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, $x^{a}\in \sqrt{in(I)}=in(I)$.
Best Answer
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, $x^{a}\in \sqrt{in(I)}=in(I)$.