Fiber product of local artinian rings with a fixed residue field

Question

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.

Answer 1

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 $k$.