= Solution
Let $S$ be a <multiplicative subset>, containing one and closed under products. On $A\times S$, use
$$
(a,s)\sim(b,t)\quad\Longleftrightarrow\quad u(ta-sb)=0\text{ for some }u\in S.
$$
The relation is reflexive and symmetric. For transitivity, if $u(ta-sb)=0$ and $v(rb-tc)=0$, then multiplication and addition give $uvt(ra-sc)=0$. Thus it is an <equivalence relation>. Write $a/s$ for the equivalence class. The <localization of a ring> $A_S=S^{-1}A$ has operations
$$
\frac as+\frac bt=\frac{at+bs}{st},\qquad \frac as\frac bt=\frac{ab}{st},\qquad -\frac as=\frac{-a}{s}.
$$
They are well defined. For example, replacing $a/s$ by $b/t$ in its sum with $c/r$ gives cross-numerator difference $r^2(ta-sb)$, and in its product with $c/r$ gives $rc(ta-sb)$. The same annihilating element $u$ kills both differences. Replacing the other input works identically. Associativity and distributivity then follow by putting all terms over the product of their denominators and using the <ring> identities in $A$. Addition and multiplication are commutative, with identities $0/1$ and $1/1$, and the displayed negative is an additive inverse. Hence \b[$S^{-1}A$ is a commutative ring]. If $0\in S$, the fraction relation identifies all elements, producing the zero <ring>.
The canonical <ring homomorphism> $\iota:A\to S^{-1}A$, $a\mapsto a/1$, makes each $s\in S$ a <unit>, with inverse $1/s$. The <universal property of localization> states that any <ring homomorphism> $h:A\to C$ sending $S$ into the <units> factors uniquely through $\iota$. Define the factor by
$$
\boxed{\widetilde h(a/s)=h(a)h(s)^{-1}.}
$$
If $u(ta-sb)=0$, then $h(u)$ is invertible, so $h(t)h(a)=h(s)h(b)$, proving well-definedness. The fraction operations prove that $\widetilde h$ is a <ring homomorphism>. Its value is forced by $a/s=\iota(a)\iota(s)^{-1}$, so it is unique. This property characterizes the <localization> up to a unique compatible <isomorphism>.
For a <prime ideal> $\mathfrak p$, its complement is a <multiplicative subset>, and the <localization at a prime ideal> is \b[$A_{\mathfrak p}=(A\setminus\mathfrak p)^{-1}A$]. Fractions with numerator outside $\mathfrak p$ are <units>; those with numerator in $\mathfrak p$ form the unique <maximal ideal> $\mathfrak pA_{\mathfrak p}$. Thus it is a <local ring>.
For the nilpotent question, suppose $a\ne0$ and $a^n=0$. Its <annihilator> is a proper <ideal>, so it lies in some <maximal ideal> $\mathfrak m$. The fraction criterion says $a/1=0$ in $A_{\mathfrak m}$ only if $sa=0$ for some $s\notin\mathfrak m$, which is impossible by the choice of $\mathfrak m$. Thus $a/1$ is a nonzero <nilpotent element> in that <localization>. This proves <reducedness is detected by prime localizations>: \b[if every prime localization is reduced, $A$ has no nonzero nilpotent elements].
For the domain question take $A=k\times k$ with $k$ a <field>. Its only <prime ideals> are $\mathfrak p_1=0\times k$ and $\mathfrak p_2=k\times0$: a <prime ideal> must contain one of the orthogonal <idempotents> $(1,0),(0,1)$, and the corresponding quotient is $k$. Localizing at $\mathfrak p_1$ makes $(1,0)$ invertible and kills $(0,1)$, giving $A_{\mathfrak p_1}\cong k$; similarly $A_{\mathfrak p_2}\cong k$. But $(1,0)(0,1)=0$ with both factors nonzero. Therefore \b[domain prime localizations do not force $A$ to be an integral domain], as expressed by <domain prime localizations do not imply a domain>.
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