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$\DeclareMathOperator\Pic{Pic}$For a commutative ring with unity $R$, if $R_{\mathrm{red}}$ is semi-normal then we have an equivalent criterion which states that $\Pic(R) \cong \Pic(R[t])$. Here $\Pic(R)$ is the Picard group of $R$. So naturally it is well understood that $\Pic(R) \ncong \Pic(R[s])$ when $R = k[t^2,t^3]$ a cusp, which is a domain that is not semi-normal.

With the help of conductor ideals I was able to show that $\Pic(R) \cong k$ but I am unable to explicitly compute $\Pic(R[s])$ will the same approach via conductor ideals work, for any suggestion or guidance I am thankful.

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  • $\begingroup$ Am I missing something here? If $R = k[t^2,t^3],$ then $R[t] = k[t],$ so you're just looking at the Picard group of $k[t].$ $\endgroup$
    – Stahl
    Commented Mar 25, 2023 at 0:32
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    $\begingroup$ @Stahl I suspect there's a notational clash, and OP means to take the polynomial ring over $R$, which probably should be denoted by $R[T]$ or some other letter. $\endgroup$
    – Wojowu
    Commented Mar 25, 2023 at 0:33
  • $\begingroup$ @Wojowu Ah yes, that would make more sense. $\endgroup$
    – Stahl
    Commented Mar 25, 2023 at 0:34
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    $\begingroup$ @Stahl yes, there was a notational clash, I have fixed it $\endgroup$
    – user443060
    Commented Mar 25, 2023 at 1:48

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