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 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 $I^c=(r_1,\dots ,r_m)$; their images generate $I$. This proves that $R_P$ is Noetherian.

The extended ideal $P{R}_P$ 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 $P{R}_P$ and cannot be invertible. Thus the nonunits are precisely $P{R}_P$, making it the unique maximal ideal. ThereforeTo 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 $x\ne 0$, its annihilator $\mathrm{Ann}(x)$ 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 $\mathrm{Ann}(x)\subseteq Q$, a contradiction. Hence $x=0$, and $R$ is semiprime. This argument does not need Noetherianity.

Primeness is not detected in the same way. Take $R=k\times k$. Its two prime ideals are $P_1=0\times k$ and $P_2=k\times 0$, 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: $e_1,e_2\ne 0$ and $e_1{e}_2=0$.