In this English translation of Serres FAC [working on p.38], for $X = K^r$ in the Zariski topology, $K$ algebraically closed, we define a locally closed subspace $Y$ as usual: the intersection of Zariski open with Zariski closed. Let $\mathscr{F}(X)$ denote the sheaf of germs of functions on $X$ with values in $K$. The sheaf $\mathscr{F}(X)$ is defined on p.10, In short, it is the sheaffification of $\mathscr{F}(X)_{U}$ being the functions $f: U \to K$, however we check that the presheaf was a sheaf to begin with thus we know that $\Gamma(U,\mathscr{F}(X)) \cong \mathscr{F}(X)_{U}$.

We have defined the structure sheaf $\mathscr{O}_X$ by specifying the stalks $\mathscr{O}_x$ to be the localizations $(K[x_1,\dots,x_r])_{\mathfrak{m}}$ for $\mathfrak{m}$ the maximal ideal of polynomials vanishing at $x$. These stalks can be embedded in the stalks $\mathscr{F}(X)_x$ and we identify them with their isomorphic images, and check that these images determine a subsheaf of $\mathscr{F}(X)$ using the definition of subsheaf on p.12. This subsheaf is denoted $\mathscr{O}_X$.

This much (I think) I understand. We can interpret the sections of $\mathscr{O}_X$ over an open subset $V$ as functions $f: V \to K$ such that for every point $y \in V$, there is a neighborhood on which $f$ can be expressed as a rational function regular at $X$. To see this note that $$\Gamma(V, \mathscr{O}_X) \hookrightarrow \Gamma(V, \mathscr{F}(X)) \cong \mathscr{F}(X)_V$$ so we can identify $\Gamma(V, \mathscr{O}_X)$ with its isomorphic image in $\mathscr{F}(X)_V$. Then let $s$ be any section in $\Gamma(V, \mathscr{O}_X)$, and let $f$ be the element of $\mathscr{F}(X)_V$ that maps to it. By definition $f$ is a function $f: V \to K$. Now the correspondence to map $f \mapsto s$ is to define \begin{align*} s&: V \to \mathscr{O}_X,\\ s&: y \mapsto [f]_y, \hspace{1em} \text{for each } y \in V, \end{align*}

where $[f]_y$ is the image of $f$ in the direct limit. Since we know this image must be in $\mathscr{O}_y$, then we know there has to be some neighborhood containing $y$ on which $f$ agrees with a rational function regular at $y$, which is what we hoped to check.

Question 1: At this point can I determine my global sections $\Gamma(X, \mathscr{O}_X)?$

By the above a global section will be a function $f:X \to K$ such that for every $y \in X$, I get an open neighborhood $U_y$ containing $y$ such that $f|_{U_y} = P(x)/Q(x)$ where $Q(x)$ does not vanish on $U_y$. Note the collection of $U_y$ sets forms an open cover. Also since $X$ is irreducible we know that every two nonempty open sets have a nonempty intersection. Pick two arbitrary open sets from the cover, $U_{y_1}, U_{y_2}$. Then $f|_{U_{y_1} \cap U_{y_2}}$ is given by $P_1(x)/Q_1(x)$ but is also given by $P_2(x)/Q_2(x)$. Since $K$ is algebraically closed, if these rational functions agree point wise, they must also agree as elements of the field of fractions $K(x_1, \dots, x_r)$. This means that $f$ restricts to the same rational function over every element of the open cover, so we have $f = P(x)/Q(x)$, thus $Q(x)$ is constant, so $f$ is just some polynomial. We have $\Gamma(X, \mathscr{O}_X) = K[x_1, \dots, x_r]$. What we just showed is one direction of the proof, but the other direction is easy, clearly any polynomial can be a section.

Question 2: I don't get how Serre extends this idea to a locally closed subspace.

For a locally closed subspace $Y$, defined at the top, let $\mathscr{F}(Y)$ be the sheaf of germs of functions on $Y$. Serre says "the operation of restriction defines a canonical homomorphism for each point $x \in Y$: $$\epsilon_x : \mathscr{F}(X)_x \to \mathscr{F}(Y)_x.$$

The idea is we are going to use this morphism to put the structure sheaf on $Y$, but let's pause here because I do not understand the map. The elements of the domain stalk are equivalences classes that look like this:

$$[f] = \{ \text{ functions that agree on some open set } U \subset X \text{ containing } x \}.$$

Elements of the codomain stalk look like this: $$[f] = \{ \text{ functions that agree on some open set } U \subset Y \text{ containing } x \}.$$

and since $Y$ is locally closed we can write $Y = O \cap C$ where $O$ is open in $X$ and $C$ is closed in $X$. Then $U$ being open in $Y$ means we can write $U = W \cap Y = W \cap (O \cap C)$ where $W$ is open in $X$. But since $W \cap O$ is just some open set in $X$, $U$ open in $Y$ means $U$ is locally closed in $X$. Now note that open sets and closed sets in $X$ are also locally closed in $X$. What this means is that functions who agree on some open set in $X$ will also agree on some locally closed set in $X$, thus will agree on some open set in $Y$. So I think how this map works, i.e. to determine an image, pick an element $[f] \in \mathscr{F}(X)_x$. The representative $f$ is some function $f: V \to K$ where $V$ is open in $X$. So send $[f] \mapsto [f|_{Y \cap V}]$? This is independent of the choice of representative.

I feel like I am missing some big ideas with this map? Or maybe I am over complicating it? To help illuminate it, I tried to prove some properties. I showed the map is surjective (trivially). If $[f]$ is any element of $\mathscr{F}(Y)_x$ then $f$ is a function $f: (U \cap Y) \to K$ where $U$ is open in $X$. Extend $f$ to all of $U$ by 0 and call the map $f^*$. Then $\epsilon_x([f^*]) = [f]$. I tried to prove injectivity but I think it is not in general injective? What it would mean to not be injective is that we have $[f] \neq [g]$ in $\mathscr{F}(X)_x$, so two functions $f,g$ on open subsets of $X$ containing $x$, such that there does not exist an open subset of $X$ on which $f,g$ agree. However, we can find a set $U \cap Y$ where they agree? This seems plausible considering the nature of the Zariski topology, open sets are too big.

The reason it is so important I understand this map is because we then put the structure sheaf on a locally closed subset $Y$ by specifying its stalks be the images $\epsilon_x(\mathscr{O}_x)$ which we denote $\mathscr{O}_x,Y$ and use these stalks to form a subsheaf $\mathscr{O}_Y$ of $\mathscr{F}(Y)$ analogous to how we did on $X$. But I feel like I cannot understand the sheaf $\mathscr{O}_Y$ because I don't understand the transfer of data that occurs by $\epsilon_x$.


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