# Imbedding of a reduced ring into a direct sum - Why reduced?

This pertains to Ex. 1.13 (self-studier) in Reid's "Undergrad. Commutative Algebra":

If $A$ is a reduced ring and has finitely many minimal prime ideals $P_i$ then

$A\hookrightarrow \bigoplus_{i=1}^n A/P_i$; moreover, the image has nonzero intersection with each summand.

I "know" how to solve the problem, showing that the kernel of the map is zero, and designating $a_i \notin P_i$ while in $\bigcap P_j$ with $j\not= i$.

This has naively ignored the stipulation of "a reduced ring," which is my question:

What would the presence of nilpotents in $A$ do? My guess is that it would make the kernel of the map non-trivial. If that is correct, I would please appreciate help in seeing how this comes about.

If that guess is not correct, then I would again appreciate even more help.

Thanks.

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Your guess is correct. The nilradical of a (commutative) ring $A$ (with identity) is the intersection of all the prime ideals. But because every prime ideal of $A$ contains a minimal prime, by Zorn's Lemma, the nilradical is actually the intersection of the minimal primes. So the kernel of the map $A\rightarrow\prod_{i\in I}A/\mathfrak{p}_i$, where the $\mathfrak{p}_i$ are the minimal primes, is exactly the nilradical. So it is zero if and only if $A$ is reduced.
@Andrew This is a great answer, but I wanted to pull out the critical part for your question: you can see that if $x$ is nilpotent, then $x$ is in every prime ideal, right? That would mean $x$ i mapped to zero in the mapping you have chosen. – rschwieb Oct 26 '12 at 18:03
@andrew If $a^n=0\in P$, then $a\in P$... use the definition of primeness and induction. – rschwieb Oct 26 '12 at 18:03