want to consider. (I mark your message by > and snip those parts I don't refer to.) >You have to decide what you mean by "larger ring" and indeed, what you >mean by "polynomial"! By "polynomial" I meant polynomial over a field; I simply forgot to write it but sent a correction. By "larger ring" I meant a ring containing another ring (or an isomorphic ring). If I understand your answer correctly this can in general not be found but another algebraic structure (R-algebra) is needed to split the polynomial. >The classical result you mentioned results from the following simpler >result (which you may have confused with the existence of splitting >fields): No, I don't think I confused them, but of course the result you mention is the more fundamental one. >If f is any irreducible polynomial in F[X] then there is >an extension field in which f has a root. Indeed the proof is easy: >the quotient ring K=F[X]/(f) is a field, contains F, and contains >a root of f (namely, X !). Ok...; I don't remember I have seen it so easily expressed before. [ snip, which I think I understood although perhaps not in all details, but on which I have no further questions for the present ] >As an example, I consider the linear (!) quaternionic equation >f(X)=i*X-X*i-1. If there is an R-algebra in which this equation has a >solution x, then we have 0=i*0+0*i=i*f(x)+f(x)*i=i*(iX-Xi-1)+(iX-Xi-1)*i >=(-X-iXi-i)+(iXi+X-i)=(-2i). This contradiction shows that R cannot >embed in the new ring. In fact (R not having any proper ideals) the >homomorphism R --> R_f is zero in this case! This shows that my request for a "larger ring" (in the above sense) cannot be fulfilled, so to get something interesting another type of enlargement is required. Right? >If your interest is specifically in quaternions "H", ... At least to begin with, yes. Then, having graduated in theoretical physics, I would also like some more mathematical background for some of the mathematical methods of theoretical physics. The 2x2-matrices used in quantum mechanics ( s_z = [ [1,0] , [0,-1] ] etc ) form a ring isomorphic to the ring of quaternions (H), I think. [ snip ] >Finally, let me remark that if your interest is not only to have >the domain R over which f is defined be the quaternions but >to have the larger domain S be a division ring as well, you are >basically stuck. There are no division rings containing H which are >finite-dimensional H-modules. Ok. But I am very satisfied to get some result for quaternions. >I have a dim memory that every equation of the simpler form > r_n X^n + r_(n-1) X^(n-1) +...+ r_1 X + r_0 = 0 >(with each r_i in H) is solvable, and indeed that many of the >usual statements of Galois theory apply to division algebras when >restricted to polynomials of this special form, but I cannot >remember the source of this information, so perhaps I ought not >go out on a limb. You know how dim memories are. This is then an affirmative (partial) answer to my original question. Your example "iX - Xi = 1" is not of this type but can you mention some example which _is_ (together with an extension of H, and roots in this extension)? BTW I got another answer, which may interest you, bearing your last paragraph in mind: ################################# From: kauta Subject: Re: quaternion equations [ snip ] ...theorem by Wedderburn (or Albert): Let D be a divison algebra finite dimensional over its center F. Then any polynomial over F can be factored (not necessarily uniquely) in D. [ snip ] ################################# Regards and thanks again, [sig deleted - djr] ============================================================================== Date: Tue, 7 Nov 95 15:51:36 CST From: rusin (Dave Rusin) To: [Permission pending] Subject: Re: quaternion equations >>You have to decide what you mean by "larger ring" and indeed, what you >>mean by "polynomial"! > >By "polynomial" I meant polynomial over a field; I simply forgot to write it >but sent a correction. My concern is whether you intended to include examples like iX-Xi-1 or not. Some of what we know from the commutative case _does_ carry over if your consider _only_ polynomials of the form a0 + a1 X + a2 X^2 + ... (i.e., "all the X's on the right"). >By "larger ring" I meant a ring containing another ring (or an isomorphic ring). >If I understand your answer correctly this can in general not be found >but another algebraic structure (R-algebra) is needed to split the polynomial. Right. An R-algebra is a ring S together with a homomorphism of rings F: R --> End(S). This is a fancy way of saying that you have a way of multiplying elements of R and elements of S together in a way which respects the ring structures. In particular, there is a subring of S which is a homomorphic image of R (namely F(S)*1_S). For example, every ring is Z-algebra: n*r makes sense for any integer n and any ring element r. The subring in question is then the set { n*1_S : n in Z} of multiples of the identity element. If the characteristic of the ring is not zero, this is a proper quotient Z/mZ of Z, not Z itself, so you wouldn't call the ring an extension of Z. If you stick to division algebras, though, then there are no proper homomorphic images of R, so that every R-algebra is in fact an extension ring of R. (Well, I'm assuming all my rings have unit elements.) The only other possibility is to have a ring S such that all products r*s are zero, that is, such that F: R --> End(S) is the zero homomorphism. For example, Z/pZ is a module over the complex numbers if you agree that c*[n mod p] is zero for every complex number c. Do you consider this to be a C-module structure? I'm not sure I would allow it (usually people insist that ring homomorphisms carry identity elements to identity elements). If you want to say that every polynomial equation has a solution in some R-algebra, you have to be prepared for this as your answer! (Actually, if you start to use non-commutative rings, you have to work a little harder to define R-algebras, since you need to discuss both right- and left- multiplications by elements of R. I'll skip this.) More generally, I was concerned because you didn't specify which kind of extension ring you wanted to solve your equation in. For example, the equation 2*X-1=0 has no solutions in Z, but it does in extension rings. The smallest such extension would be Z[X]/(2X-1), which is isomorphic to the ring Q_2 of rational numbers whose denominators are powers of 2. This ring is much larger than Z , and rather a complex extension of it (It's an infinitely-generated Z-module, but not a free Z-module; you can take as generators the numbers x_n = 1/2^n, but then you have infinitely many relations x_n = 2 x_(n+1).) If you started with a ring without zero divisors, do you want your extension to have that property? If you start with a division ring, do you want your extension ring to have that property? You have to decide in advance what question you really want answered! >>As an example, I consider the linear (!) quaternionic equation >>f(X)=i*X-X*i-1. If there is an R-algebra in which this equation has a >>solution x, then we have 0=i*0+0*i=i*f(x)+f(x)*i=i*(iX-Xi-1)+(iX-Xi-1)*i >>=(-X-iXi-i)+(iXi+X-i)=(-2i). This contradiction shows that R cannot >>embed in the new ring. In fact (R not having any proper ideals) the >>homomorphism R --> R_f is zero in this case! > >This shows that my request for a "larger ring" (in the above sense) cannot >be fulfilled, so to get something interesting another type of enlargement is >required. Right? Yeah, there is a ring S containing elements "i" "X" and "1" in which iX-Xi-1=0, and there is a homomorphism H --> S sending the quaternion i to this new "i" and the quaternion 1 to this new "1". It's an H-algebra as described above (and created in the previous post). The only problem is that "i" and "1" have to equal zero! You could for example take S=Z and X=17 (or any other integer); Z is an H-algebra just as I described it earlier as a C-algebra. Then of course iX-Xi will equal "1" (which is the quaternion 1 multiplying the integer 1) since every product with a quaternion is zero. The analysis you quoted shows there is no other way to make an H-algebra in which iX-Xi-1=0 except to have products be zero. (It's a little more cumbersome to say what's really true since H is non-commutative, but as I noted above I'm glossing over those technical issues right now.) >At least to begin with, yes. Then, having graduated in theoretical physics, >I would also like some more mathematical background for some of the >mathematical methods of theoretical physics. The 2x2-matrices used in >quantum mechanics ( s_z = [ [1,0] , [0,-1] ] etc ) form a ring isomorphic >to the ring of quaternions (H), I think. This doesn't look quite right. You can form the quaternions with complex 2x2 matrices, but this matrix can't be in the collection. (If it were, then (1+s_z)*(1-s_z)=0, which can't happen in a division ring!) Perhaps you mean [[0,1],[-1,0]] to start with. >>I have a dim memory that every equation of the simpler form >> r_n X^n + r_(n-1) X^(n-1) +...+ r_1 X + r_0 = 0 >>(with each r_i in H) is solvable, and indeed that many of the >>usual statements of Galois theory apply to division algebras when >>restricted to polynomials of this special form, but I cannot >>remember the source of this information, so perhaps I ought not >>go out on a limb. You know how dim memories are. > >This is then an affirmative (partial) answer to my original question. Your >example "iX - Xi = 1" is not of this type but can you mention some example >which _is_ (together with an extension of H, and roots in this extension)? Good question! You want a polynomial equation which is solvable in an extension ring of H but not in H itself. Unfortunately, the only way I know to make equations which are not solvable in H is of the above type. I think that all "right-polynomial" equations are solvable in H (I just spent an unproductive hour looking for equations of the type aX^2+bX+c=0 but as far as I can tell they are all solvable in H, for all choices of a,b,c in H.) If you want something really unsolvable in H, it may have to be of the iX-Xi=1 type, which in turn have no solutions in _extensions_ of H either. Thus in an appropriate sense, H is "algebraically closed". All finite-dimensional extensions of H can be embedded in M_n(H), so if you think you have an equation which is solvable in a finite-dimensional extension, then you can solve it in quaternionic matrices (You can view these as 2n x 2n complex matrices; the quaternions will be the subalgebra of all block matrices of the form [ [aI, bI], [cI, dI ] where d=a-conjugate, c= - (b-conjugate). ) >...theorem by Wedderburn (or Albert): > >Let D be a divison algebra finite dimensional over its center F. Then any >polynomial over F can be factored (not necessarily uniquely) in D. (and then by induction on the degree it factors into linear factors?) I'm not entirely sure what this means. Certainly there is nothing quoted here which prohibits D=F, so unless F is algebraically closed, there can be F-polynomials which do not factor. More generally, the existence of this factorization only means f(X)=f1(X)*f2(X) _formally_, i.e., treating X as a commuting variable. That's not quite the same as saying the polynomial has a root. For example, in H the polynomial f(X)=X^2+1 factors as (X-i)*(X+i); since f(j)=j^2+1=0, does this mean (j-i)*(j+i)=0 (and thus j= +- i, as H has no zero divisors)? Not at all -- if you wish to view (X-i)*(X+i) as a quaternionic _function_, it expands to X^2 +(iX-Xi) + 1; this only simplifies to X^2+1 if iX-Xi=0. The problem is that when treated formally here, the X is assumed to commute with all the other coefficients, whereas when treated as a function, this will in general only be true when X is in the center (F). In short, if a polynomial has a linear factor (over a division ring) then it has a root; but if it has a root, that doesn't mean it factors. This whole topic is interesting to me, although I am a little out of practice; since I deal usually in commutative ring, it is easy to slip and forget places where one has assumed commutativity. I hope I haven't led you astray. dave ============================================================================== Date: Wed, 15 Nov 1995 22:43:45 +0100 To: rusin@math.niu.edu (Dave Rusin) From: [Permission pending] Subject: Re: quaternion equations [ I snip much here, and just keep what I write about, but be sure I read the rest too! ] >My concern is whether you intended to include examples like iX-Xi-1 or not. >Some of what we know from the commutative case _does_ carry over >if your consider _only_ polynomials of the form a0 + a1 X + a2 X^2 + ... >(i.e., "all the X's on the right"). Ok, then I understand the question. I simply didn't think of this consequence of non-commutativity. >More generally, I was concerned because you didn't specify which kind of >extension ring you wanted to solve your equation in. Well, that was no accident. I simply asked for _any_ result this kind. And what you have written about is interesting enough, also when the answer is in the "negative", as for quaternions. >>This shows that my request for a "larger ring" (in the above sense) cannot >>be fulfilled, so to get something interesting another type of enlargement is >>required. Right? > >This doesn't look quite right. You can form the quaternions with complex >2x2 matrices, but this matrix can't be in the collection. (If it were, then >(1+s_z)*(1-s_z)=0, which can't happen in a division ring!) Perhaps you mean >[[0,1],[-1,0]] to start with. You are right, this is wrong. But I think there is an easy remedy. The spin matrices are usually defined s_x = [[0,1],[1,0]] , s_y = [[0,-i],[i,0]] , s_z = [[1,0],[0,-1]] Linear combinations (with real coefficients) of these, and the identity matrix, represent the quantum mechanical spin operators for particles with spin 1/2 (e g electrons). Now let I = -i*s_x , J = -i*s_y , K = -i*s_z Then I*J = -J*I = K , I^2 = -1 , etc. I think linear combinations of these (and the identity matrix) with real coefficients form a ring isomorphic to the ring of quaternins (H). Correct? >I think that all "right-polynomial" equations are solvable in H >(I just spent an unproductive hour looking for equations of the type >aX^2+bX+c=0 but as far as I can tell they are all solvable in H, for all >choices of a,b,c in H.) If you want something really unsolvable in H, >it may have to be of the iX-Xi=1 type, which in turn have no solutions in >_extensions_ of H either. Thus in an appropriate sense, H is >"algebraically closed". Really interesting, since H is a "natural" generalization (although no field) of C (the complex numbers)! And this is one more answer (or conjecture) to my original question (although "negative"). >>...theorem by Wedderburn (or Albert): I received a more precise formulation (from kauta ): ***************** The result by Wedderburn implies the following: Let D be an F-central division algebra, and pick a in D. Let f(x) be the minimal polynomial of a over F (of degree n say). Then f(X) has factorization in D[X] of the form f(X) = (X - a_1)(X - a_2)...(X-a_n), with each a_i = d_i a d_i^{-1} for some d_i in D\{0}. together with a reference: c) J.H.M. Wedderburn, On division algebras, Trans. Amer. Math. Soc. 22(1921), 129-135, Lemma 4. ************************** >In short, >if a polynomial has a linear factor (over a division ring) then it has a root; >but if it has a root, that doesn't mean it factors. I must think more about this, but is yet one answer to my question. Thanks very much for spending so much time to help me with these matters! Regards, [sig deleted - djr]