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Is this statment always true? $$(a)\subsetneq (b)\Rightarrow \text{Ann}_R b\subsetneq \text{Ann}_R a$$

If it is false, can you please provide an example? Also what is the largest class of rings that make the above statement is true.

UPDATE: to make it more interesting, let $a,b$ be zero divisors.


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2 Answers 2

up vote 4 down vote accepted

Let's suppose our ring $R$ is reduced, i.e. has no non-zero nilpotents (and commutative with identity).

Then the annihilator of any element is a radical ideal (if $x^n a = 0,$ then $(xa)^n = 0,$ thus $xa = 0$).

Hence the annihilator of $a$ is precisely the intersection of those prime ideals $\mathfrak p$ in the support of $a$, i.e. such that $a$ has non-zero image in the localization $R_{\mathfrak p}$.

Now suppose also that $R$ is Noetherian. Then the support of any element is a union of irreducible components of Spec $R$. (This is a general property of reduced Noetherian rings.) So your question amounts to asking: under what conditions on $R$ (now assumed Noetherian and reduced) does one have that any two elements $a$ and $b$ supported on the same set of components of Spec $R$ generate the same ideal.

Since $a$ and $a^2$ always have the same support, one finds that $a$ and $a^2$ would have to generate the same ideal for all $a \in R$. From this it is not hard to deduce that $R$ is forced to be a finite product of fields.

So at least for reduced Noetherian rings, your desired property holds only if the ring is a product of fields. (It is certainly a very restrictive property.)

General remark: One can often gain insight into these kinds of questions by working geometrically on the Spec of the ring, where you can "see" the meaning of the various algebraic concepts in geometric terms.

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Let $R=\mathbb{Z}$, which is an integral domain. We have $(4)\subsetneq (2)$, but $\text{Ann}(2)=(0)=\text{Ann}(4)$.

As requested, here is an example where $a$ and $b$ are zero-divisors: $(2x)\subsetneq(2)$ in $(\mathbb{Z}/4\mathbb{Z})[x]$, but $\text{Ann}(2x)=(2)=\text{Ann}(2)$.

I think I've distilled what worked about the example above. Let $R$ be any ring with at least one non-zero zero-divisor $r\in R$ and at least one non-zero-divisor, non-unit $s\in R$. Then $x(rs)=(xr)s=0$ if and only if $xr=0$, so that $\text{Ann}(rs)=\text{Ann}(r)$, but $$(rs)=(r)(s)\subsetneq (r)(1)=(r).$$ If $R$ is not of the above kind, either there are no zero-divisors (i.e. $R$ is an integral domain), or every element is either a zero-divisor or a unit (i.e. $R$ is some total ring of fractions). So any ring $R$ that isn't one of these things will work. Though actually if $R$ is an integral domain then I suppose the statement that $(a)\subsetneq (b)\Rightarrow\text{Ann}(b)\subsetneq\text{Ann}(a)$ for all zero-divisors $a$ and $b$ is vacuously true.

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Thanks Zev. Any example when both $a$ and $b$ are zero divisors? –  user9077 Apr 22 '11 at 6:31
@user9077: I've added one. –  Zev Chonoles Apr 22 '11 at 6:40
Dear Zev, The claim that $(rs) \neq (r)$ in your final displayed equation is not true in general. E.g. suppose that $R = \mathbb Z \times \mathbb Z,$ that $r = (2,0),$ and that $s = (1,2)$. Then $r$ is non-zero zero-divisor, while $s$ is a non-zero divisor non-unit. Note that $rs = r$, and so also $(rs) = (r)$. Regards, –  Matt E Apr 28 '11 at 4:42
@Matt E: You're right, that doesn't quite work. Is there anything further I can assume about $R$ or $s$ to make the statement $(rs)\subsetneq (r)$ necessarily true? –  Zev Chonoles Apr 28 '11 at 4:46

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