# Number of Homomorphisms

by wattsup03
Tags: homomorphisms, number
 P: 6 Hi, I am trying to calculate the number of homomorphisms from one field to another: a) F2 ---> F3 b) Q[X]/(X7 - 3) ---> Q[X]/(X8 + 4X5 - 6X + 2) c) F7 [X] / (X2 + X - 1) ---> F7[X] / (X2 + 1) d) Q( 21/4 ) ---> C Attempt at a solution a) I'm pretty sure there are no homomorphisms between F2 and F3 because if there was a homomorphism f, then f(1+1) = f(0) which does not equal f(1) +f(1) = 2 b) I think I need to see how many roots there are of X7 - 3 in Q[X]/(X8 + 4X5 - 6X + 2) since there is a bijection between that and the number of homomorphisms? c) Similarly here d) For this one I think the answer is four (I'm really not sure) because there is a bijection between K-Homomorphisms and the roots of the minimal polynomial of 21/4 in C, which would be 4. And over fields K-homomorphisms are ring homomorphisms? In all honesty I am pretty stuck, and if anyone could give me any advice that would be fantastic. Thanks in advance.
 P: 1 a) I agree b) I am thinking that a homomorphism f will be uniquely defined by f(x) Prove it. But 0 = f(0) = f(x^7 - 3) = f(x)^7 - f(3) = f(x)^7 - 3 and how many solutions does this have? But I suppose we have to bear in mind that this zero is k(x^8 + 4x^5 - 6x + 2) for any k c) similar d) Using this same method would seem to imply the identity is the only homomorphism, but there are 4 K-Homomorphisms as you said, so we are in trouble here This is still a work in progress for me but I hope it helps
 P: 6 Thanks meandonlyme , for b) could I put x7 - 3 into Q[X]/(X8 + 4X5 - 6X + 2) (which is x7 - 3 still) and then calculate the number of roots in there: 1 since it is over Q. So there is one root and hence one homomorphism. Similarly for c) and d)

 Related Discussions Linear & Abstract Algebra 9 Calculus & Beyond Homework 1 Linear & Abstract Algebra 3 Linear & Abstract Algebra 1 Linear & Abstract Algebra 0