Pushout results in a nodal curve

I'm looking at this MathOverflow post about how every non-normal variety results from "undue gluing". I'm working through the first example where Karl Schwede makes the following claim:

Let $X=\mathbb{A}^1$ be the affine line, let $Y = \{P_1,P_2 \}$ be a subvariety of two points, and let $Z = \{Q\}$ be a singleton. The pushout of the diagram $X\leftarrow Y \rightarrow Z$ is a nodal curve.

Now, this diagram corresponds to the diagram of rings $$ k[x]\rightarrow k\oplus k \leftarrow k $$ where $k[x] \rightarrow k\oplus k$ is the map $f\mapsto (f(P_1),f(P_2))$ and $k \to k\oplus k$ is the diagonal map $a\mapsto (a,a)$. The pushout of affine schemes corresponds to the fibred product, so we have the ring $$ k[x] \times_{k\oplus k} k = \{ (f,a) \; : \; f(P_1)=f(P_2)=a \} $$ Now I'm stuck. How does one realize this ring as the coordinate ring of a nodal curve?

Solution 1:

As Sasha suggested in the comments, suppose $P_1=1$ and $P_2=-1$. A polynomial $f\in k[x]$ takes the value $a$ at $P_1$ and $P_2$ iff it's of the form $(x^2-1)p+a$ for some $p\in k[x]$, so we need to find a way to generate the whole subalgerba of these polynomials.

I claim that $x^2-1$ and $x(x^2-1)$ suffice: given $(x^2-1)p+a$, if $\deg p>0$ is even, then we can pick $c\in k$ and $n\in\Bbb Z_{>0}$ so that $(x^2-1)p+a-c(x^2-1)^n=(x^2-1)p'+a$ is also an element of our subalgebra with $\deg p'<\deg p$, while if $\deg p>0$ is odd, we can pick $c\in k$ and $n\in\Bbb Z_{>0}$ so that $(x^2-1)p+a-cx(x^2-1)^n=(x^2-1)p'+a$ is also an element of our subalgebra with $\deg p'<\deg p$. By induction, this shows that $x^2-1$ and $x(x^2-1)$ generate $k[x]\times_{k\oplus k} k$, or $k[x]\times_{k\oplus k} k\cong k[x^2-1,x(x^2-1)]$.

To find a relation between $t=x^2-1$ and $u=x(x^2-1)$, we can take the resultant of $t-(x^2-1)$ and $u-(x(x^2-1))$ with respect to $x$ to get that $u^2-t^3-t^2=0$ (you could also get this by playing around - the relation is of low degree, so it's not completely out of the question). As $u^2-t^3-t^2$ is irreducible, we see that $k[x^2-1,x(x^2-1)]\cong k[t,u]/(u^2=t^3+t^2)$ which is the coordinate algebra of the nodal cubic.

As far as intuition goes, one often sees this procedure done in reverse: given the nodal cubic, compute a normalization. It turns out that the normalization map is a map from $\Bbb A^1$ which identifies two points, so if you've seen this and then someone comes around asking you how to glue two points on $\Bbb A^1$, you might already have some idea of what the answer could be. One could of course make other choices for the $P_i$ (and one must if you want to do this in characteristic two), but $P_i=\pm 1$ is pretty convenient from the standpoint of computations.