= Solution
{c}
Let $T=R\setminus P$. Since $P$ is a <prime ideal>, $T$ is a <multiplicative subset> containing $1$ and excluding $0$. The <localization at a prime ideal> is
$$
\boxed{R_P=T^{-1}R.}
$$
Concretely, its elements are equivalence classes of pairs $(r,s)\in R\times T$, written $r/s$, with
$$
\frac r s=\frac {r'}{s'}\quad\Longleftrightarrow\quad
u(s'r-sr')=0\text{ for some }u\in T.
$$
This extra multiplier is essential when $R$ has <zero divisors>. Addition and multiplication are
$$
\frac r s+\frac {r'}{s'}=\frac{rs'+r's}{ss'},\qquad
\frac r s\frac {r'}{s'}=\frac{rr'}{ss'}.
$$
The equivalence criterion is compatible with these operations: after multiplying by the multipliers witnessing changes of representatives, the corresponding numerators agree. The <commutative ring> axioms follow from those of $R$; the identity is $1/1$.
The canonical <ring homomorphism> $\iota:R\to R_P$, $r\mapsto r/1$, supplies the <algebra over a commutative ring> structure, with scalar action $a(r/s)=ar/s$. It also has the <universal property of localization>: whenever a <ring homomorphism> $\psi:R\to A$ sends $T$ to <units>, the unique extension is $r/s\mapsto\psi(r)\psi(s)^{-1}$.
For later use, $R_P$ is a <local ring> with <maximal ideal> $PR_P$. Reduction of numerators and denominators gives a surjective <ring homomorphism> $R_P\to\operatorname{Frac}(R/P)$ with kernel $PR_P$, so this <ideal> is proper and maximal. Every fraction outside it has numerator outside $P$ and is a <unit>, with inverse $s/r$. Thus it is the unique <maximal ideal>.
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