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$\def\bbA{\mathbb{A}} \def\spec{\operatorname{Spec}} \def\sO{\mathcal{O}} \def\sA{\mathcal{A}}$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.

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    $\begingroup$ 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. $\endgroup$
    – KReiser
    Commented Jul 31, 2023 at 20:38
  • $\begingroup$ @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. $\endgroup$ Commented Jul 31, 2023 at 22:02
  • $\begingroup$ 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!). $\endgroup$ Commented Jul 31, 2023 at 23:00
  • $\begingroup$ @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 ^^ $\endgroup$ Commented Aug 1, 2023 at 7:21

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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 $\mathcal{A}(Y)$.

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  • $\begingroup$ 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 $I(Y)=(y(1+x^2)-1)$? $\endgroup$ Commented Aug 1, 2023 at 7:29
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    $\begingroup$ $Y$ 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 $y(1+x^2)-1$. $\endgroup$ Commented Aug 1, 2023 at 13:11

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