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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.

user26857
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14201302
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2 Answers2

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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 $f\in I$.

user26857
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Casteels
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  • Thank you, this is incredibly helpful. Since this is a form of induction, is the implicit base case just 1 (since this is always the smallest monomial in any term ordering and we know it is in $\sqrt{I}$ and in I)? – 14201302 Feb 28 '15 at 15:24
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    Right it's induction but the base case is $0$. In fact $1$ is not in those ideals unless those ideals are the entire ring. – Casteels Feb 28 '15 at 15:31
  • Ah, I see, thanks. Where does zero come in a term ordering then? I'm not sure I've ever seen that addressed – 14201302 Feb 28 '15 at 15:35
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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, $I=\sqrt{I}$.

user26857
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Wágner Badilla
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  • Why $\operatorname{in}(I)=\operatorname{in}(\sqrt{I})$ implies $I=\sqrt{I}$? – user26857 Mar 25 '22 at 05:40
  • En general. Let $I,J$ be two ideals in $S=K[x_1,\ldots,x_n]$. If $in(I)=in(J)$, then $I=J$. Proof. Let $G$ be a Gröbner basis of $J$. Then $in(J)=(in(g) | g \in G)$. Thus, $in(I)=(in(g) | g \in G)$. This implies that $G$ is a Gröbner basis of $I$ as well. As $G$ is a Gröbner basis of $I$, $G$ generates to $I$. As $G$ is a Gröbner basis of $J$, $G$ generates to $J$. Therefore, I=J. – Wágner Badilla Mar 25 '22 at 17:32
  • In general, your claim is wrong. The ideals $I=(x^2)$ and $J=(x^2-y)$ in $k[x,y]$ with $x>y$ are a counterexample. If one assumes that $I\subseteq J$, then you are right; see also here. Fortunately this is the case in the present question since $I\subseteq\sqrt I$. – user26857 Mar 25 '22 at 20:31
  • Sorry, I forgot to write $J \subseteq I$ – Wágner Badilla Mar 25 '22 at 21:18