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

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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!

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    @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.2012-06-06