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Let $ d >0 $ be an integer, and let $ I \subset K[x_1,...,x_n] $ be the monomial ideal $$I = ( x_1^{a_1}x_2^{a_2}...x_n^{a_n} : \ \sum_{i=1}^n a_i =d; \ a_i < d\ \forall i).$$

(a) Compute the saturation $ \widetilde{I} $.

(b) The smallest integer $ k $ such that $ I : m^k = I : m^{k+1} $ is called the saturation number of $ I $. What is the saturation number of $ I $?

The saturation $ \widetilde{I} $ of $ I $ is the ideal $ I : m^{\infty} = \bigcup_{k=1}^{\infty} I : m^k$, where $ m = (x_1,...,x_n) $.

daisy
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2 Answers2

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Notice first that the saturation distributes over finite intersections, and for a primary monomial ideal $Q$, $\tilde{Q}=Q$ if $\sqrt{Q}\neq m$ and $\tilde{Q}=R$ if $\sqrt{Q}=m$. Thus given a primary decomposition $I=\bigcap_{i=1}^{r}Q_{i}$, $\tilde{I}=\bigcap\{Q_{i}\mid \sqrt{Q_{i}}\neq m\}$. On the other hand, $I$ is of Veronese type indexed by $d$ and $(d-1,d-1,\cdots, d-1)\in\mathbb{N}^{n}$, and we have a primary decomposition $I=m^{d}\cap (\bigcap_{i=1}^{n}(x_{1},\cdots, x_{i-1}, x_{i+1}, \cdots, x_{n}))$ by Observation 9, page 8 of the following link. So $\tilde{I}=\bigcap_{i=1}^{n}(x_{1},\cdots, x_{i-1}, x_{i+1}, \cdots, x_{n})$. Moreover, since $m^{a}:m=m^{a-1}$ for each $a\in\mathbb{N}$ (You can show first that $m^{a}:x_{i}=m^{a-1}$ for each $1\le i\le n$, and use Proposition 1.2.2 of Monomial Ideals by Herzog and Hibi), we have that $I:m^{a}=m^{d-a} \cap \tilde{I}$ for each $a\in\mathbb{N}$. Finally, since $m^{r}\supset \tilde{I}$ if and only if $r\le2$, the saturation number of $I$ is $d-2$.

user572516
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Allow me to slightly change the notation and denote the ideal of interest by \begin{equation*} I_d = \Big\langle x_1^{a_1}\dots x_n^{a_n}:\sum_{i=1}^na_i = d \ \text{ and }\ a_i < d\ \text{ for all } i \Big\rangle \end{equation*} in order to highlight the dependence on $d$. To compute the saturation of $I_d$, the main observation is that

  • $I_2:m = I_2$
  • $I_d:m = I_{d-1}$ for $d \geq 3$

Both of these equalities can be proven using double inclusion. To prove the right-to-left inclusions, we can directly verify that each generator of $I_2$ is in $I_2:m$ and that each generator of $I_{d-1}$ is in $I_d:m$ when $d \geq 3$. To prove the left-to-right inclusions, it first helps to notice that $I_d:m$ is a monomial ideal (since both $I_d$ and $m$ are), so we can restrict our attention to the monomials in $I_d:m$. These must have total degree at least $d-1$ and cannot be powers of single variables (since $I_d$ does not contain such monomials), so they must belong to $I_{d-1}$.

With this in hand, we can look at the ascending chain of ideals \begin{equation*} I_d \subseteq I_d:m \subseteq I_d:m^2 \subseteq \cdots \subseteq I_d:m^{d-3} \subseteq I_d:m^{d-2} \subseteq I_d:m^{d-1} \subseteq \cdots \end{equation*} and repeatedly apply the property $I:JK = (I:J):K$ to compute each term successively. Doing so, we obtain \begin{equation*} I_d \subseteq I_{d-1} \subseteq I_{d-2} \subseteq \cdots \subseteq I_3 \subseteq I_2 \subseteq I_2 \subseteq \cdots \end{equation*} Therefore, the saturation of $I_d$ is $I_2$ with saturation number $d-2$.

Hugues Bellemare
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