Let $(X,\mathcal{O}_X)$ be a scheme, and $Y$ a Zariski closed subset. For each $U\subset X$ open we defined the ideal of $\mathcal{O}_X(U)$, $I(U)$ to be:
$$I(U)=\{s\in \mathcal{O}_X(U): \forall x\in Y\cap U, s_x\in \mathfrak{m}_x \}$$
where $s_x$ is the germ of $s$, and $\mathfrak{m}_x$ is the unique maximal ideal in the stalk of $\mathcal{O}_X$ at $x$.
I want to show that $Y$, when endowed with the subspace topology, and equipped with the sheaf $\mathcal{O}_Y(Y\cap U)=\mathcal{O}_X(U)/I(U)$ is a scheme. I first want to show that that this well defined, i.e. if $U$ and $W$ are open, and $Y\cap U=Y\cap W$ then:
$$\mathcal{O}_X(U)/I(U)\cong \mathcal{O}_X(W)/I(W)$$
Note that since $U\cap W\cap Y=U\cap Y=W\cap Y$, it suffices to prove the statement for $Z=U\cap W$.
We define the map:
$$\phi:\mathcal{O}_X(U)\longrightarrow \mathcal{O}_X(Z)/I(Z)$$
by:
$$\phi(s)=\pi_Z\circ\theta^U_Z(s)$$
where $\pi_Z$ is the quotient map $\mathcal{O}_X(Z)\rightarrow I(Z)$, and $\theta^U_Z$ is the restriction map. One easily sees that if $s\in I(U)$, then since $s_x=\theta^U_Z(s)_x$ that $\theta^U_Z(s)\in I(Z)$, so $s\in \ker \phi$. Moreover, $s\in \ker\phi$, then $\theta^U_Z(s)\in I(Z)$, so by the same argument, $s_x\in \mathfrak{m}_x$ for all $x$, hence $s\in I(U)$. It follows that $\phi$ descends to a unique injective ring morphism:
$$\psi:\mathcal{O}_X(U)/I(U)\longrightarrow \mathcal{O}_X(Z)/I(Z)$$
To show that $\psi$ is a ring isomorphism, it suffices to show that $\phi$ is surjective.
My partial attempt is as follows :
Suppose that $[h]\in \mathcal O_X(Z)/I(Z)$, now if $h\in \text{im } \theta^U_Z$, then we are done. Suppose that $h\notin \text{im }\theta^U_Z$, then we need to find an $s\in \mathcal O_X(U)$ such that $h-\theta^U_Z(s)\in I(Z)$, implying that $h_x-s_x\in \mathfrak{m}_x$ for all $x\in Z\cap Y=U\cap Y$. We have that $U$ is an open subscheme, so choose a cover of affine opens such that $U_i\cong \operatorname {Spec}A_i $ for rings $A_i$. Now $Z$ is a subscheme of $U$, so we can cover $Z$ with $Z\cap U_i$, and each $Z\cap U_i$ is a subscheme of an affine scheme, so we can choose a refinement of the cover $\{Z\cap U_i\}$ by $\{V_{ij}\}$ where each $V_{ij}\cong \operatorname{Spec}(A_i)_{f_j}$, where $f_j\in A_i$, and $(A_i)_{f_j}$ is the localization of $A_i$ at $f_j$.
Let $x\in U_i$, and identify $x$ with $\mathfrak{p}\subset A_i$, then for some $V_{ij}$ we have that $x\in V_{ij}$. Let $h|_{V_{ij}}=\frac{a_i}{f_j^k}$ for some $a_i\in A_i$, then:
$$(h_i)_x=\frac{a_i}{f_j^k}\in A_{\mathfrak p}$$
where $A_{\mathfrak p}$ is the localization of $A$ with respect to the multiplicatively closed set $A-\mathfrak p$. The unique ideal is given by:
$$\mathfrak{m}_x=\left\{\frac{p}{r}\in A_{\mathfrak p}: p\in \mathfrak p\right\}$$, so we want to find a $b_i\in \mathfrak O_X(U_i)\cong A_i$ such that:
$$(h_i)_x-(b_i)_x=\frac{a_i}{f_j^k}-\frac{b_i}{1}=\frac{p}{r}$$
for some $p/r$.
However, I can't even seem to find a $b_i$ that can do this in this simplified situation. My idea was hoping that once I found a $b_i$, it would satisfy $(b_i)_x-h_x$ for all $x\in U_i$, and then each $b_i$ would glue together to give $b\in \mathcal O_X(U)$, but since I can't even find a $b_i$ that satisfies this for a single $x$, I'm not sure if this approach is viable. Is this approach correct, or am I missing another way of showing that the quotient is surjective? If it is, how do I find such a $b_i$?
Also, I think there is a totally different approach to putting a scheme structure on $Y$ which may be technically easier, but I would really like to stay with this approach for now.
Best Answer
You are barking up the wrong tree: $\mathcal{O}_X(U)/I(U)\cong\mathcal{O}_X(W)/I(W)$ is not true. Here is a counterexample: let $X=\Bbb P^2_k$ for a field $k$, $Y=V(z)\cong\Bbb P^1$, $U=\Bbb P^2 \setminus [1:0:0]$, and $W=D(x)$. Then $\mathcal{O}_X(U)=k$ and $I(U)=0$ so $\mathcal{O}_X(U)/I(U)=k$, while $\mathcal{O}_X(W)=k[\frac{y}{x},\frac{z}{x}]$ and $I(W)=(\frac{z}{x})$ so $\mathcal{O}_X(W)/I(W)=k[\frac{y}{x}]$.
Your main error here is the claim that $U\mapsto \mathcal{O}_X(U)/I(U)$ is a sheaf on $Y$. It is only a presheaf on $X$; what you need to do here is take the sheafification of the presheaf $U\mapsto \mathcal{O}_X(U)/I(U)$ on $X$ and then consider the restriction of that sheaf to $Y$ (i.e. the inverse image sheaf along the inclusion of the closed subspace $Y\to X$). Then you can continue in proving that this defines a closed subscheme. (For reference, you are asking about the construction of the reduced induced subscheme structure, which is well-covered in the literature.)