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Consider $\newcommand{\Spec}{\operatorname{Spec}}V(xy) \subset \Spec \mathbb C[x,y]$ as a closed subscheme. We think of $V(xy)$ as the union of the two axis in $\mathbb A_\mathbb C^2$. But why?

The category of $\mathbb C$-schemes can be embedded into a larger category of sheaves on $\newcommand{\Aff}{\operatorname{Aff}}\Aff_\mathbb C = \mathbb C\operatorname{-Alg}^{\operatorname{op}}$ equipped with some Gorthendieck topology (etale, Zariski, flat, I don't know which is best). The correspondig sheaf topos is famously a model of some of the axioms synthetic differential geometry in the sense that we have infinitesimals. The affine line has (in the internal language) elements which are not not zero, but which aren't zero either and which square to zero. In the internal language the subspace $V(xy)$ is precisely$$V(xy) = \{(x,y):\mathbb A_\mathbb C^2|\,xy=0\}$$The space $V(x)$ on the other hand has the description $V(x) = \{(x,y):\mathbb A_\mathbb C|x=0\}$. The join of $V(x)$ and $V(y)$ in the sheaf topos $\operatorname{Sh}(\operatorname{Aff},\tau)$ is for tautological reasons $$V(x)\vee_{\operatorname{Sh}(\operatorname{Aff},\tau)} V(y) = \{(x,y):\mathbb A_\mathbb C^2\,|\, x=0 \,\vee\, y=0 \}$$ But this is not the same as $V(xy)$ if the infinitesimal object $D= \{x:\mathbb A_\mathbb C\,|\,x^2=0\}$ of the affine line is non-trivial. In other words: $V(xy)$ seems a little to thick at the origin to be the union of $V(x)$ and $V(y)$.

I have asked a related question on math overflow.

Context: I am working with the functorial approach to algebraic geometry. Here you take the category of commutative rings, take its opposite $\Aff$ and equip it with a Grothendieck-topology. Then you look at functors $X:\text{cRing}\to \text{Set}$, equivalently presheaves on $\Aff$. A scheme $X$ in the usual locally ringed space style gives rise to a functor $h_X = \operatorname{Hom}_{\operatorname{Sch}}(-,X):\operatorname{Sch}^{op}\to \operatorname{Set}$ which in turn can be restricted along the inclusion $\operatorname{cRing} \to\operatorname{Sch}^{op}$ to give presheaf on the opposite of the category of rings. It turns out that this construction is fully faithful, and it makes the category of schemes a full subcategory of $\operatorname{Pr}(\Aff)$. So one can try to identify the category of schemes directly in $\operatorname{Pr}(\Aff)$ and develop scheme theory completly from this perspective. This is possible and not hard. The category of schemes one gets is embedded into the larger category of sheaves on the Zariski site. The right words to look for online are "gros Zariski topos" for example.

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  • $\begingroup$ $V(xy)$ is reduced, so it cannot be "less thick". $\endgroup$ Sep 30, 2022 at 12:48
  • $\begingroup$ @red_trumpet Yes, I know that. But how does it fit together with the internal language of Sh(Aff,Zar)? Also what about the morphism $\text{Spec }\mathbb C[x]/(x^2) \to \text{Spec }\mathbb C[x,y]/(xy)$ (induced by the map $\mathbb C[x,y]/(xy) \to \mathbb C[x]/(x^2)$ which sends both $x$ and $y$ to $x$). It is monic, and it looks to me like a "diagonal little tangent vector at the origin". How do you picture it geometrically? $\endgroup$
    – Nico
    Sep 30, 2022 at 13:00
  • $\begingroup$ I my mind the functor $V(x)\vee_{\text{Sh(Aff,Zar)}}V(y)$ is less thick (I am working in the functor of points approach). It is somehow properly contained in the subscheme $V(xy)$. $\endgroup$
    – Nico
    Sep 30, 2022 at 13:02
  • $\begingroup$ I'm not entirely sure what you mean by $V(x) \vee_{\operatorname{Sh(Aff, Zar)}} V(y)$. Is that the disjoint union of $V(x)$ and $V(y)$? So just two copies of $\mathbb A^2$? That would not be a subscheme of $\mathbb A^2$ anymore, but admits a map to $\mathbb A^2$ $\endgroup$ Sep 30, 2022 at 13:04
  • $\begingroup$ @red_trumpet Maybe I should change my question to: Why is $V(xy)$ the geometrically correct union in the context of Sh(Aff,Zar) as a model of (a part of) synthetic differential geometry. $\endgroup$
    – Nico
    Sep 30, 2022 at 13:05

2 Answers 2

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Well, if you are restricting your attention to schemes, there is simply not any better option. One way of saying this is that $V(xy)$ is the smallest closed subscheme that contains both $V(x)$ and $V(y)$. This just amounts to the algebraic fact that any element of $\mathbb{C}[x,y]$ which is divisible by both $x$ and $y$ must be divisible by $xy$. Another way to make this precise is to say that $V(xy)$ is the pushout of $V(x)$ and $V(y)$ over their intersection $V(x,y)$ in the category of schemes. So, if you restrict your attention to the the world of schemes, there just does not exist any better candidate to call the union of the two lines.

Your observation that this is not true in the larger sheaf topos is closely related to the fact that the algebraic fact above is not stable under base-change: if you map $\mathbb{C}[x,y]$ to some other $\mathbb{C}$-algebra, then the images of $x$ and $y$ may no longer have the property that any element divisible by both $x$ and $y$ is divisible by $xy$. In the context of schemes, this is the fact that the operation of taking unions of closed subschemes is not stable under base-change. Explicitly, for instance, if you pull back to the diagonal line $V(x-y)$ in $\mathbb{A}^2$, then $V(x)$ and $V(y)$ both become just the reduced point at the origin, but $V(xy)$ becomes a non-reduced thickened version.

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  • $\begingroup$ thank you! If I understand you correctly then you are saying that $V(xy)$ is not only the smallest closed subscheme which contains them, but also the smallest subscheme over all? $\endgroup$
    – Nico
    Sep 30, 2022 at 15:49
  • $\begingroup$ Yes, that is true. $\endgroup$ Sep 30, 2022 at 16:35
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By definition $$ V(xy)=\{\mathfrak{p}\in\mathrm{Spec}\mathbb{C}[x,y]\mid xy\in\mathfrak{p}\}; $$ easily one proves: $$ V(xy)=V(x)\cup V(y). $$ On the other hand, $V(xy)$ is interpretable as the set of the zeroes of the regular function $xy$ on $\mathrm{Spec}\mathbb{C}[x,y]$.

If one restricts to maximal spectrum of $\mathbb{C}[x,y]$, by Hilbert's Nullstellensatz Theorem, $\mathrm{Spec}_{max}\mathbb{C}[x,y]=\{(x-a,y-b)\in\mathrm{Spec}\mathbb{C}[x,y]\mid a,b\in\mathbb{C}\}$ and one figures out that $V_{max}(xy)=V(xy)\cap\mathrm{Spec}_{max}\mathbb{C}[x,y]$ is the union of tha axes "$x$" and "$y$" of $\mathbb{A}^1_{\mathbb{C}}$.

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  • $\begingroup$ I am not talking about $V(xy)$ as a topological space, but as a scheme. The maximal points which you mention (and also the generic points) do not capture the infinitesimals of scheme theory, but it is exactly those about which I am worried. $\endgroup$
    – Nico
    Sep 30, 2022 at 12:24
  • $\begingroup$ Have you read the second paragraph of my question? Unfortunately I already know the things which you state in your answer. :/ $\endgroup$
    – Nico
    Sep 30, 2022 at 12:26
  • $\begingroup$ You can think to $V(xy)$ as $\mathrm{Spec}\mathbb{C}[x,y]_{\displaystyle/(xy)}=\mathrm{Spec}\mathbb{C}[x,y]_{\displaystyle/(x)+(y)}$ which is the schematic union of $V(x)$ and $V(y)$. (Do you know this?) $\endgroup$ Sep 30, 2022 at 12:39
  • $\begingroup$ Yes, it should be $\text{Spec}\,\mathbb C[x,y]/(x)\cap (y)$ though. But I wonder why the schematic union is defined the way it is. See my related question on math overflow! $\endgroup$
    – Nico
    Sep 30, 2022 at 12:46
  • $\begingroup$ There is also a notion of union in the larger category of sheaves on the Zariski site, and it does not agree with the schematic union apparently. $\endgroup$
    – Nico
    Sep 30, 2022 at 12:47

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