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        riddles >> putnam exam (pure math) >> algebra
        
(Message started by: judy on Mar 24^th, 2008, 7:46pm)

Title: algebra
Post by judy on Mar 24^th, 2008, 7:46pm

Hey,

I'm new on this forum and I have two questions. Can you please help me?

1) M_k(F) is isomorphic to subring of M_e(F) if and only if e/k ( kle) F is a field.

2)Describe the degree 2 irreducible representation of D_4= <a,b: a^2=b^4=1, a^-1ba=b^-1>

Title: Re: algebra
Post by Obob on Mar 24^th, 2008, 10:04pm

(1) I don't know what you mean here. I take it that by M_k(F) you mean the ring of k x k matrices over the field F. And I think you are wanting to show that M_k(F) is isomorphic to a subring of M_e(F) if and only if k <= e. It is important to know what is meant by "isomorphism." In particular, do isomorphisms have to preserve the identity element? And does a subring with an identity element have to have the same identity element as the larger ring? Does the isomorphism have to be F-linear?

Without at least the assumption that the isomorphism is F-linear, this problem is quite difficult (and I'm not positive it is even true, since funny things can happen like having a subfield of a field be isomorphic to the whole field). If you do assume it is F-linear, the problem becomes trivial by dimension counting.

(2) D₄ is the group of symmetries of a square. Put a square in the real plane, centered around the origin (in fact, we should be able to put the square anywhere in the plane so long as none of its vertices is the origin). A symmetry of the square uniquely determines an invertible linear map of the plane. Send the symmetry to this linear map.

Title: Re: algebra
Post by judy on Mar 24^th, 2008, 10:47pm

ohh yes I have to write my question again. Sorry I didnt know how to use this symbols. I think it's ok now

M_k(F) is isomorphic to subring of M_e(F) if and only if k l e (e divide k) F is a field.

here M_k(F) means k-by-k full
matrix algebra over F

Title: Re: algebra
Post by Obob on Mar 25^th, 2008, 12:01am

Can you state what the implied definition of isomorphism is? I imagine it is "isomorphism of k-algebras with unit."

Title: Re: algebra
Post by judy on Mar 25^th, 2008, 6:25am

I think it's about Wedderburn-Artin theory and I have to check it under endomorphism.

Title: Re: algebra
Post by Obob on Mar 25^th, 2008, 11:27am

That doesn't answer my question. And what do you mean by "check it under endomorphism"? Until you have figured out what the correct definition of morphism is, the problem isn't really tractable.

Title: Re: algebra
Post by judy on Mar 26^th, 2008, 5:29am

I can't understand what you mean. I think here I dont need unit. I have to find an endomorphism on F and I need a maximal ideal. Don't need F is linear.

Title: Re: algebra
Post by Obob on Mar 26^th, 2008, 9:25am

What I mean is this: an isomorphism of M_k(F) with a subring of M_e(F) is some kind of injective map f from M_k(F) to M_e(F). For it to be called an isomorphism it has to satisfy some properties besides just being a function. For starters,

(1) f has to preserve addition. f(M+N)=f(M)+f(N) for all M, N in M_k(F).

(2) f has to preserve multiplication. f(MN)=f(M)f(N) for all M, N in M_k(F).

Now the question is what other axioms we require f to satisfy. In the context of Wedderburn theory, it is almost certainly the case that f must be F-linear:

(3) f is F-linear: for all t in F and M in M_k(F), we have f(tM)=tf(M).

Given the result you are trying to prove, f also has to preserve the identity matrix (or unit):

(4) f preserves the identity matrix: f(I_k)=I_e.

Without the hypothesis that f preserves the identity matrix, it is true that M_k(F) is isomorphic to a subring of M_e(F) if and only if k <= e, which is not the result we are trying to prove. If we do assume f preserves the identity matrix, then it is at least easy to show that if k divides e then M_k(F) is isomorphic to a subring of M_e(F): we can take a k x k matrix and put e/k copies of it along the diagonal of an e x e matrix; then since block diagonal matrices multiply block by block, this map is an isomorphism.

Endomorphisms of the field F play no role in Wedderburn theory, since one requires maps to be F-linear so that the endomorphisms of F are just given by multiplication with elements of F. I can't see how maximal ideals possibly play into this problem.

Title: Re: algebra
Post by Eigenray on Mar 28^th, 2008, 9:11pm

My non-commutative algebra is a bit rusty I'm afraid, but I can at least prove it in this case:

Suppose there is an irreducible (over F) polynomial f(x) in F[x] of degree k. Take A http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif M_k(F) to be the companion matrix of f, so that A has minimal polynomial f(x). Then the image http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(A) in M_e(F) will still have minimal polynomial f(x). Putting http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(A) into rational canonical form, we see that the characteristic polynomial of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(A) must be a power of f(x), proving k | e.

This argument works if F for any finite field http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbf.gif_q, any global field (e.g., http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif), and any non-Archimedean local field (e.g. http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif_p). (In the latter two cases, F is the field of fractions of a domain with a non-zero prime ideal, so we can use Eisenstein's criterion to show it has irreducible polynomials of any degree.)

But it does not work for, say, algebraically closed fields like http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbc.gif, since by definition there are no irreducible polynomials of degree > 1. Maybe we can show that http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif restricts to a map from M_k(F') http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif M_e(F'), where F' is the prime subfield of F? Maybe a different approach altogether is called for. But I think the characteristic polynomial of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(A) should be a power of that of A (they both have the same irreducible factors, but a priori the exponents might be different).