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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 isConcretely, its elements are equivalence classes of pairs $(r,s)\in R\times T$, written $r/s$, withThis extra multiplier is essential when $R$ has zero divisors. Addition and multiplication areThe 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 $P{R}_P$. Reduction of numerators and denominators gives a surjective ring homomorphism $R_P\to \mathrm{Frac}(R/P)$ with kernel $P{R}_P$, 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.

The equivalence relation defining the localization at a prime ideal givesThereforeEquivalently, all zero divisors of $R$ must lie in $P$. To prove necessity, if $s\notin P$ and $sr=0$ with $r\ne 0$, then $r/1=0$ in $R_P$, contradicting the hypothesis that the map is injective. Conversely, if every denominator annihilates no nonzero element, the displayed kernel of a ring homomorphism is zero, so the map is injective.

An integral domain automatically satisfies this criterion at every prime ideal, but the criterion itself does not require $R$ to be an integral domain.

Yes: $R$ can have nonzero zero divisors. Let $R=k\times k$, a direct product of rings, where $k$ is any field. The nonzero idempotents $e_1=(1,0)$ and $e_2=(0,1)$ satisfy $e_1{e}_2=0$, so both are zero divisors.

There are exactly two prime ideals:Indeed, any prime ideal must contain $e_1$ or $e_2$, since their product is zero. If it contains $e_1$, it contains $k\times 0$ and corresponds to a prime ideal of the quotient ring $k$, whose only prime ideal is zero; hence it equals $P_2$. The other case gives $P_1$.

For the localization at a prime ideal $P_1$, every denominator $(u,v)\notin P_1$ has $u\ne 0$, andis an isomorphism. It is surjective using constant first-coordinate fractions. It is injective because a numerator with first coordinate zero is annihilated by $e_1\notin P_1$, so its fraction is zero. Interchanging the coordinates proves $R_{P_2}\cong k$.

Thus the localization at a prime ideal is a field in every case, although $R$ has zero divisors. The obstruction to using the previous part's argument is that the two nonzero factors of a zero product can survive at different prime ideals.

False. In the polynomial ring $R=k[x,y]$ over a field, takeThe ideal $P$ is a maximal ideal, with quotient ring $R/P\cong k$. AlsoIn this last ring, an element with nonzero constant term is a unit, and an element with zero constant term is a multiple of $y$ and has square zero. Thus every zero divisor is a nilpotent element. This is equivalent to $I$ being a primary ideal: a zero product in the quotient ring with one nonzero factor forces the other factor to be nilpotent. Its radical of an ideal is $P$, so $I$ is $P$-primary.

However, $y\notin I$, so $I\ne P$. Moreover $x\in I$ while $x\notin P^2$, since every polynomial in $P^2$ has no term of total degree one. For every $n\ge 2$, $P^n\subseteq P^2$, so $I\ne P^n$. Hence

First suppose $X$ is integral. Induct on its dimension of a scheme. The previous two parts handle all $j>1$ and also handle $j=1$ when $h^0(X,mD)$ eventually vanishes. Otherwise choose $r>0$ with a nonzero global section of $rD$. On an integral variety it defines an effective Cartier divisor $E$, possibly empty, and multiplication by the section givesThus $h^1(X,mD)\le h^1(X,(m-r)D)+h^1(E,mD)$. For $n\ge 2$, the induction hypothesis bounds the last term by $O(m^{n-2})$; summing on each residue class modulo $r$ gives $O(m^{n-1})$. For $n=1$, $E$ is zero-dimensional and its positive-degree sheaf cohomology vanishes, so the same recurrence is bounded.

For a general projective scheme, a nonzero section can be a zero divisor, so that argument requires an additional step. The general cohomology growth for nef twists supplies it: for every coherent sheaf $\mathcal{F}$ with support dimension $d$,Its proof uses Fujita vanishing, an ample section avoiding the associated points of $\mathcal{F}$, and induction on support dimension. Taking $\mathcal{F}={\mathcal{O}}_X$ gives the required $O(m^{n-1})$ estimate for every $j\ge 1$, including nonreduced and reducible schemes; degrees above $n$ vanish.

Pass to $A=R/I$. It is enough to prove that the zero ideal of $A[t]$ is primary. The zero ideal of $A$ is primary, so every zero divisor of $A$ is nilpotent element.

Suppose $fg=0$ in $A[t]$ with $g\ne 0$. By McCoy theorem, some nonzero $a\in A$ satisfies $af=0$. Hence every coefficient of $f$ is a zero divisor and therefore nilpotent. There are only finitely many coefficients, so the ideal they generate is nilpotent; consequently some power of $f$ is zero. This proves that $(0)$ is primary in $A[t]$, and the coefficientwise quotient of a polynomial ringshows that $I^e$ is primary.