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soc -- compute the $\text{Soc}_\mathfrak{p}^*(I)$, where $I$ is a monomial ideal

Synopsis

Description

This method computes the module $\text{Soc}_\mathfrak{p}^*(I)$. Let $\mathcal{R}(I)=\bigoplus_{k\ge0}I^k$ be the Rees algebra of $I$. For the prime $\mathfrak{p}\in\text{Ass}^\infty(I)$, we set $$X_\mathfrak{p}=\begin{cases}S&\text{if}\ \mathfrak{p}\in\text{Max}^\infty(I),\\ \prod\{\mathfrak{q}\in\text{Ass}^\infty(I):\mathfrak{p}\subsetneq\mathfrak{q}\}&\text{otherwise}.\end{cases}$$ Let $\mathcal{R}'=\bigoplus_{k\ge0}I^{k+1}$. We set $$\text{Soc}_\mathfrak{p}^*(I)=\frac{(\mathcal{R}':_{\mathcal{R}(I)}\mathfrak{p}\mathcal{R}(I))}{(\mathcal{R}':_{\mathcal{R}(I)}(\mathfrak{p}+X_\mathfrak{p}^\infty)\mathcal{R}(I))}.$$ Here $X_\mathfrak{p}^\infty=\bigcup_{k\ge0}(\mathfrak{p}:\mathfrak{m}^k)$ is the saturation of $\mathfrak{p}$ with respect to the maximal ideal $\mathfrak{m}=(x_1,\dots,x_n)$. As proved by Conca, for all $k\gg0$ we have $$\text{Soc}_\mathfrak{p}^*(I)_{(*,k)}=\frac{(I^{k+1}:\mathfrak{p})}{I^{k+1}:(\mathfrak{p}+X_\mathfrak{p}^\infty)},$$ and the initial degree of this module is the $\text{v}_\mathfrak{p}$-number of $I^{k+1}$.

i1 : S = QQ[x_1..x_3];
 
i2 : I = ideal(x_1*x_2,x_1*x_3,x_2*x_3)

o2 = ideal (x x , x x , x x )
             1 2   1 3   2 3

o2 : Ideal of S
 
i3 : p = ideal(x_1,x_2)

o3 = ideal (x , x )
             1   2

o3 : Ideal of S
 
i4 : soc(I,p)

o4 = subquotient (| x_3 x_1x_2 |, | x_2x_3 x_1x_3 x_1x_2 x_3y_1 |)

           QQ[x ..x , y ..y ]                              /      QQ[x ..x , y ..y ]      \
               1   3   1   3                               |          1   3   1   3       |1
o4 : -------------------------------module, subquotient of |------------------------------|
     (- x y  + x y , - x y  + x y )                        |(- x y  + x y , - x y  + x y )|
         1 3    2 2     1 3    3 1                         \    1 3    2 2     1 3    3 1 /

See also

Ways to use soc:

For the programmer

The object soc is a method function.