# Annihilating ideals in a ring with finitely many minimal prime ideals.

Let $R$ be a commutative ring with unity and with finitely many minimal primes $P_1 , P_2 , \dots , P_n$ . Then an ideal $I$ is an annihilating ideal if and only if $I$ is contained in at least one minimal prime.

Note: An ideal $I$ is known as an annihilating ideal, if there exists a non-zero ideal $J$ such that $IJ=0$.

I am unable to do the proof. It seems like this statement only holds for reduced rings but in the paper that I am reading the statement is given for any arbitrary ring.

For the forward implication, it is obvious in case of the zero ideal. If we take any non-zero ideal $I$ then I don't know how to proceed, though. I tried to go by contradiction by assuming $I$ not to be contained in any of the minimal primes.

Also for the backward implication , If $I$ is contained in any minimal prime ideal $P_i$ then it will be annihilating if $P_i$ is annihilating but I have read somewhere a note that all the minimal primes need not be annihilating.

• I am reading "The annihilating ideal graph of commutative rings 2" by M Behboodi. You can find this at the end of page 7 Commented Aug 4, 2016 at 11:54

I do not think your claim is correct. Let $R = k[x,y]/(x^2,xy)$. (You may also work with the completion of $R$ at the maximal ideal $(x,y)R$.) Since $(x^2, xy) = (x) \cap (x^2,y)$, there are two associated primes $(x), (x,y)$ and only one minimal prime $(x)$. The ideal $m = (x,y)$ is $(0:x)$. In other words, $x m = 0$ in $R$. But $m$ is not a minimal prime.

I believe that "minimal primes" needs to be replaced by "associated primes".

For the proof, I believe the following observation is helpful thought I mostly work in the Noetherian case, you may want to ask if it works for the general case. (I do not recall the "right" definition for an associated prime in the general case.)

If $IJ =0$, then for any nonzero $x \in J$, $Ix = 0$. In other words, $I \subset \operatorname{Ann}_R x$. Hence $I$ is contained in an associated prime of $R$. Also, remember that (in a Noetherian ring), $P$ is an associated prime if $P = \operatorname{Ann} x$ for some $x$ in $R$.

• Can you tell whether this will be true in case of reduced rings ? Commented Aug 4, 2016 at 11:51
• The paper I have mentioned in the comment is using this result for reduced rings. Commented Aug 4, 2016 at 11:55
• In a reduced Noetherian ring, the set Min(R) and Ass(R) are the same. Commented Aug 4, 2016 at 19:46

Let $R$ be a reduced ring and let $I$ be an annihilating ideal of $R$.

$I$ $\subset$ $Z(R)$ $=$ $\cup_{i=1}^n P_i$

Hence , by Prime avoidance lemma , we have

$I \subset P_i$ for some $i=1,2,....,n$

Conversely let $I \subset P_i$ for some $i=1,2,...., n$.

Since $R$ is a reduced ring , $\cap_{i=1}^n P_i=0$.

Thus $(P_i) (\cap_{j=1,j \neq i}^n P_n) \subset \cap_{j=1}^n P_j=0$ .

This implies $Ann(P_i)=\cap_{j=1,j \neq i}^n P_n$.

Note that $\cap_{j=1,j \neq i}^n P_n \neq 0$ , as if $\cap_{j=1,j \neq i}^n P_n =O \subset P_i$ then

by Theorem 1.11(ii)of the book "Introduction to commutative algebra" by Atiyah and MacDonald

$P_j \subset P_i$ for some j, which is a contradiction.

Thus $0 \neq Ann(P_i) \subset Ann(I)$ implies $I$ is an annihilating ideal.