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I shall quote proposition 11.3 of Eisenbud: Commutative algebra

If $R$ is a reduced noetherian ring then an element $x\in K(R)$ belongs to $R$ if and only if the image of $x$ in $K(R)_P$ belongs to $R_P$ for every prime $P$ associated to a nonzero-divisor in $R$. Proof. Suppose that $\frac{a}{u}\in K(R)$, with $a,u\in R$ and $u$ is a nonzero-divisor. If $\frac{a}{u}$ is not contained in $R$, then $a\notin (u)$. And there is an associated prime $P$ of $(u)$ such that $a\notin (u)_P\subset R_P$. Thus, $\frac{a}{u}\notin R_P$.

I understand this proof but cannot figure out why $R$ must be reduced and first thought $R$ does not need to be reduced. Is there any reason $R$ must be reduced?

Thanks.

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    $\begingroup$ The identification $K(R_P) = K(R)_P$ is needed in the statement and the last step of the proof relies on it; this may not hold if $R$ is not reduced, see Exercise 3.15. If $R$ is reduced, then $R_P \subset K(R_P) = K(R)_P \subset K(R)$, so we might just say that $\bigcap_P R_P = R$ for $P$ in the specified range. $\endgroup$
    – Luc Guyot
    Commented May 7, 2019 at 14:30
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    $\begingroup$ As the last inclusion in the chain above is only an inclusion of $R$-modules, i.e., $K(R)_P = \bigoplus_{\mathfrak{p} \subseteq P} K(R/\mathfrak{p}) \subseteq \bigoplus_{\mathfrak{p}} K(R/\mathfrak{p}) = K(R)$, where $\mathfrak{p}$ ranges in the set of minimal primes, the intersection $\bigcap_P R_P$ is to be understood as an intersection of submodules and not of rings. (This is probably why the proposition 11.3 is not phrased in terms of intersection). $\endgroup$
    – Luc Guyot
    Commented May 8, 2019 at 7:47
  • $\begingroup$ Ah, I missed that! Thx! $\endgroup$
    – Bin
    Commented May 8, 2019 at 10:11

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