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The kernel is defined as the set that maps all elements to zero.

$$\text{ker}\ \alpha = \alpha^{-1}(\mathbf{0})=\{ \mathbf{u} \in U \ : \alpha \mathbf{u}=\mathbf{0}\}$$

Suppose that $\alpha$ is homomorphism between two rings, then we have not one binary operation but two; with two identities: One with respect to addition, the other one with respect to multiplication. But we could also define another kernel in an analogous way (for the other identity):

$$\text{ker}^{'}\ \alpha = \alpha^{-1}(\mathbf{1})=\{ \mathbf{u} \in U \ : \alpha \mathbf{u}=\mathbf{1}\}$$

Do any useful properties arise from defining the fibre over the multiplicative identity?

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Sure, you can forget the addition of your ring and get a monoid $(R, \times, 1)$, something like a group but where elements aren't necessarily invertible.

A morphism of rings $f : R \to S$ is in particular a morphism of monoids (i.e. $f(ab) = f(a) f(b)$), and what you're considering is the kernel of that morphism of monoids. So it has all the properties that kernels of monoid morphisms have.

I don't really think you can say anything more specific than that, it's a notion that plays pretty poorly with the addition of $R$ ($1+1$ isn't $1$ very often, for example), so the ring structure is not very relevant, I believe. Compare that with the usual kernel, which is the kernel of the group morphism $(R,+,0) \to (S,+,0)$; since $0 \times x = 0$ for all $x$, the multiplication does play interestingly with the kernel.

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