Primary ideal exercise

Question

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!

Answer 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.