# Fiber product of local artinian rings with a fixed residue field

Let $$k$$ be a finite field and suppose $$A,B,C$$ are Artinian local rings with residue field $$k$$. Suppose we have local homomorphisms $$f \colon A \to C, g \colon B \to C$$ which induce the identity on residue fields. Apparently the fiber product $$A \times_C B$$ is supposed to again be an Artinian local ring with residue field $$k$$, but I'm not sure why the residue field of the fiber product is also $$k$$.

Letting $$m = \{(a,b) \in A \times_C B : f(a) \in m_C\}$$ denote the ideal in $$A \times_C B$$, we see that projection onto either coordinate and then reducing gives a map $$A \times_C B \to C/m_C \cong k$$ with kernel $$m$$, thus $$(A \times_C B)/m$$ is a field since $$k$$ is finite. Moreover, since $$f$$ and $$g$$ induce the identity on residue fields, any element of $$A \times_C B$$ outside of $$m$$ is a unit, thus $$A \times_C B$$ is local. But I'm not sure why the map $$(A \times_C B)/m \to C/m_C$$ must be surjective.

Notice that an Artinian local ring $$R$$ with residue field $$k$$ is necessarily augmented: the ring map $$R \to k$$ has a section $$k \to R$$ which is also a ring map (because the maximal ideal is nilpotent). In particular, an Artinian local ring with residue field $$k$$ is an augmented $$k$$-algebra, and a local ring map which is the identity on residue fields is the same thing as an augmented $$k$$-algebra morphism.

Now the surjection is easy to see: if $$\lambda \in k$$, then it's clear that $$(\lambda,\lambda) \in A \times_C B$$ is sent to $$\lambda$$ under the natural projection to $$C/\mathfrak{m}_C$$. Note: nowhere was it necessary to use that $$k$$ was finite - this is true for any field $$k$$.