Showing that if the initial ideal of I is radical, then I is radical.

Question

I need to show that given a term order $<$, and an ideal $I$, if $\mathrm{in}_<(I)$ is radical, then $I$ is radical.

Any help or hints would be appreciated as I'm not really sure where to start, since I know a few different facts about radical ideals.

Answer 1

From the context, I will assume your ideals are in some polynomial ring $k[x_1,\ldots,x_n]$. Of course you always have $I\subseteq \sqrt{I}$ so it's the reverse inclusion that you need to prove.

Choose $f\in\sqrt{I}$ and suppose that we already know $g\in I$ for any $g\in\sqrt{I}$ with smaller initial term than $f$.

By definition, $f^m\in I$ for some positive integer $m$. On the other hand, $\mathrm{in}(f^m)=\mathrm{in}(f)^m$ and, since $\mathrm{in}(I)$ is radical, we have $\mathrm{in}(f)\in \mathrm{in}(I)$. Therefore there is an $h\in I$ with $\mathrm{in}(h)=\mathrm{in}(f)$. Thus $g=f-h\in\sqrt{I}$ has smaller initial term than $f$ and so $g\in I$. Since $g\in I$ and $h\in I$ we conclude that $f\in I$.

Answer 2

First, we prove $\operatorname{in} \left(\sqrt{I}\right) \subseteq \sqrt{\operatorname{in}(I)}$. A generator of $\operatorname{in} \left(\sqrt{I}\right)$ has the form $\operatorname{in}(f)$ with $f \in \sqrt{I}$. Thus, there exists $n \in \mathbb{N}$ such that $f^n \in I$. Then, $\operatorname{in}(f)^n=\operatorname{in}(f^n) \in \operatorname{in}(I)$. Hence, $\operatorname{in}(f) \in \sqrt{\operatorname{in}(I)}$.

On the other hand, since $I \subseteq \sqrt{I}$, $\operatorname{in}(I) \subseteq \operatorname{in}(\sqrt{I})$. Then, we have $\operatorname{in}(I) \subseteq \operatorname{in}(\sqrt{I})\subseteq \sqrt{\operatorname{in}(I)}=\operatorname{in}(I)$, and so $\operatorname{in}(I)=\operatorname{in}(\sqrt{I})$. Therefore, $I=\sqrt{I}$.