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Consider the following diagram of algebraic varieties:

$$\mathbb{A}^0 \to \mathbb{A}^1 \rightrightarrows \mathbb{A}^2$$

Here the first arrow is the inclusion of the origin into the line, and the next two are the inclusion of the line into the plane as the X and the Y axes.

Does this diagram have a colimit in the category of schemes?

(The first arrow giving inclusion of the point is not relevant; it just so happens that it was there when I met this question.)

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    $\begingroup$ No, I meant the question as asked: take the plane, and try to glue the entire X axis to the entire Y axis. $\endgroup$
    – Vivek Shende
    Commented Apr 2, 2021 at 12:35
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    $\begingroup$ Ah, so you basically want to know whether $\operatorname{Spec} R$ where $R := \{f\in k[x,y] \colon f(t,0)=f(0,t)\}$ works as a colimit in the category of schemes (it is the colimit in the category of affine schemes). $\endgroup$
    – Gro-Tsen
    Commented Apr 2, 2021 at 12:43
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    $\begingroup$ It is finitely generated because it contains the ring of symmetric polynomials $S:=k[x+y,xy]$, and $k[x,y]$ is a finite module over $S$. After a quick check, I think $R$ is generated by $xy^2$ as an $S$-algebra. $\endgroup$
    – Laurent Moret-Bailly
    Commented Apr 2, 2021 at 13:12
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    $\begingroup$ PS: I've become convinced there are too many answers being left in comments so, if no one else gets around to turning this into an answer in the next day or so, I'll write up a CW version. $\endgroup$
    – David E Speyer
    Commented Apr 2, 2021 at 13:52
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    $\begingroup$ As an aside, I asked Macaulay 2 for an equation for the surface with coordinate ring $R = k[xy, x+y, xy^2]$, it is $a^3 + c^2 = abc$ where $a=xy$, $b = x+y$, $c=x^2 y$. And here you can see the surface for yourself: math3d.org/6cuumoSv $\endgroup$
    – Vivek Shende
    Commented Apr 2, 2021 at 16:46

2 Answers 2

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We can rewrite the coequalizer as the pushout of the diagram $$ \begin{array}{ccc} X & \to & \mathbb A^2 \\ \downarrow & & \\ \mathbb A^1 & & \\ \end{array} $$ where $X$ is the union of the $x$- and $y$-axis, and the vertical map quotients by the involution swapping the two components.

The category of affine schemes has all pushouts: they are given by fibered product of coordinate rings. But a pushout in the category of affine schemes is not necessarily a pushout in the category of all schemes. A sufficient condition for a pushout of affine schemes to be a pushout in the category of schemes can be found in a paper of Karl Schwede ("Gluing schemes and a scheme without closed points", Theorem 3.4): it suffices that one leg of the pushout is a closed immersion. So we are fine.

As noted by Gro-Tsen in a comment, the pushout can be written explicitly as $\mathrm{Spec}(R)$ where $R = \{ f \in k[x,y] : f(t,0)=f(0,t)\}$.

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  • $\begingroup$ Which leg of the pushout is a closed immersion? $\endgroup$
    – Piotr Achinger
    Commented Apr 5, 2021 at 16:38
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    $\begingroup$ The horizontal leg: $X$ is here $V(xy) \subset \mathbb A^2$, not the disjoint union of the two axes. But you are right that if you directly translate the coequalizer to a pushout you get $\mathbb A^1 \sqcup \mathbb A^1$ rather than $X$ in the upper corner, and a (small) additional argument is needed for why the pushout is the same. $\endgroup$
    – Dan Petersen
    Commented Apr 5, 2021 at 17:21
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Dan Petersen has already answered the hard part -- how to show that the affine pushout $\{ f \in k[x,y] : f(t,0) = f(0,t) \}$ is also the scheme push out. I write to record a discussion in the comments about how to see that this ring is finitely generated and obtain an explicit list of generators.

Let $R = k[x,y]$ and let $S = k[x,y]^{S_2} = k[b,c]$ where $b=x+y$ and $c = xy$. Let $A = \{ f \in k[x,y] : f(t,0) = f(0,t) \}$. Then clearly $A$ is an $S$-submodule of $R$. Since $S$ is noetherian and $R$ is generated as an $S$-module by $1$ and $x$, this shows that $A$ is finitely generated as an $S$-module.

If the characteristic of $k$ is not $2$, one can get explicit generators easily. $A$ is invariant under switching the generators $x$ and $y$, so $A$ splits into positive and negative eigenspaces, call them $A_+$ and $A_-$, for this switch. The positive eigenspace is just $S$. The negative eigenspace $A_-$ is $xy(x-y) S$, since it is easy to see that anything in $A_-$ is divisibly by $xy(x-y)$, and the quotient is in $A_+$. So $A = S \oplus xy(x-y) S$ and the ring is generated by $b = x+y$, $c = xy$ and $f=xy(x-y)$, with the defining relation $f^2 = c^2 (b^2-4c)$.

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    $\begingroup$ A generating set that works uniformly in characteristic $2$ is $x+y, xy, x^2y$: to get $x^a y^b$ for all $a,b>0$ it suffices to handle the case when $b=1$ and $a=1$. The case $b=1$ follows by induction since $x^a y = (x+y) x^{a-1} y + (x^2y) x^{a-2}$ and the case $a=1$ follows by symmetry since $xy^2 = xy (x+y)-x^2y$. This has defining relation (with the same $b,c$ and with $g=x^2y$) $g^2 = c ( b g - c^2)$. $\endgroup$
    – Will Sawin
    Commented Apr 5, 2021 at 16:15
  • $\begingroup$ Oh, sorry, I didn't check the comments - this was pointed out (without proof) by Laurent Moret-Bailly and the equation was found by Vivek Shende. $\endgroup$
    – Will Sawin
    Commented Apr 5, 2021 at 16:55

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