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Let $A$ be a ring, $x$ a nonzero element of $A$ and consider the annihilator of $x$, i.e $Ann(x)$.

Now let $S$ denote the collection of all prime ideals of $A$ containing $Ann(x)$. It can be shown using Zorn's lemma that $S$ contains minimal elements (with respect the inclusion $\subseteq$).

My question is about the following argument:

Let $P$ be a minimal prime ideal in the set of all prime ideals that contain $Ann(x)$ and suppose $z \in P$. My question is: why does it follows that $z \in \sqrt{Ann(x)}$ ?

We know that $\sqrt{Ann(x)}$ consists of all prime ideals of $A$ that contain $Ann(x)$.

So let $P$ be a minimal prime ideal in the set of all prime ideals that contain $Ann(x)$ and assume $z \in P$.

Now let $J$ be any prime ideal of $A$ that contains $Ann(x)$. We need to show that $z \in J$.

Since $P$ is minimal with respect the inclusion we can't have that $J$ is properly contained in $P$. Why does it follows then that $P$ must contain $J$ ?

Can you please explain?

amWhy
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user6495
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2 Answers2

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Unless I'm doing something silly, I think the claim is incorrect.

Let $A=\mathbb{Z}/36\mathbb{Z}$, and let $x=6+36\mathbb{Z}$.

The annihilator of $x$ consists of all $m+36\mathbb{Z}$ such that $6m$ is divisible by $36$, equivalently, such that $m$ is divisible by $6$; that is, $\mathrm{Ann}(6+36\mathbb{Z}) = (x)$.

The prime ideals of $A$ are the images of the prime ideals of $\mathbb{Z}$ that contain $36$, namely $(2+36\mathbb{Z})$ and $(3+36\mathbb{Z})$; they are both minimal elements of the set of all prime ideals that contain $(x)$. Now take $z=2+36\mathbb{Z}$. Then $z$ is not in the radical of the annihilator: if $2^n+36\mathbb{Z} \in 6+36\mathbb{Z}$, then $9|2^n-6$ and $4|2^n-6$. The latter requires $n=1$, in which case the former cannot hold.

Arturo Magidin
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  • @Arturo Magidin: thank you, let me refer you to http://dangtuanhiep.files.wordpress.com/2008/09/papaioannoua_solutions_to_atiyah.pdf , page 31, 4.9. I don't understand that part of the solution. – user6495 May 26 '11 at 21:08
  • @user6495: As far as I can tell, the proposed solution is incorrect, confusing "minimal" with "minimum." – Arturo Magidin May 26 '11 at 21:18
  • @Arturo Magidin: yeah that's what I thought , thanks. – user6495 May 26 '11 at 21:19
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    @user6495: The moral is: don't trust Atiyah--MacDonald solution sets that you find on the web! This is not the only one out there with mistakes. (In fact, I don't know of one without at least some mistakes.) Regards, – Matt E May 27 '11 at 09:00
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    @Matt E In fact, I would say that there is a bigger moral: one should not even see Atiyah and Macdonald solution sets let alone trust them! However, this is just my opinion ... – Amitesh Datta Jun 01 '11 at 11:47
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    @Amitesh: Dear Amitesh, I agree, this is probably the best moral to draw! Regards, – Matt E Jun 01 '11 at 13:16
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It doesn't follow and, in fact, your statement is not true. Consider the ring $R=\mathbb{Z}/6\mathbb{Z}$, which has minimal prime ideals $P=(2)$ and $Q=(3)$. Then $P,Q$ are both minimal primes containing ${\rm Ann}(1)=\{0\}$. However, none of $2,4\in P$ or $3\in Q$ are in $\sqrt{{\rm Ann}(1)}=\{0\}$.

George Lowther
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