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Suppose $A$ is a reduced commutative ring. Is $A[x]\setminus A$ is multiplicatively closed?

rschwieb
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messi
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1 Answers1

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Let $A$ be any reduced commutative ring with nonzero zero divisors, say $ab=0$ for some nonzero $a$ and nonzero $b$.

Since $(ax)(bx)=0$, $A[x]\setminus A$ is not multiplicatively closed.

So, you need to assume $A$ is at least a domain. Once that is true, then $A[x]\setminus A$ is obviously multiplicatively closed: just consider the degrees of polynomials in question!

rschwieb
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  • great, thanks. That was a nice counterexample, it was quite simple. – messi Jun 06 '12 at 00:22
  • @zac Glad you liked it... and don't forget to accept answers you like in the next couple of days. It helps show people that you are worth spending the time on. – rschwieb Jun 06 '12 at 00:24
  • Btw i was trying to generalize the question "Reduced ring is integrally closed in polynomial ring" which you answered. But again you gave a cute counterexample. – messi Jun 06 '12 at 00:56
  • @zac Frustratingly, the questions I've asked have no answers or I've already accepted answers, so I can't check for sure. I believe you should be able to click a checkmark next to my answer, and turn it green. – rschwieb Jun 06 '12 at 01:00