= Solution
In a commutative <ring>, if $x^m=0$, then the principal <ideal> $(x)$ satisfies $(x)^m=0$. Thus a <semiprime ring> has no nonzero <nilpotent elements>. Conversely, if $I^m=0$ for an <ideal> $I$, then every $x\in I$ satisfies $x^m=0$; absence of nonzero <nilpotent elements> forces $I=0$. Hence \b[commutative semiprimeness is exactly reducedness].
For a <prime ideal> $P$, let $S=R\setminus P$. Every <ideal> $I$ of $R_P=S^{-1}R$ is the extension of its contraction $I^c$: if $r/s\in I$, multiply by the <unit> $s/1$ to get $r/1\in I$, so $r\in I^c$. If $R$ is <Noetherian>, write $I^c=(r_1,\ldots,r_m)$; their images generate $I$. This proves that $R_P$ is <Noetherian>.
The extended <ideal> $PR_P$ is proper. Indeed $1=p/s$ would imply $u(s-p)=0$ for some $u\notin P$, and then $us=up\in P$, contradicting primeness and $u,s\notin P$. A fraction with numerator outside $P$ is invertible, with inverse $s/r$. A fraction with numerator in $P$ lies in the proper <ideal> $PR_P$ and cannot be invertible. Thus the nonunits are precisely $PR_P$, making it the unique maximal <ideal>. Therefore
$$
\boxed{R_P\text{ is a Noetherian local ring with maximal ideal }PR_P.}
$$
To prove reducedness is detected by prime localizations, let $x$ be a <nilpotent element> of $R$. Its image in every $R_Q$ is zero because these localizations are semiprime, hence reduced. If $x\ne0$, its <annihilator> $\operatorname{Ann}(x)$ is a proper <ideal>; choose a maximal <ideal> $Q$ containing it. The zero criterion for <localization> says $x/1=0$ in $R_Q$ only if $sx=0$ for some $s\notin Q$, which would put $s$ in $\operatorname{Ann}(x)\subseteq Q$, a contradiction. Hence $x=0$, and $R$ is semiprime. This argument does not need Noetherianity.
Primeness is not detected in the same way. Take \b[$R=k\times k$]. Its two <prime ideals> are $P_1=0\times k$ and $P_2=k\times0$, and its corresponding localizations are both isomorphic to the <field> $k$. To see these are all the <prime ideals>, the orthogonal <idempotents> $e_1,e_2$ have product zero, so a <prime ideal> contains one of them and is then the kernel of the other projection. Both localizations are prime <rings>, but $R$ is not prime: $e_1,e_2\ne0$ and $e_1e_2=0$.
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