Non-separable $\mathbb{A}^2$-form is trivial Suppose $A$ is a finitely generated $\mathbb{Q}$-algebra and $A \otimes_{\mathbb{Q}} \mathbb{R} \cong \mathbb{R}[X, Y]$. Then is $A \cong \mathbb{Q}[X, Y]$? Here $\mathbb{R}[X, Y]$ is a two variable polynomial ring over $\mathbb{R}$. 
I know this is true for separable algebraic extension, but I don't know for general case. Any help or suggestion is highly appreciated.
 A: Here by the "Lefschetz principle", I really mean a few of the "limit theorems" that are worked out more carefully in EGA IV, Sections 8 and 11.3.
Let $Q$ be a commutative unital ring (e.g., the field $\mathbb{Q}$).  Let $A$ be a commutative, unital $Q$-algebra.  For every directed system $S_\bullet = (S_i)_{i\in I}$ of commutative, unital $Q$-algebras $S_i$ with colimit $S = \varinjlim S_i$, there is an induced set map,
$$
h^A_{S^\bullet}:\varinjlim \text{Hom}_{Q-\text{alg}}(A,S_i) \to \text{Hom}_{Q-\text{alg}}(A,S).
$$
Proposition. The $Q$-algebra $A$ is finitely presented if and only if the set map $h^A_{S^\bullet}$ is a bijection for every  filtering directed system $S_\bullet$ of commutative, unital $Q$-algebras.
Proof. If $G$ is a finite set of generators of $A$, then $Q$-algebra morphisms from $A$ to a $Q$-algebra are uniquely determined by their values on $G$, from which injectivity of $h^A_{S^\bullet}$ follows.
Assume that $A$ is finitely presented, and let $P$ be a finite set of generators for the ideal of relations among $G$.  Then for every $Q$-algebra morphism from $A$ to $S$, there exists an $i\in I$ such that the images of the finitely many elements of $G$ lift to $S_i$.  For such a lift, the finitely many elements of $P$ map to elements of $S_i$ that further map to $0$ in $S$.  Since the system $S_\bullet$ is filtering, there is a larger $j\in J$ such that the images of these finitely many elements are already $0$ in $S_j$.  Thus, the $Q$-algebra morphism from $A$ to $S$ factors through a $Q$-algebra morphism from $A$ to $S_i$.
For the opposite direction, consider the filtering directed system of all $Q$-algebra morphisms from finitely presented $Q$-algebras to $A$.  The identity map from $A$ to itself factors through one of these $Q$-algebra morphisms if and only if $A$ is finitely presented. QED
Corollary.  Let $A$ and $B$ be finitely presented $Q$-algebras.  Let $(R_i)_{i\in I}$ be a filtering directed system of $Q$-algebras with colimit $R$.  Every $R$-algebra isomorphism from $A\otimes_Q R$ to $B\otimes_Q R$ is the base change of an $R_i$-algebra isomorphism from $A\otimes_Q R_i$ to $B\otimes_Q R_i$ for some $i\in I$.
Proof.  By the previous step, the two inverse isomorphism between $A\otimes_Q R$ and $B\otimes_Q R$ factor though $R_i$-algebra homomorphisms between $A\otimes_Q R_i$ and $B\otimes_Q R_i$.  Moreover, the compositions of these two morphisms and the identity morphisms become equal after base change from $R_i$ to $R$.  Thus, using the injectivity part of the proposition, they become equal after base change from $R_i$ to $R_j$ for some larger $j\in I$.  QED
Now, in your case, take $B$ equals $\mathbb{Q}[X,Y]$.  Take $(R_i)_{i\in I}$ to be the collection of all finitely generated $\mathbb{Q}$-subalgebras of $\mathbb{R}$.  This is a filtering directed system of rings whose colimit equals $\mathbb{R}$.  Thus, by the corollary, every $\mathbb{R}$-algebra isomorphism of $A\otimes_{\mathbb{Q}}\mathbb{R}$ and $\mathbb{R}[X,Y]$ is the base change of an $R_i$-algebra isomorphism $A\otimes_{\mathbb{Q}}R_i$ and $R_i[X,Y]$.
Since $R_i$ is an integral domain that is finitely generated over $\mathbb{Q}$, the quotient of $R_i$ by any maximal ideal is a field $K$ that is a finite extension of $\mathbb{Q}$ (by the Nullstellensatz).  Every finite field extension of $\mathbb{Q}$ is separable.  Thus, forming the base change of your isomorphism by the $\mathbb{Q}$-algebra homomorphism $R_i\to K$, there exists a finite, separable extension $K/\mathbb{Q}$ and a $K$-algebra isomorphism of $A\otimes_{\mathbb{Q}} K$ and $K[X,Y]$.
