Closed subschemes of affine scheme In Mumford's Red Book (p. 106, 2nd edition, Theorem 3) it is proved that any closed subscheme $(f,f^\sharp):Y→X, f$ the inclusion, of an affine scheme $X=\operatorname{Spec}R$ is of the form $\operatorname{Spec}R/I$ with $I=\operatorname{ker}f^\sharp(X)$.
The proof runs as follows:
- may assume that the map $f^\ast$ obtained by taking global sections is injective
- prove that $f$ is surjective (and thus a homeomorphism)
- prove that the map of sheaves is injective, which is done by checking this on the stalks, i.e. $R_P\to \mathcal{O}_{Y,y}$ with $P=f(y)$ is injective for all $y \in Y$.
Let $a \in R$ be in the kernel of this map and define the open set $U=\{y′\in Y\vert f^\ast (a)=0\in \mathcal{O}_{Y,y′}\}$. Then $f(U)$ contains some $(\operatorname{Spec}R)_t$ with $t\in R \setminus P$. Mumford then says that this means, $f^\ast (a)$ vanishes on the principal open $Y_{f^\ast (t)}$. However, I do not understand how one can make this conclusion without knowing that $f^\sharp$ is an isomorphism. By definition of $U$ (and the fact that sheaves are in particular separated presheaves), $f^\ast (a)$ vanishes on $U=f(U)$ and thus on $(\operatorname{Spec}R)_t=X_t$. If I knew that $f(Y_{f^\ast (t)})=X_t$, I'd be done; however I do not see a way of proving this equality or the inclusion $Y_{f^\ast (t)} \subseteq U$ not using the fact that  $f^\sharp$ is an isomorphism.
Help is appreciated.
 A: I am a bit confused by the notation used (I have never used Mumford or read any books by him). If you'll permit me, I can provide a proof of this in what I feel to be a slightly more straightforward fashion. 
Let $ \mathscr{I} = \ker f^\#$ be the sheaf of ideals and take $I = \mathscr{I} (X)$. We then have an exact sequence
$$ 0 \to I \to R \to \mathcal{O}_Y (Y) $$
Now any closed immersion $f$ identifies $Y$ with the support of $\mathscr{I}$ as a closed subscheme of $X$. Let $g \in R$ and let $h = f^\# (X)(g) \in \mathcal{O}_Y (Y)$. By tensoring with the above sequence with $R_g$ which is localization, which is an exact functor, we get
$$ 0 \to I_g = I \otimes_R R_g \to R_g \to \mathcal{O}_Y(Y)_h = \mathcal{O}_Y (Y_h) = f_* \mathcal{O}_Y (D(g))$$
Therefore, $\mathscr{I} (D(g)) = I_g$. Now for the $i: \textrm{Spec } R/I \to \textrm{Spec } R$ induced by the projection map, we see that $\mathscr{I} = \ker i^\# $ and so indeed, $Y \cong \textrm{Spec } R/I$.
A: By definition $y\in U=\{y^{\prime}\in Y\mid\mathcal{O}_{Y,y^{\prime}}\ni0=f^{*}_{y^{\prime}}(a)\}$ and it is open as affirmed, then $f(y)=[P]\in f(U)$ is an open neighbourhood of $[P]$; by definition there exists an element $t\in R\setminus P$ such that $D_R(t)=Spec(R_t)\subseteq f(U)$.
On the other hand:
$$
U\supseteq f^{-1}(D_R(t))=f^{-1}(\{[Q]\in Spec R\mid t\notin Q\})=f^{-1}(\{[Q]\in Spec R\mid t\not=0\in R_Q\})=\{y^{\prime}\in Y\mid f^{*}_{y^{\prime}}(t)\not=0\in\mathcal{O}_{Y,y^{\prime}}\}=Y_{f^{*}(t)},
$$
by definition $f^{*}_{y^{\prime}}(a)=0$ for any $y^{\prime}\in Y_{f^{*}(t)}$; then (this is the trick!):
$$
Y_{f^{*}(t)}=\{y^{\prime}\in Y\mid f^{*}_{y^{\prime}}(at)=0\in\mathcal{O}_{Y,y^{\prime}}\}\Rightarrow at=0\in R_P\iff a=0\in R_P
$$
because $t\in R\setminus P$.
Is it all clear?
