Recent content by jakelyon

You mean that I map all even to 0 and all odd to 1?

I have seen that there exists two group homomorphisms from Z to Z/2Z. However, I cannot seem to understand this. I mean I know that there exists a trivial grp hom. which sends all of Z to 0 in Z/2Z. But I cannot think of anymore. Any help?

Dear Landau, Thanks for clearing things up for me. However, is two the order of Hom(Z_2, Z_2) or are there more?

I am thinking of group homomorphisms, so I know that it must map identities to identities. But I didn't think that it could only map identities to identities, thus not ruling out the second homomorphism. But then again, I am not 100% sure this is correct.

Can anyone tell me how many endomorphisms there are for Z/2Z? I think it is two: 0 --> 0 and 1 --> 1 0 --> 0 and 1 --> 0

Thanks for replying daveyinaz, but I am not sure I am following. However, I been doing some reading: (2x) is the ideal consisting of all linear combinations of 2x (with integer coefficients). Now, by moding Z[x] out by (2x) it is "like" sending x to 0. So, if I am correct, then Z[x]/(2x) =...

Can anyone explain, in detail, why/why not Z[X]/(2x) is isomorphic to Z/2Z? I know that every element in Z[x] can be written as a_0 + a_1 x + a_2 x^2 + ... with a_i in Z and only finitely many a_i's are nonzero. Now, does (2x) = (2, 2x, 2x^2,...)? Also, the quotient is "like" taking 2x=0, or...