I shall quote proposition 11.3 of Eisenbud:Commutative algebra

If $R$ is a reduced notherian 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.

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.