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Non-seperable $\mathbb{A}^2$-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 seperable field extension, but I don't know for general case. Any help or suggestion is highly appreciated.