= Solution
For a <commutative ring> $R$ and a <multiplicative subset> $S$ containing $1$, the <universal property of localization> specifies a pair $(R_S,\iota)$ with $\iota:R\to R_S$ a unital <ring homomorphism>, such that every $\iota(s)$, $s\in S$, is a <unit>, and any unital <ring homomorphism> $f:R\to T$ into a commutative <ring> that sends $S$ to <units> factors uniquely through $\iota$. Thus there is a unique $\widetilde f:R_S\to T$ satisfying $\widetilde f\iota=f$. The property determines the pair up to a unique compatible <isomorphism>.
For existence, take pairs $(r,s)\in R\times S$ and declare
$$
(r,s)\sim(r',s')\quad\Longleftrightarrow\quad u(s'r-sr')=0\text{ for some }u\in S.
$$
The extra multiplier is essential when $R$ has <zero divisors>. This is an <equivalence relation>: for transitivity, multiply the two cross-multiplication equations by appropriate denominators and add them, using the product of their annihilating multipliers. Write the class as $r/s$ and define
$$
\frac rs+\frac{r'}{s'}=\frac{rs'+r's}{ss'},\qquad\frac rs\frac{r'}{s'}=\frac{rr'}{ss'},\qquad\iota(r)=\frac r1.
$$
Multiplying the defining equivalence equations verifies that these operations are well defined. They give a <commutative ring>, and $s/1$ has inverse $1/s$. The required map is necessarily
$$
\widetilde f(r/s)=f(r)f(s)^{-1}.
$$
It is well defined because applying $f$ to $u(s'r-sr')=0$ and cancelling the units $f(u),f(s),f(s')$ equates the two images. Its addition and multiplication laws follow directly from the fraction formulas. This establishes existence and the <universal property of localization>. If $0\in S$, this construction gives the zero <ring>, which is allowed as a localization.
For a <prime ideal> $P$, its complement $S=R\setminus P$ is multiplicatively closed, and <localization at a prime ideal> means $R_P=S^{-1}R$. The map
$$
R_P\longrightarrow\operatorname{Frac}(R/P),\qquad r/s\longmapsto\overline r/\overline s
$$
is well defined and surjective, with <kernel of a ring homomorphism> $PR_P$: a numerator has zero image exactly when it belongs to $P$. Since $R/P$ is an <integral domain>, its <field of fractions> is a nonzero <field>, so $PR_P$ is a proper <maximal ideal>. Every fraction $r/s$ with $r\notin P$ is a <unit>, with inverse $s/r$. Every fraction with $r\in P$ belongs to this proper ideal and cannot be a <unit>. Thus the nonunits are exactly $PR_P$, and every proper ideal lies within it. Consequently
$$
\boxed{R_P\text{ is a local ring with unique maximal ideal }PR_P.}
$$
Now suppose each $R_P$ is <reduced>. If $x\in R$ is <nilpotent>, its image $x/1$ is <nilpotent> in every $R_P$, hence is zero. If $x\ne0$, its <annihilator> $\operatorname{Ann}_R(x)$ is a proper ideal, so choose a <maximal ideal> $P$ containing it. Maximal ideals in a <commutative ring> are <prime ideals>. The <vanishing criterion in a module localization> would give $sx=0$ for some $s\notin P$. But then $s\in\operatorname{Ann}_R(x)\subseteq P$, a contradiction. Hence \b[$R$ is reduced]. This is <reducedness is detected by prime localizations>; no finiteness hypothesis is needed.
The stronger local-domain condition does not force $R$ to be an <integral domain>. Take $R=k\times k$ for a <field> $k$. Its only <prime ideals> are $P_1=0\times k$ and $P_2=k\times0$: a <prime ideal> must contain one of the orthogonal <idempotents> $(1,0)$ and $(0,1)$, and the remaining quotient is the <field> $k$. At $P_1$, inverting $(1,0)$ kills the second component, so $R_{P_1}\cong k$; similarly $R_{P_2}\cong k$. Both localizations are <integral domains>, but
$$
(1,0)(0,1)=0
$$
is a product of nonzero elements in $R$. Therefore \b[the answer is no], as illustrated by <domain prime localizations do not imply a domain>.
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