I'm currently working on exercises in Hartshorne's Algebraic Geometry and coming up with some questions on Exercise II.1.21. I'll briefly restate the exercise and ask questions along with the description.
Let $X$ be a variety over an algebraically closed field $k$ and let $\mathcal{O}_X$ be the sheaf of regular functions on $X$.
Question 1: Can we define the sheaf of regular functions $\mathcal{O}_X$ on an arbitary algebraic sets $X$ (NOT necessarily be irreducible)? Maybe we can define a regular function on an open subset $U$ of $X$ just as the functions $f: U \rightarrow k$ that are regular at every point $p \in U$ (the same as the definition of regular functions on a variety) and collect them in a set $\mathcal{O}_X(U)$. It seems that $[U \mapsto \mathcal{O}_X(U)]$ remains to be a sheaf?
(a) Let $Y$ be a closed subset of $X$. For each open set $U \subset X$, let $\mathcal{I}_Y(U)$ be the ideal containing $\mathcal{O}_X(U)$ consisting of those regular functions which vanish at all points of $Y \cap U$. Show that the presheaf $[U \mapsto \mathcal{I}_Y(U)]$ is a sheaf. ......
Question 2(EDITED): I have "proved" this. In the "proof", I haven't used the assumption that "$Y$ is closed in $X$". So is it true that $\mathcal{I}_Y$ is a (well-defined) sheaf for any subset $Y$ of $X$ (NOT necessarily be closed)? Here is my "proof":
Attempt 2 (Proof): $\mathcal{I}_Y: [U \mapsto \mathcal{I}_Y(U)]$ is a presheaf, with the restriction map given by the usual restriction of regular functions.
For the first sheaf condition, let $U$ be any open subset of $X$, and $\{ U_i\}_{i \in I}$ be one of its open cover. Let $s \in \mathcal{I}_Y(U)$, such that $s\vert_{U_i} = 0 \in \mathcal{I}_Y(U_i)$ for all $i \in I$, which means that $s$ is the zero regular function defined on $U_i$. Then since $\{ U_i\}_{i \in I}$ is an open cover of $U$, $s$ is the zero regular map defined on the entire $U$, which clearly vanishes at all points of $U$. So, $s = 0 \in \mathcal{I}_Y(U)$.
For the second sheaf condition, with the same notation, let $s_i \in \mathcal{I}_Y(U_i)$ for all $i \in I$ satisfying the compatibility condition, i.e. $s_i \vert_{U_{ij}} = s_j \vert_{U_{ij}}$, for all $i,j \in I$. Then we can glue then together to obtain a regular function $s \in \mathcal{O}_X(U)$, such that $s\vert_{U_i} = s_i$ for all $i \in I$. Now we shall show that $s \in \mathcal{I}_Y(U)$. Let $P \in Y \cap U$. Since $\{ U_i\}_{i \in I}$ is an open cover of $U$, there exists $i_0 \in I$, such that $P \in Y \cap U_{i_0}$. Now since $s\vert_{U_{i_0}} = s_{i_0}$, we have $s(p) = s \vert_{U_{i_0}} (p) = s_{i_0}(p) = 0$, where the last equality holds because $s_{i_0} \in \mathcal{I}_Y(U_{i_0})$. Here $p$ is arbitarily chosen, and hence $s$ vanishes at all points $P \in Y \cap U$. So, $s \in \mathcal{I}_Y(U)$. $\blacksquare$
(b) If $Y$ is a subvariety, then the quotient sheaf $\mathcal{O}_X / \mathcal{I}_Y$ is isomorphic to $i_{\ast}(\mathcal{O}_Y)$, where $i: Y \rightarrow X$ is the inclusion, and $\mathcal{O}_Y$ is the sheaf of regular functions on $Y$.
I have proved this:
Attempt (Proof): Let $i: Y \rightarrow X$ be the inclusion, and for any open subset $U$ of $X$, $$ i_{\ast}(\mathcal{O}_Y)(U) := \mathcal{O}_Y(i^{-1}(U)) = \mathcal{O}_Y(U \cap Y). $$ So, we can define a sheaf morphism $i^{\sharp}: \mathcal{O}_X \rightarrow i_{\ast}(\mathcal{O}_Y)$ as $$ i^{\sharp}(U): \mathcal{O}_X(U) \rightarrow i_{\ast}(\mathcal{O}_Y)(U)= \mathcal{O}_Y(U \cap Y), \quad f \mapsto f\vert_{U \cap Y}. $$ We can check that this is indeed a sheaf morphism. By the construction, the sheaf $\mathcal{I}_Y$ (it is indeed a sheaf, as shown in (a)) is the kernel of the morphism $i^{\sharp}$.
Now, we need to show the surjectivity of $i^{\sharp}$. To do that, we turn to the maps induced on stalks: $$ i^{\sharp}_p: (\mathcal{O}_X)_p \rightarrow (i_{\ast}(\mathcal{O}_Y))_p = \mathcal{O}_Y(\cdot \cap Y)_p. $$ There are two cases. (1) When $P \not\in Y$, then there exists an open neighborhood of $P$, not intersecting $Y$. (Here maybe we are using $Y$ is closed) So $(i_{\ast}(\mathcal{O}_Y))_p=0$, and hence $i^{\sharp}_p$ is surjective. (2) When $P \in Y$, consider a germ $t_p \in (i_{\ast}(\mathcal{O}_Y))_p$ represented by $\langle t, U \rangle$, where $t \in i_{\ast}(\mathcal{O}_Y)(U) = \mathcal{O}_Y(U \cap Y)$, and $P \in U$. This germ also represents an element in $(\mathcal{O}_X)_p$. Hence $i^{\sharp}_p$ is surjective.
Combining the two paragraphs above, we obtain that $i_{\ast}(\mathcal{O}_Y) = \mathcal{O}_X / \mathcal{I}_Y$ as desired. Furthermore, we obtain a short exact sequence: $$ 0 \longrightarrow \mathcal{I}_Y \longrightarrow \mathcal{O}_X \longrightarrow i_{\ast}(\mathcal{O}_Y) \longrightarrow 0 . \quad \quad (\star) $$ $\blacksquare $
In (c), Hartshorne asked
(c) Now let $X = \mathbb{P}^1$ and let $Y$ be the union of two distinct points $P , Q \in X$. Then there is an exact sequence of sheaves on $X$: $$ 0 \longrightarrow \mathcal{I}_Y \longrightarrow \mathcal{O}_X \longrightarrow i_{\ast}(\mathcal{O}_P) \oplus i_{\ast}(\mathcal{O}_Q) \longrightarrow 0 \quad \quad (\star \star)$$ ......
Question 3: Some online solution manuals claimed that this exact sequence can be obtained directly using the exact sequence $(\star)$, by applying $X = \mathbb{P}^1$ and $Y = \{ P , Q \}$. But it seems that $Y= \{ P , Q \}$ is not irreducible, hence not a subvariety of $X$. So why can we use (b) directly?
Attempt 3 (The source of my questions in this post): However, in the proof of (b) above, it seems that we did NOT use the fact that $Y$ is irreducible, except that when we define the sheaf $\mathcal{O}_Y$, $Y$ need to be irreducible. This is the source of my Question 1. i.e. if we can throw away the restriction that $Y$ is irreducible and define $\mathcal{O}_Y$ as well (described in Question 1), then $\mathcal{O}_Y$ is well-defined (for merely a closed set), then we can use the short exact sequence $(\star)$ directly, although $Y = \{P, Q \}$ is not irreducible.
Here, If all my attempts above turn out to be wrong, then I hope to prove (c) directly:
Question 4: If we cannot use the result of (b) directly, I'm trying to throw away what we have proved so far and prove the sequence $ (\star \star)$ is exact directly. A tricky point is to constuct a surjective sheaf morphism $$ \beta: \mathcal{O}_X \rightarrow i_{\ast}(\mathcal{O}_P) \oplus i_{\ast}(\mathcal{O}_Q). $$ And I am wondering how to construct this map?
Attempt 4: My attempt is to define on each open set $U \subset X$ the map $$ \beta(U): \mathcal{O}_X(U) \rightarrow i_{\ast}(\mathcal{O}_P)(U) \oplus i_{\ast}(\mathcal{O}_Q)(U) $$ by sending $f$ to $(f,0), (0,f), (f,f)$ or $(0,0)$ if $P \in U, Q \not\in U$; $P \not\in U, Q \in U$; $P \in U, Q \in U$ and $P \not\in U, Q \not\in U$ accordingly, and check that the induced map on stalks $\beta_p$ are surjective for all $p \in X$, but I haven't managed to do so.
Sorry for such a long post and thank you all for commenting and answering them :)
After some discussion with my classmates, I have figured some of the questions out. So here I hope to share my understandings. Sorry if there are any mistakes.
For Question 1: we can indeed define the sheaf of regular functions on any algebraic sets (not necessarily irreducible). A regular function on an open subset $U$ of an algebraic set $X$ can be defined as a function $f: U \rightarrow k$ such that $f$ is regular at every point of $U$. Then we may check that this indeed make $\mathcal{O}_X$ a sheaf.
For Question 2: Indeed we do not need the fact that $Y$ is closed in the proof of (a) in the post. However, we need the closedness of $Y$ to establish the exact sequence of (b). The exact sequence is vital and hence we insist on putting the requirement of closedness at the begining.
For Question 3: Since the answer of Question 1 is positive, we can directly apply $X = \mathbb{P}^1_k$ and $Y=\{P, Q \}$ in the exact sequence of (b).
For Question 4: By the above explanation of Question 3, we do not need to answer Question 4. Yet if we really hope to dig out what the map $\beta$ is, we shall go back in the proof of (b) (as listed in the post) and write it out in the context of (c).
Sorry for any possible mistakes and misunderstandings!