
#1
Nov211, 05:02 AM

P: 6

Hi,
I am trying to calculate the number of homomorphisms from one field to another: a) F_{2} > F_{3} b) Q[X]/(X^{7}  3) > Q[X]/(X^{8} + 4X^{5}  6X + 2) c) F_{7} [X] / (X^{2} + X  1) > F_{7}[X] / (X^{2} + 1) d) Q( 2^{1/4} ) > C Attempt at a solution a) I'm pretty sure there are no homomorphisms between F_{2} and F_{3} 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 X^{7}  3 in Q[X]/(X^{8} + 4X^{5}  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 KHomomorphisms and the roots of the minimal polynomial of 2^{1/4} in C, which would be 4. And over fields Khomomorphisms 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. 



#2
Nov211, 08:03 AM

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 KHomomorphisms as you said, so we are in trouble here This is still a work in progress for me but I hope it helps 



#3
Nov211, 09:10 AM

P: 6

Thanks meandonlyme ,
for b) could I put x^{7}  3 into Q[X]/(X^{8} + 4X^{5}  6X + 2) (which is x^{7}  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) 


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