Let $K$ be a field, and let $L/K$ be an **algebraically closed** field extension (i.e. the only elements of $L$ that are algebraic over $K$ are already in $K$). Let $R$ be a $K$-algebra that is an integral domain. Does it follow that $R \otimes\_K L$ is an integral domain? I'm particularly interested in the case where $R$ is a finitely generated $K$-algebra.


My question is closely related to [this question](https://mathoverflow.net/q/203801/19045), where Will Sawin gives a yes answer when $L$ is purely transcendental over $K$. Also (at least when $R$ is finitely generated as a $K$-algebra), the answer seems to be yes when $K$ is algebraically closed, according to a recent preprint. Furthermore, the answer is typically 'no' if $L$ has algebraic elements over $K$, even when $R$ itself is a field. For instance, let $f$ be an irreducible polynomial in $K[t]$ that has a root in $L$, and let $R := K[t]/(f)$.



Is a base-change of an integral domain by an extension of its base field without algebraic elements still a domain?
No. Let $K = \mathbf{F}\_p(s,t)$, $A = K[x,y]/(sx^p + t y^p - 1)$. One checks $A$ is a domain, even Dedekind, so we can define $L = {\rm{Frac}}(A)$. By exploring residue fields of $A$ at maximal ideals and using that $A$ is Dedekind one shows with some thought that $K$ is algebraically closed in $L$. But for the field $R = K[s^{1/p}, t^{1/p}]$ of degree $p^2$ over $K$ we have $R \otimes\_K L = L[x,y]/(h^p)$ for $h = s^{1/p}x + t^{1/p}y - 1$, so this is non-reduced. In this case $A$ is a domain of finite type over $K$ that is a geometrically irreducible $K$-algebra that is geometrically everywhere non-reduced. This example is due to MacLane, but I don't know a literature reference for it.


