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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 $R$-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!

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    $\begingroup$ You introduced the polynomials $f_i$ and the ring $S$ in your first paragraph, but did not use it there. Is there a type somewhere? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Oct 15, 2012 at 3:32
  • $\begingroup$ @Mariano Suarez-Alvarez : Thank you very much! I edited it. $\endgroup$
    – Knot
    Commented Oct 15, 2012 at 3:56
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    $\begingroup$ (1) Shouldn't M be a bigraded S-module ? (2) What are the categories mapped by your functor ? (3) Which kind of isomorphism do you expect in the 2nd question ? $\endgroup$
    – Ralph
    Commented Oct 15, 2012 at 13:29
  • $\begingroup$ @Ralph: Thank you, I have confused with $R$ and $S$. The 2nd question is in fact a results of a lemma in a paper, that I do not think it is trivial but I have not proved it yet, so I decided to post it here. Thank you very much! P.s : The functor is from the category of bigraded modules to the category of bigraded module, I think. $\endgroup$
    – Knot
    Commented Oct 15, 2012 at 13:40
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    $\begingroup$ It seems to me that your functor is from the category of bigraded S-modules to the category of graded R-modules, but I didn't check details yet. $\endgroup$
    – Ralph
    Commented Oct 15, 2012 at 13:51

1 Answer 1

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Question 1 has been answered by Pablo in his comment above.

Question 2: $S(-a,-b)^{(n)} = S^{(n-b)}(-a)$ is just a rewrite of the definition.

Set $N = S(-a,-b)$. Then $$N_p^{(n)} = N_{(p,n)} = S_{(p-a,n-b)} = \bigoplus_{i,j} kX_1^{i_1}\cdots X_r^{i_r}T_1^{j_1}\cdots T_s^{j_s} \cong \bigoplus_j \bigoplus_i kX_1^{i_1}\cdots X_r^{i_r}\quad (\ast)$$ (as $k$-vector spaces) where the sum is taken over all tuples $i,j$ such that $j_1+\cdots j_s = n-b$ and $i_1+ \cdots + i_r = (p-a)-(p_1j_1 + \cdots + p_sj_s)$. By using
$$\bigoplus_i kX_1^{i_1}\cdots X_r^{i_r}=R_{(p-a)-(p_1j_1 + \cdots + p_sj_s)}=R(-a -p_1j_1 - \cdots - p_sj_s)_p$$ we find $$N^{(n)} \cong \bigoplus_j R(-a -p_1j_1 - \cdots - p_sj_s)$$ as graded $k$-vector spaces and since $R$ acts in $(\ast)$ on the monomials $X^i$ it's clear that this is also an isomorphism of graded $R$-modules.

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  • $\begingroup$ @Ralph: Thank you very much for answering my question. May I ask : Why in $(*)$ the component containging the variables $T1,...T_s$ disappear? $\endgroup$
    – Knot
    Commented Oct 16, 2012 at 1:50
  • $\begingroup$ It doesn't disappear, just the monomial $T^j$ disappears. More precisely, (for fixed i) I'm using the (k-vector space) isomorphism $$\bigoplus_j kX^iT^j \to \bigoplus_j kX^i,\;X^iT^j \mapsto (0,...,0,X^i,0...,0),$$ with $X^i$ placed in position $j$. It's the same as writing $\bigoplus_j kT^j \xrightarrow{\sim} \bigoplus_j k,\;\sum_j a_jT^j \mapsto (a_j)_j$ $\endgroup$
    – Ralph
    Commented Oct 16, 2012 at 6:51

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