Question

Suppose $F$ is a field and $A$ the $F$-algebra $F[X]\times_{(F\times F)} F$ given by the missing corner of a cartesian square in $F$-algebras \begin{equation} F~\xrightarrow{\Delta} ~F\times F~ \xleftarrow{f} ~F[X], \end{equation} $f(X)=(0,1)$. Geometrically this is $\mathbb{A}^1_F$ with $0$ and $1$ identified.

$A$ looks like the node $F[X,Y]/(X^2+X^3-Y^2)$. Is $A$ actually isomorphic to a quotient of $F[X,Y]$?

Answer 1

A is generated by all polynomials $g(x)$ such that $g(0)=g(1)$. In particular it contains $x^2-x$ and $x^3-x^2$, and those are algebra generators. They satisfy $a^3+ab-b^2=0$, so $A \cong F[a,b]/(a^3+ab-b^2)$.]

If you define your map $f$ by $f(X) = (1,-1)$ instead, you get $a^3+a^2-b^2$ on the nose.