In proof of the properness of projective space I'm reading the Gortz's Algebraic Geometry, Theorem 13.40, and stuck at understanding some statement :


(If needed, I'll upload more details for each step in the proof.)
Why the underlined statement is true?
My first attempt is as follows :
For the underlined statement, it suffices to show that $D(h) \subseteq S-f(Z)$. Let $ \mathfrak{q} \in D(h)$. We show that $\mathfrak{q} \in S-f(Z)$. Note that by reversing the argument of the thrid paragraph of the proof of Theorem 13.40, it suffices to show that $(A_d/I_d)\otimes_R \kappa(\mathfrak{q}) = (A_d/I_d)\otimes_R \operatorname{Frac}(R/\mathfrak{q})=0$ ( Or, C.f. Liu's Algebraic Geometry and arithmetic curves, p.108, Theorem 3.30 ). Note that since $\mathfrak{q} \in D(h)$, $\mathfrak{q} \not\ni h $. So, $[h] := h +\mathfrak{q}$ is invertible in $\operatorname{Frac}(R/\mathfrak{q})$. Now fix an elementary element $(p(T_0, \cdots , T_n) + I_d) \otimes \frac{r+q}{r'+q} \in (A_d/I_d)\otimes_R \operatorname{Frac}(R/\mathfrak{q})$. Then note that
$$ (p(T) + I_d) \otimes \frac{r+\mathfrak{q}}{r'+\mathfrak{q}} = (p(T) + I_d ) \otimes \frac{[h][r]}{[h][r']} \overset{\mathrm{?}}{=} (h+I_d)(p(T) + I_d) \otimes \frac{[r]}{[h][r']} = (h \cdot p(T) + I_d) \otimes \frac{[r]}{[h][r']} \overset{\mathrm{ hA_d \subseteq I_d}}{=} ( 0 + I_d ) \otimes \frac{[r]}{[h][r']}=0$$
So $(A_d/I_d)\otimes_R \operatorname{Frac}(R/\mathfrak{q})=0$. But the equality marked by question mark is really true? If so, why? I am not sure certainly it is.
Or is there any ohter method to prove that $D(h) \subseteq S-f(Z)$ ?
Can anyone helps?
 A: I think this just follows from the definition of the tensor product.
If $M,N$ are $R$-modules, then $m\otimes (rn)=(rm)\otimes n$ inside $M\otimes_R N$ for all $r\in R$, $m\in M$ and $n\in N$, more or less by definition. Now in your case, $M=A_d/I_d$ and $N=\operatorname{Frac}(R/\mathfrak{q})$, and $m=p(T)$ and $n=\frac{[r]}{[r']}$. As you correctly pointed out, we have
$$
n=\frac{[h][r]}{[h][r']}=h\cdot\frac{[r]}{[h][r']}
$$
where $h\cdot-$ denotes the action of $h\in R$ on an element of $N$. This last equality just follows from the $R$-module structure of $\operatorname{Frac}(R/\mathfrak{q})$. Hence you are allowed to move the $h$ to the otherside:
$$
(p(T)+I_d)\otimes\left(h\cdot\frac{[r]}{[h][r']}\right)=(h\cdot(p(T)+I_d))\otimes\frac{[r]}{[h][r']}=(hp(T)+I_d)\otimes\frac{[r]}{[h][r']}=0.
$$
Small detail: I don't think it makes sense to write $h+I_d$, as $I_d$ isn't an $R$-submodule of $R$. But the $R$-action on an element of $A_d/I_d$ is simply given by $R$-action on the representant.
