A particular closed subscheme

Question

Look at the following definition:

Let $(X,\mathscr O_X)$ be a scheme. A closed subscheme of $(X, \mathscr O_X)$ is a scheme $(Z, \mathscr O_Z)$ such that:

1. $Z$ is a closed subset of $X$ where $i:Z\longrightarrow X$ is the immersion.
2. the sheaf $i_\ast\mathscr O_Z$ is isomorphic (as sheaf on $X$) to $\mathscr O_X/\mathscr I$ where $\mathscr I$ is a sheaf of ideals of $\mathscr O_X$

Now consider tha case $X=$Spec$R$, and let $\mathfrak a\subseteq R$ an ideal. The canonical projection $\pi: A\rightarrow A/\mathfrak a$ induces a homeomorphism $^a\pi$ between Spec$A/\mathfrak a$ and $V(\mathfrak a)\subseteq X$. If $\mathscr O_{V(\mathfrak a)}:=\,^a\pi_\ast\mathscr O_{\textrm{Spec}A/\mathfrak a }$, then we have that $(V(\mathfrak a), O_{V(\mathfrak a)})$ is an affine scheme where clearly $V(\mathfrak a)$ is a closed subset of $X$.

Now I want to prove that $(V(\mathfrak a), O_{V(\mathfrak a)})$ is a closed subscheme in the sense of the above definition, but I have problems to find the sheaf of ideals $\mathscr I$ such that $\mathscr O_{V(\mathfrak a)}\cong \mathscr O_X/\mathscr I$.

Many thanks in advance.

Answer 1

You have $i:\mbox{Spec}(A/a)\to\mbox{Spec}(A)$, given by the projection $A\to A/\frak{a}$. You know that the sheaf $\mathcal{O}_{\mbox{Spec}(A)}$ is globally $A$, and on each principal open set it's just a localization of $A$. So basically, if we look at only principal open sets $D(f)$, we get that the map $\mathcal{O}_{\mbox{Spec}(A)}\to i_*\mathcal{O}_{\mbox{Spec}(A/\frak{a})}$ is $A_f\to (A/a)_\overline{f}$ (remember that $i_*\mathcal{O}_{\mbox{Spec}(A/\frak{a})}(D(f))=\mathcal{O}_{\mbox{Spec}(A/\frak{a})}(D(\overline{f}))$, where $\overline{f}$ is the image of $f$ in $A/\frak{a}$). But this is just the localization of the map $A\to A/\frak{a}$, and so is surjective.

Answer 2

Let me give, as a somewhat weird exercise, a very formal answer.

On any ringed space $(X,\mathcal O)$ a sheaf ideal $\mathcal I \subset \mathcal O$ is determined by the collection of ideals $\mathcal I_x \subset \mathcal O_x \:(x\in X)$. That collection of $\mathcal I_x$'s is not arbitrary, but the point is that it determines $\mathcal I$.
For $U\subset X$ open , a section $s\in \Gamma(U,\mathcal I)$ is then a section $s\in \Gamma(U,\mathcal O)$ such that for all $x\in U$ one has $s_x\in \mathcal I_x$.

And now back to your problem, where $X=\text {Spec}(A)$ .
If $\mathfrak a \subset A$ is an ideal, the corresponding sheaf of ideals $\mathcal I_{\mathfrak a}=\mathcal I\subset \mathcal O_X=\mathcal O$ is characterized by the demand that for $x=[\mathfrak p]\in X$ one has: $$ \mathcal I_x= \mathfrak a_ \mathfrak p \subset A_{\mathfrak p}= \mathcal O_x $$ where $\mathfrak a_ \mathfrak p $ is the set of formal fractions $\frac{a}{\pi}$ with $a\in \mathfrak a$ and $\pi\in A\setminus \mathfrak p$ .