1

I have an exercise about the properties of primary ideal. It's Exercise 15.17 of "Step in commutative algebra", R. Y. Sharp.

Let $(A,\mathfrak{m})$ be a local ring and $I$ be a proper ideal of $A$. Prove that the following statements are equivalent:

  1. $I$ is $\mathfrak{m}$-primary;
  2. $\operatorname{Var}(I)=\{\mathfrak{m}\}$;
  3. $\operatorname{Ass}(I)=\{\mathfrak{m}\}$;
  4. $\operatorname{length}(A/I)<+\infty$;
  5. $\sqrt{I}=\mathfrak{m}$;
  6. There exist $h\in\mathbb{N}$ such that $\mathfrak{m}^h\subseteq I$.

I can prove that: $(1)\Leftrightarrow (2), (1)\Leftrightarrow (5)$. I think in statement $(3)$ it must be: $\operatorname{Ass}(A/I)=\{\mathfrak{m}\}$. If that, $(1)\Leftrightarrow (3)$ is ok. So please check for me about statement $(3)$. And help me solve statements $(4)$ and $(6)$.

Thanks!

Rachel
  • 529

1 Answers1

1

Because Sharp's book isn't my favorite I don't know what he denotes by $\operatorname{Ass}(I)$, but if you are used with the notation for modules then you are right, it must be $\operatorname{Ass}_A(A/I)$.

If $\mathfrak m^h\subseteq I$, then $\sqrt I=\mathfrak m$, so $I$ is $\mathfrak m$-primary. Conversely, if $I$ is $\mathfrak m$-primary then $\sqrt I=\mathfrak m$, so every generator of $\mathfrak m$ at some power is in $I$, and thus $\mathfrak m$ at some power is contained in $I$.

$\operatorname{length}_A(A/I)<\infty\Leftrightarrow A/I$ artinian, and therefore $\mathfrak m/I$ at some power is $(0)$, so $\mathfrak m$ at some power is contained in $I$. One can also prove the converse, but I'll take an easier way showing that $(5)\Rightarrow(4)$: if $\sqrt I=\mathfrak m$, then the only prime contining $I$ is $\mathfrak m$, so $A/I$ is a noetherian ring of Krull dimension zero, that is, artinian.

user26857
  • 52,094
  • If $I$ is $\mathfrak{m}$-primary then $\sqrt I=\mathfrak{m}$, so $\forall x\in\mathfrak{m}: (x)=\mathfrak{m}, \exists n \in\mathbb{N}: x^n\in I$. So $\exists h\in\mathbb{N}: \mathfrak{m}^h\subseteq I$. Right? And please show me $(2)\Rightarrow (1)$. Thanks again! – Rachel Sep 30 '14 at 03:47
  • @Rachel If $x\in\mathfrak m$, then $x\in\sqrt I$ and by definition there is $h\ge 1$ such that $x^h\in I$. There is no reason to think that $(x)=\mathfrak m$! But $\mathfrak m$ is finitely generated, so every generator at some power is in $I$ and since any element of $\mathfrak m$ is a linear combination of generators it follows that there is $t\ge 1$ such that any element of $\mathfrak m$ at the $t$th power is in $I$, so $\mathfrak m^t\subseteq I$. – user26857 Sep 30 '14 at 08:56
  • $(2)\Rightarrow (1)$: The only prime ideal containing $I$ is $\mathfrak m$. But the radical of $I$ is the intersection of all primes containing $I$, that is, $\mathfrak m$. It follows that $I$ is $\mathfrak m$-primary. (Actually, (1) and (5) are trivially equivalent.) – user26857 Sep 30 '14 at 08:59
  • Ok, the last thing i can't understand: why $\mathfrak{m}$ is finitely generated? – Rachel Sep 30 '14 at 09:13
  • @Rachel Your ring should be noetherian, otherwise it is called quasi-local; see Definition 8.26. – user26857 Sep 30 '14 at 09:31