From: Dave Rusin Date: Thu, 22 Oct 1998 21:02:47 -0500 (CDT) To: edgar@math.ohio-state.edu Subject: Re: linear algebra Newsgroups: sci.math.research In article you write: >(1) Let A be an algebra over a field F of dimension n. If B > is a subalgebra, must the dimension of B be a divisor of n? Appropriate for the class: take M to be a 3x3 matrix whose minimal polynomial is X^3. Then let A be the subalgebra of M_3(F) generated by I and M, and B the subalgebra generated by I and M^2. Their dimensions are 3 and 2 respectively. Or let A = F^3 and B = F^2 + (0). On the other hand, if A and B are simple algebras over F then yes, [B:F] | [A:F]. dave ============================================================================== From: Dave Rusin Date: Thu, 22 Oct 1998 21:23:49 -0500 (CDT) To: edgar@math.ohio-state.edu Subject: linear algebra: correction. I wrote in haste. >>(1) Let A be an algebra over a field F of dimension n. If B >> is a subalgebra, must the dimension of B be a divisor of n? [...] >Or let A = F^3 and B = F^2 + (0). That's not really right. Usually the definition of "algebra" includes having a unit element, so that F is contained in A and B; in that case we usually assume subalgebras also share the unit, which this example doesn't. But I wanted to give you an example which shows semi-simplicity is insufficient. Take A = M_3(F) and B = M_2(F) + M_1(F); 5 doesn't divide 9. So the most general result you are likely to get still requires A and B to be simple, I suppose. Of course I am assuming you mean "associative algebra". dave