exact sequence of ideal sheaves (Hartshorne Theorem III.3.7) Let $X$ be a scheme and $Y$ a closed subscheme. Let $i : Y \hookrightarrow X$ be the inclusion.
Then, we define the ideal sheaf of $Y$, denoted $\mathcal {I}_Y$to be the kernel of the morphism $i^{\sharp} : \mathcal{O}_X \rightarrow i_{*} \mathcal{O}_Y$.
Let $P$ be a closed point of $X$, let $U$ be an affine open subet of $X$ containing $P$,and let $Y=X-U$.
Then is the following sequence exact??
$$0 \rightarrow \mathcal{I}_{Y\cup \{P\} } \rightarrow \mathcal{I}_Y \rightarrow k(P) \rightarrow 0$$
where $k(P)=\mathcal{O}_P/\mathcal{m}_P$.
 A: Ok, so first of all I'm just learning this stuff so I'm kind of shaky.  If anyone has any comments for me or notices any errors please point them out.  As far as I understand there are many different choices for ideal sheaf on $Y$ and they correspond to different subscheme structures.  Since Hartshorne does not specify a choice of subscheme structure, I think we can make our own choice (since he only uses the existence of such a SES to derive something unrelated to the subscheme structure) and think of the ideal sheaf of $Y$ as the collection of sections of $\mathcal O_X$ that are not invertible at the stalks of elements of $Y$.  We also should make the same choice for the ideal sheaf of $Y\cup\{p\}$.  
With this choice in mind it is clear that on the complement of $p$, which is open since $p$ is a closed point, the ideal sheaves agree.  Since the skyscraper sheaf is zero here, we have exactness.  Thus we only need exactness at the stalk at $p$.  Looking at the complement of $Y$ we reduce to the case where we have an affine scheme $U$ and our ideal sheaf on a closed point $p$ injecting into the sheaf of regular functions:
$$0\to I_p\to \mathcal O_U\to k(p)\to 0$$
But by our choice of ideal sheaf, $I_p$ is just the maximal ideal corresponding to $p$.  This shows exactness on stalks at $p$.  
A: Seth's answer is excellent, and i wish to complement it a bit.
Let $i_1: Y \rightarrow X$, $i_2: Y \cup \{P\} \rightarrow X$ be the inclusion maps. Suppose we give to $Y$ and $Y \cup \{P\}$ the reduced induced closed subscheme struture. That is, for every open set $Y \cap U$ of $Y$ that corresponds to an open affine set $U = \operatorname{Spec}(A)$ of $X$, we set $\mathcal{O}_Y(Y \cap U) = A / J$, where $J = \bigcap_{Q \in Y\cap U} Q$, and similarly for $Y \cup \{P\}$. Since $P$ is a closed point, the stalks $\mathcal{O}_{Y,Q}, \, \mathcal{O}_{Y \cup \{P\},Q}$ are equal for any point $Q \in Y$. This implies that $(i_{1*}\mathcal{O}_Y)_Q = (i_{2*}\mathcal{O}_{Y \cup \{P\}})_Q, \, \forall Q \in Y$. Moreover, these latter stalks are also equal for any $Q \not\in Y \cup \{P\}$, since for such a $Q$ they are equal to zero. This implies that $\mathcal{I}_{Y,Q} = \mathcal{I}_{Y \cup \{P\},Q}$ for any $Q \neq P$. Since for such a $Q$ the skyscraper sheaf $k(P)$ is zero, exactness needs to be checked only at $P$. 
Let $U = \operatorname{Spec}(A)$ be an open affine subset of $X$ containing $P$. Then $Y \cap U =\operatorname{Spec}(A/J)$, where $J = \bigcap_{Q \in Y\cap U} Q$. Moreover, $(Y \cup \{P\}) \cap U =\operatorname{Spec}(A/J\cap m)$, where $m$ is a maximal ideal of $A$ not containing $J$ that represents $P$ in $\operatorname{Spec}(A)$. Then $\mathcal{I}_Y|_{Y \cap U} = \tilde{J}$, and $\mathcal{I}_{Y \cup \{P\}}|_{(Y \cup \{P\}) \cap U} = \tilde{(J \cap m)}$. Computing the stalks at $P$ is the same as localizing at $m$, which gives $\mathcal{I}_{Y,P} = J A_m = A_m$, and $\mathcal{I}_{Y \cup \{P\},P}= J \cap m A_m = m A_m$. Their quotient is $k(P) = A_m / mA_m$. 
An interesting note: As Seth says, the specific closed subscheme structure given to $Y, Y \cup \{P\}$ is not of particular relevance for the purpose of the proof of Theorem III.3.7 in Hartshorne. For example, we may work with the inverse image sheaves $i_1^{-1} \mathcal{O}_{X},i_2^{-1} \mathcal{O}_{X}$, in which case our exact sequence will change to $0 \rightarrow \mathcal{I}_{Y \cup \{P\}} \rightarrow \mathcal{I}_{Y} \rightarrow \mathcal{O}_{X,P} \rightarrow 0$. Then it can be checked that the rest of the arguments in the proof still apply.
