I have a question about closed subschemes (that appears as a question in Eisenbud and Harris's book Geometry of Schemes.)
Let $X$ be an arbitrary scheme and let $Y, Z$ be closed subschemes of $X$. Explain what it means for $Y$ to be contained in $Z$.
My thoughts
In the affine case $X = \text{Spec}(R)$ we know that a closed subscheme $Y \hookrightarrow \text{Spec}(R)$ is can be written as $\text{Spec}(R/\mathfrak{a})$ for a uniquely determined ideal $\mathfrak{a} \subseteq R$, so I guess that in that case $\text{Spec}(R/\mathfrak{a}) \subseteq \text{Spec}(R/\mathfrak{b})$ precisely when $\mathfrak{b} \subseteq \mathfrak{a}$.
I don't know how that works for general schemes. Of course, if we want $(Y,\mathcal{O}_Y) \subseteq (Z,\mathcal{O}_Z)$ to hold, we want that $Y \subseteq Z$ as topological spaces. I suggest we denote the embeddings of topological spaces by $i : Y \hookrightarrow X$, $ \ j : Z \hookrightarrow X$ and $k : Y \hookrightarrow Z$.
Now we move on to the map of sheaves. We need a map of scheaves $\mathcal{O}_Z \twoheadrightarrow k_*\mathcal{O}_Y$. We already have surjections of sheaves $\mathcal{O}_X \twoheadrightarrow i_*\mathcal{O}_Y$ and $\mathcal{O}_X \twoheadrightarrow j_*\mathcal{O}_Z$. If we get a map $\mathcal{O}_Z \to k_*\mathcal{O}_Y$ we get its surjectivity by abstract nonsense. But when can we get such a map?
Sidenote
I am more familiar with Hartshorne's definitions then with Eisenbud's when it comes to closed subschemes. Hartshorne defines a closed immmersion of schemes $(Y,\mathcal{O}_Y) \hookrightarrow (X,\mathcal{O}_X)$ as an topological embedding $Y \stackrel{i}{\hookrightarrow} X $ with $Y \cong i(Y) \stackrel{\text{closed}}{\subseteq} X.$ A closed subscheme is an equivalence class of such embeddings, (although suprisingly those equivalences don't show up often.)