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Let $R$ be a reduced ring and $T(R)$ the total ring of fractions of $R$ (i.e. localizing $R$ at nonzerodivisors).

Any element of $T(R)$ maps naturally to an element of $T(R/P)$ since $a/b \in T(R)$ projects to $a + P / b+ P$ and $b \notin P$ since it is not a zerodivisor.

I wonder the following

If the projection of $a/b \in T(R)$ is in $R/P$ for every minimal prime $P$, then is $a/b$ necessarily in $R$?

An easy argument shows that $a^n /b \in R$ for some $R$ necessarily, but I can't see why or why not $a/b$ itself would need to be in $R$.

I guess essentially I am asking for insight into why we can or cannot understand divisibility relations of reduced rings through their factor domains.

the.myth.of.math.and.meth
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1 Answers1

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Here's a counterexample. Let $k$ be a field and $R=k[x,y]/(xy)$, whose two minimal primes are $P=(x)$ and $Q=(y)$. The element $\frac{x+y^2}{x+y}\in T(R)$ is in both $R/P$ and $R/Q$ after projecting: mod $P$ it reduces to $\frac{y^2}{y}=y$ and mod $Q$ it reduces to $\frac{x}{x}=1$. However, $x+y$ does not divide $x+y^2$ in $R$ (if $(x+y)f(x,y)=x+y^2$ mod $xy$, then $f$ must have a constant term of $1$ to get the $x$ term in the product but then there would also be a $y$ term in the product).

Eric Wofsey
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