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Let $R$ be a polynomial ring $k[x_1,...,x_n]$, let $f_1,...f_s$ be some non-zero polynomial sin $R$ of degree $p_1,...,p_s$ respectively.Define $S$ by $k[X_1,...,X_n, T_1,...T_s]$ with bigrading defined by $degX_i=(1,0)$ and $degT_j=(p_j,1)$. For a bigraded $S$-module $M=\bigoplus_{p,n\in\Bbb{N}}M_{(p,n)}$, define $M^{(n)}$ to be the graded $S$-module $\bigoplus_{p\in\Bbb{N}}M_{(p,n)}$ in which the action of $x_i$ can be understanded as $X_i$ with its obvious grading.

There are two claims that I could not prove, so I decided to post it here, and I hope you will help me to prove it.

1-The functor $(.)^{(n)}$ is an exact functor.

2-There are the isomorphisms: $S(-a,-b)^{(n)}\cong S^{(n-b)}(-a)\cong \bigoplus_{a_1+...+a_s=n-b}R{(-a_1p_1-...-a_sp_s-a)}$

Thank for reading my question!

Let $R$ be a polynomial ring $k[x_1,...,x_n]$, let $f_1,...f_s$ be some non-zero polynomial sin $R$ of degree $p_1,...,p_s$ respectively.Define $S$ by $k[X_1,...,X_n, T_1,...T_s]$ with bigrading defined by $degX_i=(1,0)$ and $degT_j=(p_j,1)$. For a bigraded $S$-module $M=\bigoplus_{p,n\in\Bbb{N}}M_{(p,n)}$, define $M^{(n)}$ to be the graded $R$-module $\bigoplus_{p\in\Bbb{N}}M_{(p,n)}$ in which the action of $x_i$ can be understanded as $X_i$ with its obvious grading.

There are two claims that I could not prove, so I decided to post it here, and I hope you will help me to prove it.

1-The functor $(.)^{(n)}$ is an exact functor.

2-There are the isomorphisms: $S(-a,-b)^{(n)}\cong S^{(n-b)}(-a)\cong \bigoplus_{a_1+...+a_s=n-b}R{(-a_1p_1-...-a_sp_s-a)}$

Thank for reading my question!

Let $R$ be a polynomial ring $k[x_1,...,x_n]$, let $f_1,...f_s$ be some non-zero polynomial sin $R$ of degree $p_1,...,p_s$ respectively.Define $S$ by $k[X_1,...,X_n, T_1,...T_s]$ with bigrading defined by $degX_i=(1,0)$ and $degT_j=(p_j,1)$. For a bigraded $R$-module $M=\bigoplus_{p,n\in\Bbb{N}}M_{(p,n)}$, define $M^{(n)}$ to be the graded $R$-module $\bigoplus_{p\in\Bbb{N}}M_{(p,n)}$ in which the action of $x_i$ can be understanded as $X_i$ with its obvious grading.

There are two claims that I could not prove, so I decided to post it here, and I hope you will help me to prove it.

1-The functor $(.)^{(n)}$ is an exact functor.

2-There are the isomorphisms: $S(-a,-b)^{(n)}\cong S^{(n-b)}(-a)\cong \bigoplus_{a_1+...+a_s=n-b}R{(-a_1p_1-...-a_sp_s-a)}$

Thank for reading my question!

Let $R$ be a polynomial ring $k[x_1,...,x_n]$, let $f_1,...f_s$ be some non-zero polynomial sin $R$ of degree $p_1,...,p_s$ respectively.Define $S$ by $k[X_1,...,X_n, T_1,...T_s]$ with bigrading defined by $degX_i=(1,0)$ and $degT_j=(p_j,1)$. For a bigraded $S$-module $M=\bigoplus_{p,n\in\Bbb{N}}M_{(p,n)}$, define $M^{(n)}$ to be the graded $S$-module $\bigoplus_{p\in\Bbb{N}}M_{(p,n)}$ in which the action of $x_i$ can be understanded as $X_i$ with its obvious grading.

There are two claims that I could not prove, so I decided to post it here, and I hope you will help me to prove it.

1-The functor $(.)^{(n)}$ is an exact functor.

2-There are the isomorphisms: $S(-a,-b)^{(n)}\cong S^{(n-b)}(-a)\cong \bigoplus_{a_1+...+a_s=n-b}R{(-a_1p_1-...-a_sp_s-a)}$

Thank for reading my question!

Let $R$ be a polynomial ring $k[x_1,...,x_n]$, let $f_1,...f_s$ be some non-zero polynomial sin $R$ of degree $p_1,...,p_s$ respectively.Define $S$ by $k[X_1,...,X_n, T_1,...T_s]$ with bigrading defined by $degX_i=(1,0)$ and $degT_j=(p_j,1)$. For a bigraded $R$-module $M=\bigoplus_{p,n\in\Bbb{N}}M_{(p,n)}$, define $M^{(n)}$ to be the graded $R$-module $\bigoplus_{p\in\Bbb{N}}M_{(p,n)}$ in which the action of $x_i$ can be understanded as $X_i$ with its obvious grading.

There are two claims that I could not prove, so I decided to post it here, and I hope you will help me to prove it.

1-The functor $(.)^{(n)}$ is an exact functor.

2-There are the isomorphisms: $S(-a,-b)^{(n)}\cong S^{(n-b)}(-a)\cong \bigoplus_{a_1+...+a_s=n-b}R{(-a_1p_1-...-a_sp_s-a)}$

Thank for reading my question!

Let $R$ be a polynomial ring $k[x_1,...,x_n]$, let $f_1,...f_s$ be some non-zero polynomial sin $R$ of degree $p_1,...,p_s$ respectively.Define $S$ by $k[X_1,...,X_n, T_1,...T_s]$ with bigrading defined by $degX_i=(1,0)$ and $degT_j=(p_j,1)$. For a bigraded $R$-module $M=\bigoplus_{p,n\in\Bbb{N}}M_{(p,n)}$, define $M^{(n)}$ to be the graded $R$-module $\bigoplus_{p\in\Bbb{N}}M_{(p,n)}$ in which the action of $x_i$ can be understanded as $X_i$ with its obvious grading.

There are two claims that I could not prove, so I decided to post it here, and I hope you will help me to prove it.

1-The functor $(.)^{(n)}$ is an exact functor.

2-There are the isomorphisms: $S(-a,-b)^{(n)}\cong S^{(n-b)}(-a)\cong \bigoplus_{a_1+...+a_s=n-b}R{(-a_1p_1-...-a_sp_s-a)}$

Thank for reading my question!