Yes.
Can you help me. Here is my work.
(a) Prove that there is no ring homomorphism Z/5Z → Z.
(b) Prove that there is no ring homomorphism Z/5Z→ Z/7Z.
(a) Axiom of ring homomorphism are
f(1) = 1
f(a+b) = f(a) + f(b)
f(ab) = f(a)*f(b)
Consequences:
f(0) = 0
f(-r) = -f(r)
Homomorphism must therefore be
f(0) = 0
f(1) = 1
f(4 mod 5) = f(-1 mod 5) = -f(1) = -1
f(4) = f(2) + f(2) = 2*f(2) = -1
f(2) =
f(2) can't exist because there is no inverse with respect to multiplication in Z.
(b)Homomorphism must therefore be
f(0) = 0
f(1) = 1
f(4 mod 5) = f(-1 mod 5) = -f(1) = -1 mod 7 = 6 mod 7
f(4) = f(2) + f(2) = 2*f(2) = 6
f(2) = = 6 * 4 = 24 = 3 mod 7
f(2+3) = f(5) = f(0) = 0 = f(2) + f(3)
f(3) = -f(2) = -3 = 4 mod 7
Recapitulate
f(0) = 0
f(1) = 1
f(2) = 3
f(3) = 4
f(4) = 6
f(1+3) = f(4) = 6 not equal to f(1) + f(3) = 1+4 = 5
Contradiction and there is no homomorphism from Z/5Z→ Z/7Z
I don't really understand one thing. In b) we are assuming that f: Z/5Z -> Z/7Z is a homomorphism, so f(1) = 1 but also f(2) = f(1+1) = f(1)+f(1) = 1+1 = 2? And by the same argument f(a mod 5) = a mod7? I feel a bit bad about this because we have to show that NO function is a ring homom. and don't think it is okay to "assume" what it the specific rule of the mapping. I am a bit confused with all that as you can see...
But could we say that f(2)=2 and f(2)= f(-3) = -f(3) = -3 = 4, but this is a contradiction because 2 is not equal to 4 in Z/7Z?