Lemma 66.15.1. Let $A$ be a ring. Let $k$ be a field. Let $\mathfrak p_ n$, $n \geq 1$ be a sequence of pairwise distinct primes of $A$. Moreover, for each $n$ let $k \to \kappa (\mathfrak p_ n)$ be an embedding. Then the closure of the image of

\[ \coprod \nolimits _{n \not= m} \mathop{\mathrm{Spec}}(\kappa (\mathfrak p_ n) \otimes _ k \kappa (\mathfrak p_ m)) \longrightarrow \mathop{\mathrm{Spec}}(A \otimes A) \]

meets the diagonal.

**Proof.**
Set $k_ n = \kappa (\mathfrak p_ n)$. We may assume that $A = \prod k_ n$. Denote $x_ n = \mathop{\mathrm{Spec}}(k_ n)$ the open and closed point corresponding to $A \to k_ n$. Then $\mathop{\mathrm{Spec}}(A) = Z \amalg \{ x_ n\} $ where $Z$ is a nonempty closed subset. Namely, $Z = V(e_ n; n \geq 1)$ where $e_ n$ is the idempotent of $A$ corresponding to the factor $k_ n$ and $Z$ is nonempty as the ideal generated by the $e_ n$ is not equal to $A$. We will show that the closure of the image contains $\Delta (Z)$. The kernel of the map

\[ (\prod k_ n) \otimes _ k (\prod k_ m) \longrightarrow \prod \nolimits _{n \not= m} k_ n \otimes _ k k_ m \]

is the ideal generated by $e_ n \otimes e_ n$, $n \geq 1$. Hence the closure of the image of the map on spectra is $V(e_ n \otimes e_ n; n \geq 1)$ whose intersection with $\Delta (\mathop{\mathrm{Spec}}(A))$ is $\Delta (Z)$. Thus it suffices to show that

\[ \coprod \nolimits _{n \not= m} \mathop{\mathrm{Spec}}(k_ n \otimes _ k k_ m) \longrightarrow \mathop{\mathrm{Spec}}(\prod \nolimits _{n \not= m} k_ n \otimes _ k k_ m) \]

has dense image. This follows as the family of ring maps $\prod _{n \not= m} k_ n \otimes _ k k_ m \to k_ n \otimes _ k k_ m$ is jointly injective.
$\square$

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