Do groups of units change base nicely, assuming the fields are algebraically closed?

Question

Let $K$ be an algebraically closed field. Let $X$ be an irreducible affine algebraic variety over $K$. Let $L/K$ be a field extension, where $L$ is also algebraically closed. Suppose the group of units of the coordinate ring $\mathscr{O}(X)$ is trivial (i.e. equals $K^\times$). Is it true that the group of units of $\mathscr{O}(X \times_K L)$ is also trivial (i.e. equals $L^\times$)?

Answer 1

Yes. Let $u$ be a unit in $\mathcal O(X \times_K L) = \mathcal O(X) \otimes_K L$ and write $u= \sum_{i=0}^n a_i b_i$ and $u^{-1}=\sum_{i=0}^n a_i b_i'$ where $a_i \in \mathcal O(X)$ and $b_i \in L$. Without loss of generality, we may assume $a_0=1$ and the $a_i$ are $K$-linearly independent (since if one is a $K$-linear combination of others we can delete it and adjust the $b_i$).

Suppose $b_j \neq 0$ for some $j>0$. The ring $K[b_0,\dots,b_n,b_0',\dots, b_n'] \subseteq L$ is finitely generated and reduced. Hence by the Nullstelensatz there is a homomorphism $f$ from this ring to $K$ sending $b_j$ to something nonzero. We have $$ 1= (id \otimes f) ( (\sum_{i=0}^n a_i b_i) ( \sum_{i=0}^n a_ib_i') = (\sum_{i=0}^n a_i f(b_i)) (\sum_{i=0}^n a_i f(b_i')) $$ so that $(\sum_{i=0}^n a_i f(b_i)) $ is a unit of $\mathcal O ( X)$, hence lies in $K$, which since $f(b_j) \neq 0$ contradicts the linear independence of the $a_i$.

So $b_j=0$ for all $j>0$ and thus $u = b_0 \in L$.

Note that the only assumption on $L$ used in the proof is that $L$ is a reduced $K$-algebra (but it was already clear that if the statement holds for algebraically closed fields then it holds for reduced rings since every reduced ring embeds into a product ic algebraically closed fields).