Number of Homomorphisms

[wattsup03]

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( 2^1/4 ) ---> 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 2^1/4 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.

 

[meandonlyme]

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

 

[wattsup03]

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)