# Isomorphic affine classical varieties over a non-algebraically closed field. Does the isomorphism preserve polynomials?

$$\def\bbA{\mathbb{A}} \def\spec{\operatorname{Spec}} \def\sO{\mathcal{O}} \def\sA{\mathcal{A}}$$Let $$k$$ be a non-algebraically closed field. Let $$X=V(I)\subset\bbA^n_k$$ be a classical affine variety (where $$I\subset k[x_1,\dots,x_n]$$ is an ideal), which is naturally a locally ringed space over $$\spec k$$ when equipped with the sheaf of regular functions. Denote $$\sA(X)=k[x_1,\dots,x_n]/I(X)$$ to the $$k$$-algebra of polynomial functions on $$X$$. Then we have a containment $$\tag{1}\label{1} \sA(X)\subset\Gamma(X,\sO_X)$$ which in general is not an equality (for instance, consider $$k=\mathbb{R}$$, $$X$$ the real affine line and $$\frac{1}{1+x^2}\in\Gamma(X,\sO_X)$$). Suppose now that we have another classical affine variety $$Y=V(J)\subset\bbA^m_k$$ and an isomorphism $$\varphi:X\to Y$$ of locally ringed spaces over $$\spec k$$. We then get an induced isomorphism $$\varphi^*:\Gamma(Y,\sO_Y)\to\Gamma(X,\sO_X)$$ of $$k$$-algebras, given by precomposition by $$\varphi$$.

I have two questions:

1. Does it hold $$\varphi^*(\sA(Y))\subset\sA(X)$$?

2. If 1 is false in general, can we always find some isomorphism $$\psi:X\to Y$$ such that $$\psi^*(\sA(Y))\subset\sA(X)$$?

(Note that if $$k$$ were algebraically closed, then 1 holds, for \eqref{1} is an equality on this case.)

I am trying to produce a counterexample for 1 given by some automorphism of the real affine. I am thinking on a map of the form \begin{align*} \rho:\mathbb{R}&\to\mathbb{R}\\ a&\mapsto\frac{f(a)}{1+a^2}, \end{align*} where $$f(x)\in\mathbb{R}[x]$$. However, I don't know if we can find $$f$$ so that $$\rho$$ is bijective and $$\rho^{-1}$$ is given by a quotient of polynomials.

Put in other words, this post is asking: is the $$k$$-algebra of polynomials functions $$\sA(X)\subset\Gamma(X,\sO_X)$$ an invariant of $$X$$ as a classical affine algebraic variety? (I guess this question is relevant in the field of real algebraic geometry, for instance.)

EDIT: E. Wofsey's counterexample implies that the answer is no.

• Why would you want to do this? Doing algebraic geometry with "classical" varieties over non-algebraically-closed fields was something we largely stopped doing for a reason. Commented Jul 31, 2023 at 20:38
• @KReiser Short answer: for fun, what else? xD. Short and more serious answer: I said that such a question might be interesting in the realm of real algebraic geometry. Long answer: After realizing one can construct $F$ here for the schematic varieties with dense rational points, now I am trying to show fully faithfulness of $F$. I can show it in the affine case. In the general case, one takes an classical affine cover and has to glue the schematized affine pieces. To make the gluing possible, I think reduced the problem to what I asked here. Commented Jul 31, 2023 at 22:02
• Do you assume that $I=I(X)$ as well? (So, $I$ is not just an ideal that cuts out $X$ but is the largest such ideal?) Without that, the natural map $\mathcal{A}(X)\to\Gamma(X,\mathcal{O}_X)$ need not be injective (for instance, $X$ could be empty!). Commented Jul 31, 2023 at 23:00
• @EricWofsey Yes, when I write $\mathcal{A}(X):=k[x_1,\dots,x_n]/I(X)$, with “$I(X)$” I mean the ideal of polynomials that valuate identically zero in all of $X$. Thank you very much for your answer, by the way ^^ Commented Aug 1, 2023 at 7:21

Consider, for instance, $$X=\mathbb{A}^1$$ and $$Y=V(y(1+x^2)-1)\subset\mathbb{A}^2$$ over $$\mathbb{R}$$. Then there is an isomorphism $$\varphi:X\to Y$$ given by $$\varphi(x)=(x,(1+x^2)^{-1})$$ whose inverse is just the first projection. This does not map $$y\in\mathcal{A}(Y)$$ to an element of $$\mathcal{A}(X)$$, and indeed $$\mathcal{A}(Y)=\mathbb{R}[x,(1+x^2)^{-1}]$$ does not embed into $$\mathbb{R}[x]$$ at all (since there are no nonconstant units in $$\mathbb{R}[x]$$).
More generally, given any $$X$$ and elements $$f_1,\dots,f_n\in\Gamma(X,\mathcal{O}_X)$$, the graph of $$(f_1,\dots,f_n):X\to\mathbb{A}^n$$ is a variety $$Y$$ that is isomorphic to $$X$$ but which has $$f_1,\dots,f_n$$ in $$\mathcal{A}(Y)$$.
• In this situation, I know how to show that $I(X)=(0)$ (so that $\mathcal{A}(X)=\mathbb{R}[x]$), but how does one show $I(Y)=(y(1+x^2)-1)$? Commented Aug 1, 2023 at 7:29
• $Y$ is dense in the closed subscheme of $\operatorname{Spec}\mathbb{R}[x,y]$ defined by $y(1+x^2)-1$ (since $Y$ is infinite and that subscheme is an irreducible curve). So, any polynomial that vanishes on $Y$ must vanish on that whole subscheme and thus be in the ideal generated by $y(1+x^2)-1$. Commented Aug 1, 2023 at 13:11