0
$\begingroup$

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 $F[X,Y]$?

$\endgroup$

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.