Past exam of the commutative algebra course of the University of Oxford / 2023 / 2 / b

Choose $f=1$, so that $R_f=R$ and $f\notin \mathfrak{p}$. For an integral domain, the canonical ring homomorphism $R\to R_{\mathfrak{p}}$ is injective: if $a/1=0$, the vanishing criterion in a module localization gives $sa=0$ for some $s\notin \mathfrak{p}$. Such an $s$ is nonzero, so the absence of zero divisors gives $a=0$. Thus

Consider the kernel of a ring homomorphismSince $R$ is a Noetherian ring, this ideal has a finite generating set of an ideal, say $K=(a_1,\dots ,a_t)$. For each $i$, choose $s_i\notin \mathfrak{p}$ with $s_i{a}_i=0$, and putThen $f\notin \mathfrak{p}$ by the prime ideal condition, and $fK=0$. If $K=0$, take the empty product $f=1$.

If $a/f^n$ maps to zero under the ring homomorphism ${\theta }_f$ from a, then $a/1=0$ in $R_{\mathfrak{p}}$ because $f/1$ is a unit there. Hence $a\in K$, so $fa=0$. The vanishing criterion in a module localization now gives $a/f^n=0$ already in $R_f$. ThereforeThis is the finite generation clears localization kernels argument: the finite generating set of an ideal permits a single element outside $\mathfrak{p}$ to annihilate the entire kernel of a ring homomorphism.