Characterization of Reduced rings I'm trying to work through Eisenbud, Exercise 11.10, which states:

A Noetherian ring R is reduced iff

*

*The localization of R at each prime of codimension 0 is regular, and

*all primes associated to 0 have codimension 0


I know I should put an attempt of the problem here but I really don't know where to start. The best I can do is state some facts that I know that might be useful:
a) Saying that a prime $p$ in $R$ has codimension $0$ is the same as saying that dim $R_p$ is $0$.
b) Primes associated to $0$ are primes in $\operatorname{Ann}_R(0)$. I Don't know what's special about codimension 0 here. I guess we could also use the fact (a) above.
I'm not sure how to proceed.
 A: First let me mention a helpful characterization of codimension 0 (= minimal) primes of a ring:
A prime $P$ is minimal iff for each $r \in P$ there exists $s \notin P$ such that $sr$ is nilpotent.
This is a good exercise which I leave to you.
First we check that a reduced Noetherian ring has $(1)$ and$ (2)$. $(1)$ is straightforward because for a reduced ring $R$ also $R_P$ is reduced for each prime $P$. Then for a minimal prime $P$, $R_P$ is a reduced ring with exactly one prime, which you can check is a field, hence regular.
For $(2)$, recall that an associated prime $P$ of a Noetherian ring is defined as a prime such that $rP = 0$ for some $r \not= 0$. In a reduced ring this entails $r \notin P$, and so $P$ is minimal by the above characterization of minimal primes. This shows $(2)$.
Next we show that $(1)$ and $(2)$ together imply reducedness.
A regular local ring of dimension $0$ is a field, so (1) says that $R_P$ is a field for every minimal prime $P$.
Since $R$ is Noetherian we get a primary decomposition $\bigcap Q_i = 0$.  We don't even need to care here that it's a "minimal" decomposition.  Just note that the radicals $P_i = \sqrt{Q_i}$ are all associated primes of $R$, which are all minimal by $(2)$.
Let $r \in R$ be nilpotent. For each minimal prime $P_i$, find $s_i \notin P_i$ such that $s_ir = 0$, using the fact that $r_i$ maps to zero in each of the localizations at the $P_i$, which are fields by $(1)$.  Thus we get a finitely generated ideal $I$ such that $I$ is not contained in any $P_i$ and $rI = 0$.
But then it is straightforward to check that $r \in Q_i$ for all $i$ (else you'd get $I \subseteq P_i$ for some $i$, absurd), so conclude $r = 0$.
Looking back at the proof, we have just showed: If a ring $R$ has a finite primary decomposition $\bigcap Q_i = (0)$ and for each $P_i = \sqrt{Q_i}$ the localization $R_{P_i}$ is reduced, then $R$ is reduced.
For me the takeaway of this exercise is: In general, a ring is reduced iff its localization at every prime is reduced. If the ring $R$ has a primary decomposition of $(0)$, then it suffices to check reducedness at the finite set of localizations associated to that decomposition.  Moreover, this is the case for all Noetherian rings.
