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



Monomial order and initial ideals
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, $x^{a}\in \sqrt{in(I)}=in(I)$.


