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the question is exactly "why a reduced ring (commutative with 1) can be embedded into a sum of integral rings?" Is this simply because in the normalization process we can have many irreducible component?

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up vote 6 down vote accepted

Consider the obvious map from your ring $R$ to the direct product $\prod_{\mathfrak p}R/\mathfrak p$ of all the quotients of $R$ by all its prime ideals $\mathfrak p$. The kernel is zero precisely because the ring is reduced.

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Just a comment to supplement Mariano's answer (left as an answer for ease of formatting):

There is a (very) basic commutative algebra fact which is being taken as known here, namely that in any commutative ring, the intersection of all prime ideals -- called the nil radical of $R$ -- is also equal to the set of all nilpotent elements of $R$. Among infinitely many other places, see Proposition 53 in Section 4 of these notes for a proof.

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