# Isomorphic as affine schemes v.s. isomorphic as affine $k$-schemes

Let $$k$$ be an Algebraically closed field. Let $$A,B$$ be integral domains which are finitely generated as $$k$$-algebras.

If $$A$$ and $$B$$ are isomorphic as rings, then are $$A$$ and $$B$$ also isomorphic as $$k$$-algebras ?

• @KReiser: That question doesn't require the field to be algebraically closed or the algebras to be finitely generated (and its answer does not answer this question). Commented Aug 31, 2019 at 2:46
• @EricWofsey I see that the linked question does not require $k$ algebraically closed nor the ring to be an fg domain, but the example of $\sigma:k\to k$ which isn't a $k$-algebra automorphism should work, no? Or have I missed something? Commented Aug 31, 2019 at 3:01
• @KReiser: Ah, that's an example of a ring-isomorphism which is not a $k$-algebra isomorphism, but not a pair of $k$-algebras that are isomorphic as rings but not as $k$-algebras (since the $k$-algebras involved are isomorphic by a different map). Commented Aug 31, 2019 at 3:02

No. For instance, let $$A$$ be the coordinate ring of an affine open subset of an elliptic curve over $$k$$ whose $$j$$-invariant $$c$$ is not an element of the prime field. Then there is an automorphism $$\alpha$$ of $$k$$ such that $$\alpha(c)\neq c$$. Let $$B$$ be the same ring as $$A$$, but considered as a $$k$$-algebra by composing the $$k$$-algebra structure of $$A$$ with $$\alpha^{-1}$$. Then $$B$$ will be an affine open subset of an elliptic curve whose $$j$$-invariant is $$\alpha(c)$$, so it is not isomorphic to $$A$$ as a $$k$$-algebra.
More generally, if you take an affine $$k$$-algebra $$A$$ and modify its $$k$$-algebra structure by some automorphism of $$k$$ to get a new $$k$$-algebra $$B$$, then $$B$$ will typically not be isomorphic to $$A$$ as a $$k$$-algebra (though it will be in some of the most basic cases like polynomial rings, and proving it isn't in any particular example takes some work).
What is true is that $$A$$ and $$B$$ always must be related by an automorphism of $$k$$ in this way. Indeed, the copy of $$k$$ inside $$A$$ is determined by the ring structure of $$A$$, as the unique maximal subfield of $$A$$. To prove this, suppose $$f\in A$$ is not in $$k$$. Considering $$f$$ as a morphism $$\operatorname{Spec} A\to\mathbb{A}^1_k$$, it is nonconstant (since $$f\not\in k$$) and thus a dominant morphism so its image contains all but finitely many points of $$\mathbb{A}^1_k$$. In particular, its image contains some point that is algebraic over the prime field. Letting $$p$$ be the minimal polynomial of this point over the prime field, then $$p(f)$$ is nonzero but not invertible in $$A$$. Thus, $$f$$ is not contained in any subfield of $$A$$.
So, any ring isomorphism $$A\to B$$ must map the copy of $$k$$ inside $$A$$ to the copy of $$k$$ inside $$B$$. That is, it restricts to an automorphism of $$k$$, so that it is a $$k$$-algebra isomorphism if you modify the $$k$$-algebra structure of $$B$$ by this automorphism of $$k$$.