Is $\mathbb{Z}[x,y]/(x^3+px+q - y^2)$ integral over $\mathbb{Z}[x]$?

Question

Let $ B = \mathbb{Z}[x,y]/(x^3+px+q - y^2)$ and $A = \mathbb{Z}[x]$. I want to know whether the ring extension $A \subset B$ is integral or not.

There can be two possibilities: $x^3+px+q$ is irreducible or is not irreducible.

It's enough to understand whether $y$ is integral. If we find a finitely generated $A$-module $C$ such that $A[y] \subset C \subset B$ then $y$ is integral. We can try to find a faithful finitely-generated $A[y]$-module $M$ then $y$ is integral as well.

I see that there is a ring of integer polynomials that are not zero on a cubic curve. Geometric interpretation gives me no clue, alas.

$B=A[y]$ and $y$ is a zero of $X^2-r$ for some $r \in A$, so the extension seems integral? — Aphelli, Aug 27 '19 at 12:47
@Mindlack Does it mean that any polynomial instead of $x^3+px+q$ can be applied and the extension still will be integral? — Lada Dudnikova, Aug 27 '19 at 13:07
It is an integral extension, but not necessarily an integral domain: these are two unrelated notions. — Bernard, Aug 27 '19 at 13:28

score 4 · Accepted Answer · answered Aug 27 '19 at 13:07

4

The element $y\in B$ is integral over $A$ by definition of $B$ as the quotient of $B:=A[y]/(y^2-r)$ with $r=x^3+px+q$. It follows that the subring of $A[y]\subset B$ is integral over $A$, and clearly $A[y]=B$.

answered Aug 27 '19 at 13:07

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Is $\mathbb{Z}[x,y]/(x^3+px+q - y^2)$ integral over $\mathbb{Z}[x]$?

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