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, 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 $R := K[t]/(f)$.

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, 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 $R := K[t]/(f)$.