I am currently working my way through Q. Liu's book "Algebraic Geometry and Arithmetic Curves". I'm puzzled by the proof that every projective morphism is proper, see below
I understand that $(B/I)_m\otimes_A\mathcal{O}_{Y,y}=0$, but I do not understand why
This implies that there exists an $f\in A$ such that $y\in D(f)$ and $f\cdot(B/I)_m=0$
Can anyone expand on how this is implied?
If $A$ is a commutative ring, $M$ is a finitely generated $A$-module, $\mathfrak{p} \subseteq A$ a prime ideal with $M_\mathfrak{p}=0$, then there is some $f \in A \setminus \mathfrak{p}$ with $f M = 0$. Namely, if $M = \langle m_1,\dotsc,m_n\rangle$ and $f_i \in A \setminus \mathfrak{p}$ satisfy $f_i m_i=0$, then $f=\prod_i f_i$ does the job.
This can also be rephrased as the equality of subsets $\mathrm{supp}(M)=V(\mathrm{Ann}(M))$ of $\mathrm{Spec}(A)$.