1) Is (Z)^m is ring homomorphic with (Z)^n? Z=integer,n not equalto m?
2)what look like of finite groups of C*,C*=non zero complex number?
2006-06-30 23:47:43 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
If you are asking if there is a ring homomorphism from Z^m to Z^n, then yes, just send everything to (zero)^n.
If you are asking if they are isomorphic, then no.
I don't understand what you mean by the second part, try rewording it a little better.
2006-06-30 23:55:26 · answer #1 · answered by Eulercrosser 4 · 0⤊ 0⤋
1) Some books say that a ring homomorphism must send the identity element to the identity element. If so, the zero map doesn't work. However you can always, for example, project onto the first coordinate, and then apply the "diagonal" map a -> (a,a,a,a,...). The composition will be a homomorphism. If you are talking about isomorphisms, then there will be none for dimension reasons. (Either tensor with R or know about bases for free Z-modules.)
2) Finite subgroups, you say? This is a hard question, because I think you really need to use the classification of finite abelian groups! You should look it up, it's interesting. After doing so, next think about how many kth roots of unity there could possibly be for various integers k. This will cut down on the possible groups quite a bit. Enjoy.
2006-07-04 00:29:14 · answer #2 · answered by Steven S 3 · 0⤊ 0⤋
2) hint: any finite subgroup af the multiplicative group of a field is cyclic
2006-07-01 18:21:49 · answer #3 · answered by Theta40 7 · 0⤊ 0⤋
C* isomo. to C*/<> and ideal
2006-07-01 07:00:42 · answer #4 · answered by s topology 1 · 0⤊ 0⤋