# Algebraic characterisation of geometrically integral affine schemes of finite type over a field

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 $$k$$?

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 $$\operatorname{Frac}(A)\otimes_k k'$$.)