0
$\begingroup$

Let $S=K[x_1,\ldots, x_n]$ polynomial ring. Let $I \subseteq S$ an ideal and $<$ be a (global) monomial order in $S$. If in$_<(I)$ a radical ideal, then in$_<(I)=$ in$_<(P_1) \;\cap$ in$_<(P_2)\cap \ldots \cap$ in$_<(P_l)$, where $P_1,\ldots, P_l$ are the minimal prime ideals of $I$.

Question

Is this true?

I think yes.

Since $in(I)$ is a radical ideal, $I$ is a radical ideal. Thus, $I=\bigcap_{i=1}^{l} P_i$, and so, $in(I)=in\left( \bigcap_{i=1}^{l} P_i \right) \subseteq \bigcap_{i=1}^{l} in(P_i)$.

Now, we show that $\bigcap_{i=1}^{l} in(P_i) \subseteq in(I)$. As each $in(P_i)$ is a monomial ideal, $\bigcap_{i=1}^{l} in(P_i)$ is a monomial ideal. Let $x^{a}$ be a generator of $\bigcap_{i=1}^{l} in(P_i)$. We have that $x^{a} \in in(P_i)$ for every $i=1,\ldots,l$. Hence, $x^{a}=in(f_i)$ with $f_i \in P_i$. We note that $g=f_1 \cdot \ldots \cdot f_l \in \bigcap_{i=1}^{l} P_i=I$. As a consequence, $in(g)=in(f_1) \cdot \ldots \cdot in(f_l)=(x^{a})^{l}$. Thus, $(x^{a})^l \in in(I)$. Therefore, $x^{a}\in \sqrt{in(I)}=in(I)$.

$\endgroup$
1
  • $\begingroup$ Your proof is correct. $\endgroup$
    – pinaki
    Mar 31, 2022 at 3:31

1 Answer 1

0
$\begingroup$

Since $in(I)$ is a radical ideal, $I$ is a radical ideal. Thus, $I=\bigcap_{i=1}^{l} P_i$, and so, $in(I)=in\left( \bigcap_{i=1}^{l} P_i \right) \subseteq \bigcap_{i=1}^{l} in(P_i)$.

Now, we show that $\bigcap_{i=1}^{l} in(P_i) \subseteq in(I)$. As each $in(P_i)$ is a monomial ideal, $\bigcap_{i=1}^{l} in(P_i)$ is a monomial ideal. Let $x^{a}$ be a generator of $\bigcap_{i=1}^{l} in(P_i)$. We have that $x^{a} \in in(P_i)$ for every $i=1,\ldots,l$. Hence, $x^{a}=in(f_i)$ with $f_i \in P_i$. We note that $g=f_1 \cdot \ldots \cdot f_l \in \bigcap_{i=1}^{l} P_i=I$. As a consequence, $in(g)=in(f_1) \cdot \ldots \cdot in(f_l)=(x^{a})^{l}$. Thus, $(x^{a})^l \in in(I)$. Therefore, $x^{a}\in \sqrt{in(I)}=in(I)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.