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Hi, 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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This seems a little too elementary for MO; see the FAQ. Perhaps you should ask on instead. – Qiaochu Yuan Feb 3 '11 at 17:10
Note that this question has been asked -- and answered, and the answer has been accepted -- on math.SE. – Pete L. Clark Feb 3 '11 at 18:46
Hint: In a reduced ring, the intersection of all prime ideals is $0$. – Martin Brandenburg Feb 3 '11 at 21:29
@OP: by sum you probably mean product. – Qing Liu Feb 7 '11 at 14:21

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