**Proof.**
The ring $A_1 \times _ A A_2$ is a $\Lambda $-algebra via the map $\Lambda \to A_1 \times _ A A_2$ induced by the maps $\Lambda \to A_1$ and $\Lambda \to A_2$. It is a local ring with unique maximal ideal

\[ \mathfrak m_{A_1} \times _{\mathfrak m_ A} \mathfrak m_{A_2} = \mathop{\mathrm{Ker}}(A_1 \times _ A A_2 \longrightarrow k) \]

A ring is Artinian if and only if it has finite length as a module over itself, see Algebra, Lemma 10.53.6. Since $A_1$ and $A_2$ are Artinian, Lemma 89.3.4 implies $\text{length}_\Lambda (A_1)$ and $\text{length}_\Lambda (A_2)$, and hence $\text{length}_\Lambda (A_1 \times A_2)$, are all finite. As $A_1 \times _ A A_2 \subset A_1 \times A_2$ is a $\Lambda $-submodule, this implies $\text{length}_{A_1 \times _ A A_2}(A_1 \times _ A A_2) \leq \text{length}_\Lambda (A_1 \times _ A A_2)$ is finite. So $A_1 \times _ A A_2$ is Artinian. Thus the only thing that is keeping $A_1 \times _ A A_2$ from being an object of $\mathcal{C}_\Lambda $ is the possibility that its residue field maps to a proper subfield of $k$ via the map $A_1 \times _ A A_2 \to A \to A/\mathfrak m_ A = k$ above.

Proof of (1). If $f_2$ is surjective, then the projection $A_1 \times _ A A_2 \to A_1$ is surjective. Hence the composition $A_1 \times _ A A_2 \to A_1 \to A_1/\mathfrak m_{A_1} = k$ is surjective and we conclude that $A_1 \times _ A A_2$ is an object of $\mathcal{C}_\Lambda $.

Proof of (2). If $f_2$ is a small extension then $A_2 \to A$ and $A_1 \times _ A A_2 \to A_1$ are both surjective with the same kernel. Hence the kernel of $A_1 \times _ A A_2 \to A_1$ is a $1$-dimensional $k$-vector space and we see that $A_1 \times _ A A_2 \to A_1$ is a small extension.

Proof of (3). Choose $\overline{x} \in k$ such that $k = k'(\overline{x})$ (see Fields, Lemma 9.19.1). Let $P'(T) \in k'[T]$ be the minimal polynomial of $\overline{x}$ over $k'$. Since $k/k'$ is separable we see that $\text{d}P/\text{d}T(\overline{x}) \not= 0$. Choose a monic $P \in \Lambda [T]$ which maps to $P'$ under the surjective map $\Lambda [T] \to k'[T]$. Because $A, A_1, A_2$ are henselian, see Algebra, Lemma 10.153.10, we can find $x, x_1, x_2 \in A, A_1, A_2$ with $P(x) = 0, P(x_1) = 0, P(x_2) = 0$ and such that the image of $x, x_1, x_2$ in $k$ is $\overline{x}$. Then $(x_1, x_2) \in A_1 \times _ A A_2$ because $x_1, x_2$ map to $x \in A$ by uniqueness, see Algebra, Lemma 10.153.2. Hence the residue field of $A_1 \times _ A A_2$ contains a generator of $k$ over $k'$ and we win.
$\square$

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