Question about kernel and homomorphism I was wondering is there any reason we take the identity e` for the kernel for ring homomorphism to be the additive identity instead of the multiplicative one?
 A: Here's a line of reasoning that leads to this definition of kernel: among groups, one has the isomorphism $G/\text{ker}\varphi\cong \text{im}\varphi$, where $\varphi$ is any homomorphism out of $G$. This is a constantly useful theorem, so we'd like something equivalent for rings. And it's the usual definition of kernel as the inverse image of zero that gets us this theorem. As commenters have mentioned, this rests on the fact that the kernel is an ideal, so that $R/\text{ker}\varphi$ is at least a ring; in contrast the inverse image of the multiplicative identity is just some multiplicative submonoid of $R$, so that the best kind of isomorphism theorem you could get out such a "kernel" would be an isomorphism of monoids. And that's not very useful.
A: One reason is that the idea kernels in general works best with groups.
For a group homomorphism $f$ the kernel tells us something interesting about what happens to elements under $f$, namely that $f(a)=f(b)$ exactly when $ab^{-1}$ is in the kernel of $f$. This relation depends on taking the inverse of either $a$ or $b$, so it doesn't work well if we try to use it with something else than a group, where elements may not have inverses in general.
Kernels work for rings, vector spaces and so forth because these structures can all be described as an abelian group with some extra structure. In the case of a ring, the additive group is the only group structure in sight, so if we want to speak of kernels, it is the identity element of that group we want the preimage of.
If we're speaking of fields we almost have a multiplicative group -- but not quite; 0 is not a member of the multiplicative group. And speaking of kernels is not very relevant in the case of fields anyway, because a field homomorphism is necessarily injective, so having an object that tells us when $f(a)=f(b)$ is not really useful.
