0
$\begingroup$

Let $\phi : X \rightarrow Y$ be a morphism of affine varieties over $\mathbb{R}$. Let $\phi^*$ be the pullback on coordinate rings $\phi^* : O_Y \rightarrow O_X$. I am trying to understand the relation between this and the scheme theoretic version. Let $\operatorname{Specmax}(X)$ be the set of maximal ideals of $O_X$. This space is bijective with, to my understanding, the points of $X$. Similarly let $\operatorname{Specmax}(Y)$ be the set of maximal ideals of $O_Y$. Is $(\phi^*)^{-1}$ a map from $\operatorname{Specmax}(X)$ to $\operatorname{Specmax}(Y)$?

This question is related to the one in a previous post previous question about affine varieties over $\mathbb{R}$.

$\endgroup$
1
  • $\begingroup$ Some typographical suggestions: you can use \operatorname{Specmax} to format $\operatorname{Specmax}$ and have it come out looking nice (I've made this upgrade for you this time). The notation $(\phi^*)^{-1}$ is kind of messy - if you have a choice, it might be a good idea to find something else. $\endgroup$
    – KReiser
    Oct 15, 2020 at 7:37

1 Answer 1

1
$\begingroup$

Yes, it is true that $\varphi$ induces a map on maximal spectra: the key statement here is that if we have a map of finitely generated algebras over a field, then the pullback of a maximal ideal is maximal. This is proven here on MSE for instance.

On the other hand, it is not true that the points of $\operatorname{Specmax} O_X$ are in bijection with points of the $\Bbb R$-variety $X$, assuming you mean "variety" in the classical sense as a subset of $\Bbb R^n$: for instance, $(x^2+1)$ is a maximal ideal of $\Bbb R[x]=O_{\Bbb A^1_{\Bbb R}}$, but it doesn't correspond to any point in $\Bbb R$. Instead, you should say that the maximal ideals of $O_X$ with residue field $\Bbb R$ are in correspondence with the points of $X$ in the sense of classical varieties. One can get the map of varieties by restricting - the image of any point in the maximal spectrum with residue field $\Bbb R$ is again a point with residue field $\Bbb R$.

All of the extra points we get, first from maximal ideals with residue field $\Bbb C$, and then from generic points, make the theory of varieties over the reals a lot nicer (you can do the same thing with an arbitrary non-algebraically closed field $k$ playing the role of $\Bbb R$ here, just substitute "finite extensions of $k$" for $\Bbb C$ in this sentence). The geometry is a lot better with them in the picture: finite maps are surjective, intersections behave appropriately, you can use stalks at generic points, etc. If you're specifically concerned with the geometry of the real points, you may be helped by investigating some semi-algebraic geometry. Here aresome expository texts on real algebraic geometry to see how all of these ideas play off against each other.

$\endgroup$
2
  • $\begingroup$ Thanks, your answer is really on point. I have some familiarity with algebraic geometry over $\mathbb{C}$, but over $\mathbb{R}$, it gets tricky. Projective completions are not useful as the projections are no longer closed necessarily, and things get semi-algebraic. I will look up the notes you have linked to. BTW, did you mean $\mathbb{R}$ and not $\mathbb{C}$ in the last paragraph? $\endgroup$
    – creet
    Oct 15, 2020 at 17:05
  • $\begingroup$ You're welcome. In the last paragraph, $k$ is playing the role of $\Bbb R$ and finite extensions of $k$ are playing the role of $\Bbb C$. $\endgroup$
    – KReiser
    Oct 15, 2020 at 18:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .