Suppose $A$ is a reduced commutative ring. Is $A[x]\setminus A$ is multiplicatively closed?
For a reduced ring $A$, must $A[x]\setminus A$ be multiplicatively closed?
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commutative-algebra
1 Answers
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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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