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Let $R$ be a domain with affine spectrum $S$ and consider the scheme $X=\mathbb A^1_R=\operatorname {Spec}R[T] $ over $S$.
Let $P\subset R[T]$ be an ideal with $P\cap R=0$ and let $Y\subset X$ be the associated subscheme. The extension of $P$ to $(\operatorname {Quot}(R))[T])$ is principal, generated by some polynomial $f$ which we may assume has coefficients in $R[T]$. What is the relationship between $Y$, the subscheme $V(f)\subset X$ and (probably) inverse images of subschemes in $S$ ?
(I have asked this question three days ago on math stackexchange but got no answer nor comment)

Edit: I'm also interested in knowing if one can say more under the assumption that $P$ is prime.

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  • $\begingroup$ $V(f)$ is the union of $Y$ and a subscheme of $X$ which is the pull back of some subscheme from $S$. $\endgroup$
    – Mohan
    Commented May 12, 2020 at 17:34
  • $\begingroup$ @Mohan: Well of course, that's the obvious guess, as I hinted in my question. I'm asking for a detailed, rigorous proof. $\endgroup$
    – lefuneste
    Commented May 12, 2020 at 17:50
  • $\begingroup$ Are you assuming $R$ is Noetherian? I was. $\endgroup$
    – Mohan
    Commented May 12, 2020 at 17:52
  • $\begingroup$ No, I was not assuming $R$ noetherian. But I would be happy to read a proof of your claim assuming noetherianness. $\endgroup$
    – lefuneste
    Commented May 12, 2020 at 20:01
  • $\begingroup$ Then, use Krull's principal ideal theroem and look at the prime decomposition of $(f)$. Set theoretically this is the union of $Y$ and the pull back of a closed subset of $S$, since all minimal primes containing $f$ have height one. $\endgroup$
    – Mohan
    Commented May 12, 2020 at 20:47

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