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Post Closed as "Duplicate" by Steven Landsburg, Max Horn, Friedrich Knop, Bugs Bunny, CommunityBot
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user443060
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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[t])$$\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[t])$$\Pic(R[s])$ will the same approach via conductor ideals work, for any suggestion or guidance I am thankful.

$\DeclareMathOperator\Pic{Pic}$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 $\Pic(R) \ncong \Pic(R[t])$ 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 $\Pic(R[t])$ will the same approach via conductor ideals work, for any suggestion or guidance I am thankful.

$\DeclareMathOperator\Pic{Pic}$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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YCor
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For$\DeclareMathOperator\Pic{Pic}$For a commutative ring with unity $R$, if $R_{red}$$R_{\mathrm{red}}$ is semi normal-normal then we have an equivalent criterion which states that $Pic(R) \cong Pic(R[t])$$\Pic(R) \cong \Pic(R[t])$. Here $Pic(R)$$\Pic(R)$ is the picardPicard group of $R$. So naturally it is well understood that $Pic(R) \ncong Pic(R[t])$$\Pic(R) \ncong \Pic(R[t])$ 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$$\Pic(R) \cong k$ but I am unable to explicitly compute $Pic(R[t])$$\Pic(R[t])$ will the same approach via conductor ideals work, for any suggestion or guidance I am thankful.

For a commutative ring with unity $R$, if $R_{red}$ 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 $Pic(R) \ncong Pic(R[t])$ 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 $Pic(R[t])$ will the same approach via conductor ideals work, for any suggestion or guidance I am thankful.

$\DeclareMathOperator\Pic{Pic}$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 $\Pic(R) \ncong \Pic(R[t])$ 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 $\Pic(R[t])$ will the same approach via conductor ideals work, for any suggestion or guidance I am thankful.

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user443060
user443060
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Picard group of a cusp

For a commutative ring with unity $R$, if $R_{red}$ 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 $Pic(R) \ncong Pic(R[t])$ 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 $Pic(R[t])$ will the same approach via conductor ideals work, for any suggestion or guidance I am thankful.