# Monomial order and initial ideals

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

• Your proof is correct. Mar 31, 2022 at 3:31

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