Definition of kernels of ring and field homomorphisms Let $\varphi:A\to B$ a ring morphism. Why do we define

$$\ker\varphi=\{x\in A\mid \varphi(x)=0\}$$

and not $$\ker\varphi=\{x\in A\mid \varphi(x)=1\} ?$$
Maybe it's a consequence of the fact that the element of $A$ are not necessarily invertible. Therefore, I ask the same question for field homomorphisms.
 A: For one, given any ring homomorphism $\varphi: A \to B$, $$\ker \varphi := \{x \in A \,|\, \varphi(x) = 0_B\}$$ is an ideal, which is exactly enables us to invoke isomorphism theorems for rings, and conclude, for example, that $\varphi$ descends to an isomorphism $$\tilde{\varphi} : A / \ker \varphi \stackrel{\cong}{\to} \operatorname{im} \varphi.$$
Now, since $\varphi(0_A) = 0_B$, (for $A$ not the trivial ring, so that $0_B \neq 1_A$) the set $$\varphi^{-1}(1_B) = \{x \in A \,|\, \varphi(x) = 1_B\}$$ doesn't contain $0_A$ and hence isn't even a subring, let alone an ideal.
On the other hand, one can consider for any ring $X$ (with identity) the set $X'$ of elements in $X$ which have a multiplicative inverse, which is a group under (the restriction of) ring multiplication. Then, for any ring homomorphism $\varphi: A \to B$ between rings with identity for which $\varphi(1_A) = 1_B$, checking definitions shows that $\varphi|_{A'}$, regarded as a map $A' \to B'$, is a group homomorphism, and its kernel is $$\ker \varphi|_{A'} := \{ x \in A' \,|\, \varphi(x) = 1_B \} = \varphi^{-1}(1_B) \cap A', $$ but this notion is distinct from the kernel of the original ring homomorphism.

As for fields, any field homomorphism $\varphi: A \to B$ is injective: Since $\ker \varphi := \{x \in A : \varphi(x) = 0_B\}$ is an ideal of $A$ (regarded as ring), it must be $\{0_A\}$ or $A$. On the other hand, by definition $\varphi(1_A) = 1_B \neq 0_B$, so the kernel cannot be all of $A$, and hence it is $\{0_A\}$.
A: Well first off, not all rings are required to have a multiplicative identity.
And even if we take the definition of rings requiring a multiplicative identity, a good reason to do this is because makes our kernel an ideal, just like how in groups the kernel is always a normal subgroup. This allows us to get the first isomorphism theorem of rings just like for groups: If $\varphi:R\to S$ is a ring homomorphism, then
$$R/\ker\varphi\cong\text{im}(\varphi)$$
As for fields, there is no reason to change the definition because a field is just a special case of a ring.
A: In universal algebra, which generalizes our study of structures like groups and rings to other sorts of structures with operations, there is a broader notion of kernel that can help illustrate the kinds of issues you're asking about. In universal algebra, the kernel of a homomorphism isn't a subset of the domain at all; it's the equivalence relation on the domain given by saying two elements are equivalent if they map to the same place. There is an analogous definition of an ideal/normal subgroup also, as an equivalence relation that plays nicely with the algebraic structure (called a congruence relation). All the expected results carry over to this more general framework: the congruence relations are exactly the relations that can be kernels of homomorphisms, and generalized versions of the isomorphism theorems hold.
The above general situation relates to the situation for groups and rings because, in those structures, you can recover a congruence relation just by looking at the equivalence class of 0. That equivalence class is what we usually call the kernel of a homomorphism in group/ring theory. However, 0 isn't special in that regard: in groups and rings, we can recover the congruence relation by looking and any one equivalence class. In that respect, we could equally well have developed ring theory by regarding the kernel of a ring homomorphism to be the inverse image of 1. There is even an analogous term for "ideal" if we want to do that: the subsets of a ring which might serve as inverse images of 1 under a homomorphism are called "filters" of the ring.
