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Let $k$ be a field, not necessarily algebraically closed, and let $A$ be a $k$-algebra of finite type. Recall that $\operatorname{Spec} (A)$ is geometrically integral (over $k$) if $\operatorname{Spec} (A) \times_{\operatorname{Spec} (k)} \operatorname{Spec} k'$ is integral for every field extension $k'$ of $k$. Translated back to algebra, that is the same as requiring that $A \otimes_k k'$ be an integral domain for every field extension $k'$ of $k$.

In particular this is so for $k' = k$, so henceforth we assume $A$ is an integral domain. Let $K = \operatorname{Frac} (A)$, the fraction field of $A$. Then $K \otimes_k k'$ is also an integral domain for every field extension $k'$ of $k$, so in particular $K$ is a separable (but possibly transcendental) field extension of $k$. Also, $A \otimes_k k'$ is an integral domain for every algebraic extension $k'$ of $k$, so $k$ is algebraically closed inside $A$.

Question. Now, suppose $A$ is a $k$-algebra of finite type with the following properties:

  • $A$ is an integral domain.
  • $k$ is algebraically closed inside $A$.
  • $\operatorname{Frac} (A)$ is a separable field extension of $k$.

Does it follow that $A \otimes_k k'$ is an integral domain for every field extension $k'$ of $k$?

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1 Answer 1

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No. For instance, let $k=\mathbb{Q}$ and $A=\mathbb{Q}[x,y]/(x^2+y^2)$. Since $x^2+y^2$ is irreducible over $\mathbb{Q}$, $A$ is a domain, and the separability condition is trivial since the characteristic is $0$. To see that $\mathbb{Q}$ is algebraically closed in $A$, note that $\operatorname{Frac}(A)$ can be identified with $\mathbb{Q}(i)(x)$ with $i$ mapping to $y/x$ and then $A$ is the subring $\mathbb{Q}[x,ix]$ which is just the subring of $\mathbb{Q}(i)[x]$ consisting of polynomials with constant term in $\mathbb{Q}$. No nonconstant polynomial is algebraic over $\mathbb{Q}$ so $\mathbb{Q}$ is algebraically closed in $A$.

However, $x^2+y^2$ factors as $(x+iy)(x-iy)$ over $\mathbb{Q}(i)$ so $A\otimes \mathbb{Q}(i)$ is not a domain.

(More generally, you need $k$ to be algebraically closed not just in $A$ but in $\operatorname{Frac}(A)$. This suffices, since it implies $\operatorname{Frac}(A)\otimes_k k'$ is a domain and thus $A\otimes_k k'$ is a domain since it is a subring of $\operatorname{Frac}(A)\otimes_k k'$.)

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