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Let $A,B$ be two $\Bbb Q$-algebras. Assume that $A \otimes_{\Bbb Q} \Bbb C \cong B \otimes_{\Bbb Q} \Bbb C$ as $\Bbb C$-algebras. Does it follows that $A \otimes_{\Bbb Q} \overline{\Bbb Q} \cong B \otimes_{\Bbb Q} \overline{\Bbb Q}$ as $\overline{\Bbb Q}$-algebras ?

Here $\overline{\Bbb Q}$ is an algebraic closure of $\Bbb Q$. (Clearly the converse holds, since $- \otimes_{\overline{\Bbb Q}} \Bbb C$ is a functor, so it preserves isomorphisms).

My feeling is that this is true if $A,B$ are finitely generated as $\Bbb Q$-algebras. We could even replace $\Bbb Q$ by any (perfect) field $k$ and considering two algebraically closed fields $K \subset K'$ containing $k$. We would have $A = k[X_1,...,X_r]/I,B=k[X_1,...,X_s]/J$ with a $K'$-algebras isomorphism $\phi : K'[X_1,...,X_r]/I^e \to K'[X_1,...,X_s]/J^e$ ($(\cdot)^e$ denotes the extension of ideals). Then I would like that $\phi$ is actually defined over a finite extension $L \supset k$, so in particular $L \subset K$ and $\phi$ is defined over $K$ and provides an isomorphism $A \otimes_k K \cong B \otimes_k K$ of $K$-algebras.

Looking at this question, I think that this is true if we replace algebras by modules.

This is wrong in larger generality. For instance, let $A = \Bbb Q[x,y] / (y^2-x^2) \cong \Bbb Q \times \Bbb Q$ and $B=\Bbb Q[x,y] / (y^2-2x^2)$. Then $A \otimes_{\Bbb Q} \Bbb Q(\sqrt 2) \cong B \otimes_{\Bbb Q} \Bbb Q(\sqrt 2) \cong \Bbb Q(\sqrt 2) \times \Bbb Q(\sqrt 2)$ as $\Bbb Q(\sqrt 2)$-algebras, but $A \not\cong B$ as $\Bbb Q$-algebras (not even as rings, since $B$ is a domain but not $A$). So we need to make base extensions by algebraically closed fields.

Alphonse
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  • Are you looking for an extended isomorphism induced by $A \otimes_{\mathbb{Q}} \mathbb{C} \cong B \otimes_{\mathbb{Q}} \mathbb{C}$? Or any isomorphism? – George Mouselli Dec 24 '17 at 11:17
  • Any isomorphism would be fine, I guess. – Alphonse Dec 24 '17 at 11:27
  • Related: https://math.stackexchange.com/questions/2403707/k-algebra-isomorphic-to-polynomial-ring-after-base-extension – user26857 Dec 24 '17 at 14:26
  • (Not completely related: https://math.stackexchange.com/questions/855287/algebraic-varieties-that-are-isomorphic-after-a-base-change?rq=1) – Alphonse Jan 06 '18 at 17:43
  • A more general question would be : if $f : A \otimes_{k} K' \to B \otimes_k K'$ is a morphism of $K'$-algebras, is there a morphism $g : A \otimes_k K \to B \otimes_k K$ of $K$-algebras such that $f = g \otimes_K id_{K'}$ ? This is true for instance if $A = k[t_1, ..., t_r, u_1^{\pm 1}, ..., u_s^{\pm 1}]$ and $B$ has an analogue form. – Alphonse Jan 08 '18 at 19:03
  • What happens if we replace $\Bbb C$ by $\overline{\Bbb Q}(\pi)$ ? – Alphonse Jan 08 '18 at 20:28
  • The most general question is : what are conditions on a ring $R$ and $R$-algebras $S,S'$ such that for any $R$-algebras $A,B$ we have $$A \otimes_R S \cong B \otimes_R S \text{ as $S$-algebras } \implies A \otimes_R S' \cong B \otimes_R S' \text{ as $S'$-algebras. }$$ – Alphonse Feb 02 '18 at 13:58
  • Possibly related: https://math.stackexchange.com/questions/2850023 – Watson Jul 15 '18 at 11:35
  • I also found exercise 2-5 in Algebraic Geometry J.S. Milne - v5.22 (2012). – Alphonse Nov 17 '18 at 16:47
  • More general : given varieties $X,Y$ and a morphism $T \to S$, when is $Hom_S(X,Y) \to Hom_T(X_T, Y_T)$ surjective (typically $S = Spec(\bar{\Bbb Q}), T = Spec(\Bbb C)$) ? – Alphonse Dec 04 '18 at 14:17
  • In general, if $K/F$ is Galois (which is not the case here), twisted forms of an $F$-algebra $A$ over $K$ form a pointed set which corresponds bijectively to $$H^1(\mathrm{Gal}(K/F) ; \mathrm{Aut}_K(A \otimes_F K)),$$ see Gille-Szamuely 2.3.3 or Berhuy, III.8.15. – Watson Dec 27 '18 at 19:32

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