Isomorphism

by Icebreaker
Tags: isomorphism
 P: n/a "Let R be the ring Zp[x] of polynomials with coefficients in the finite field Zp, and let f:R->S be a surjective homomorphism from R to a ring S. Show that S is either isomorphic to R, or is a finite ring." If S is isomorphic to R, then we're done. If S is not isomorphic to R, then by hypothesis, f is not injective. That is, ker(f) contains more than the zero element of R. ker(f) is an ideal of R, therefore there exists a surjective homomorphism p:R->R/ker(f), where p(x) = x + ker(f). My question is, if R is homomorphic to S and R is homomorphic to T, what is the relation between S and T? Are they the same? Isomorphic? In the question, R and S are given. In my answer, S is constructed.
 HW Helper Sci Advisor P: 2,588 Did you finishing solving the problem? If Ker(f) has a non-zero element, then it has some polynomial of degree k. You know then that elements of R/Ker(f) have degree no more than k, and since there are only a finite number of possible coefficients for a given term in a polynomial, you've shown that R/Ker(f) is finite, hence so is S. Certainly, S and T need not be the same or isomorphic. For every polynomial p in R, there is the ideal generated by p, (p), and there is always a surjective homomorphism from R to R/(p). But it's not true that for any polynomials p, q, that R/(p) is isomorphic to R/(q).
 P: n/a I did finish solving the problem. I was just wondering whether the answer to my question was true. Thanks anyway.
HW Helper