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Prove that a direct product of semiprime rings is a semiprime ring. Is it true for prime rings?

Help me a hint.

Thank for any insight.

Truong
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1 Answers1

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One of many equivalent definitions of "prime ring" is that for any two ideals $I,J$ of the ring, $IJ=0$ implies that $I=\{0\}$ or $J=\{0\}$.

Let $R$ and $S$ be nonzero rings and look at $R\times S$. Can you find a pair of nonzero ideals whose product is zero?

To prove that a product of finitely many semiprime rings is semiprime, you can use the following characterization of semiprime rings: a semiprime ring is one with no nonzero nilpotent right ideals.

You already know (or can easily prove) that the right ideals of $\prod_{i=1}^n R_i$ are of the form $\prod_{i=1}^n T_i$ where $T_i$ is a right ideal of $R_i$. Consider what it would mean for a right ideal of the product to be nilpotent.

rschwieb
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  • $I\times\left{ 0\right} $ and $\left{ 0\right} \times J $. Is it true? – Truong Nov 25 '13 at 15:08
  • @chuyenvien94 Lacking psychic powers, I don't know what you what $I,J$ to be :) Relate it to $R\times S$. – rschwieb Nov 25 '13 at 15:26
  • Can you make your hint more precise? – Truong Nov 25 '13 at 15:28
  • @chuyenvien94 No, I don't think that's necessary. Think about it! What you wrote suggests you have the right idea already... You just haven't explained it properly yet. – rschwieb Nov 25 '13 at 15:30
  • $I$ is an ideal of $R$ and $J$ is an ideal of $S$ – Truong Nov 25 '13 at 15:34
  • @chuyenvien94 If you pick $I$ and $J$ to be both zero ideals, then you haven't proven anything. Can you pick some more carefully? – rschwieb Nov 25 '13 at 15:35
  • I take Ix0 and Jx0 with I,J are nonzero ideals of R,S resp. – Truong Nov 25 '13 at 15:38
  • @chuyenvien94 OK: close enough! Really you should just say $R\times 0$ and $0\times S$. Those are the only nonzero ideals that you can count on. If $R$ and $S$ are simple, there is no other choice. – rschwieb Nov 25 '13 at 15:44
  • @chuyenvien94 Thanks for patiently working on it :) – rschwieb Nov 25 '13 at 15:46
  • You told "prime ring"? Evidently, direct product of prime rings can not be prime ring. It's only true for semiprime – Truong Nov 25 '13 at 15:51
  • @chuyenvien94 That's correct! You've shown the product of two nonzero rings can't ever be a prime ring. Semiprime rings are safe, though. Semiprime rings are exactly those which have no nonzero nilpotent right ideals. You can think about this and discover why a product of finitely many semiprime rings also has this property. I hope you're aware that the ideals of the direct product are just direct products of ideals of the individual rings. – rschwieb Nov 25 '13 at 15:54
  • @chuyenvien94 Sorry, I focused only on the second half of the question initially. I added more for the first half. – rschwieb Nov 25 '13 at 16:01