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I want to prove that a class of ideals are all primary ideals. It may be easier to demonstrate with an example.

Let $I=(a^2,ab,ac,b^2,ad+bc,bd+c^2)$. I want to show that $I$ is $P$-primary where $P=(a,b,c)$. Set $R=k[a,b,c,d]$. Macaulay2 has confirmed this.

Since $\sqrt I=P$, it suffices to show that $I\supseteq IR_P \cap R$, as the right-hand side is by definition the primary component of $I$, which is larger than $I$. So how to show this?

Some observations:

  1. The generators of $I$ form a Grobner basis of $I$, using the lex ordering $a>b>c>d$.
  2. If we consider $d$, which is outside of $P$, to be a constant, then the initial terms of the generators of $I$ do not involve $d$.

I feel like there is some Grobner basis theorem that concludes that I want. Maybe I am wrong as I am new to Grobner bases technique. Any help is appreciated. Thank you.

T C
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1 Answers1

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If we want to obtain an element belonging to the ideal and divisible by $d$, then we get

$$(ad + bc)c - \left( bd + c^2 \right)b = acd - b^2d,$$

that belongs to the ideal I. So if we want to get the element $x$ in the ideal $I$ that were divisible by some element $z$ that does not belong to the ideal $P$ (it is needed, for $x/z$ should belong to the intersection of $IR_P$ and $R$), then $x/z$ will also belong to $I$

Parfig
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    The statement almost proven by @Parfig is $\forall r\in R_P,\exists i\in I.,(ir\in I)$. It's trivial, just take $i=dj$ for $j\in I$ and "denominator" $d$ of $r$. However, the problem is $\forall r\in R_P,\forall i\in I.,(ir\in R\Rightarrow ir\in I)$, which doesn't seem to follow from their answer – te4 Mar 27 '25 at 12:30