Let $X=\mathbb{P}^1\_{\mathbb{Z}}$ and $Y\subset X$ be a local complete intersection of codimension two with Ideal sheaf $I\_Y$. 


(I'm mostly interested in the case where $Y$ is a single point $x$ lying over a prime $p$, i.e. $f(x)=p$, where $f: X \rightarrow Spec(\mathbb{Z})$, or a finite set of points, each lying over a different prime).


My question is: how does the restriction of $I\_Y$ to a special fiber $\mathbb{P}^1\_{\mathbb{F}\_q}$ look like (for some prime q)? That is: can we describe the sheaf $I\_{Y{|\mathbb{P}^1\_{\mathbb{F}\_q}}}=j^{\*}I\_Y$, where $j: \mathbb{P}^1\_{\mathbb{F}\_q} \rightarrow \mathbb{P}^1\_{\mathbb{Z}}$?


If $Y$ is a point $x$ lying over $p$, then we have: $0\rightarrow I\_Y \rightarrow \mathcal{O}\_X \rightarrow k(x) \rightarrow 0$, let us assume $k(x)=\mathbb{F}\_p$. 


We have to look at $0\rightarrow \mathcal{T}or^1\_\mathbb{Z}(k(x),\mathbb{F}\_q)\rightarrow I\_Y\otimes\_{\mathbb{Z}}\mathbb{F}\_q \rightarrow \mathcal{O}\_X\otimes\_{\mathbb{Z}}\mathbb{F}\_q \rightarrow k(x)\otimes\_{\mathbb{Z}}\mathbb{F}\_q \rightarrow 0$. Now if $p\neq q$ this should give $I\_Y\otimes\_{\mathbb{Z}}\mathbb{F}\_q \cong \mathcal{O}\_X\otimes\_{\mathbb{Z}}\mathbb{F}\_q=\mathcal{O}\_{\mathbb{P}^1\_{\mathbb{F}\_q}}$.


But i'm not able to see what happens for $q=p$. Do we possibly get the ideal sheaf of $x$ in $\mathbb{P}^1\_{\mathbb{F}\_p}$? Or something more complicated?


Background: Assume $E$ is a locally free sheaf of rank $2$ on $X$ sitting in an exact sequence $0\rightarrow \mathcal{O}\_X(d)\rightarrow E \rightarrow I\_Y \rightarrow 0$. Now i'm trying to understand the splitting types of $E$ on the special fibers $\mathbb{P}^1\_{\mathbb{F}\_p}$. 


Or is it easier to compute $0\rightarrow \mathcal{T}or^1\_\mathbb{Z}(I\_Y,\mathbb{F}\_q)\rightarrow \mathcal{O}\_X(d)\otimes\_{\mathbb{Z}}\mathbb{F}\_q \rightarrow E\otimes\_{\mathbb{Z}}\mathbb{F}\_q \rightarrow I\_Y\otimes\_{\mathbb{Z}}\mathbb{F}\_q \rightarrow 0$?


Edit: Using Will's answer we see that we can write $I\_Y\otimes \mathbb{F}\_q\rightarrow O\_{\mathbb{P}^1\_{\mathbb{F}\_q}}(-1)\rightarrow 0$. This shows that we have: $E\otimes \mathbb{F}\_q \rightarrow O\_{\mathbb{P}^1\_{\mathbb{F}\_q}}(-1)\rightarrow 0$. Using that the degree is constant on $Spec(\mathbb{Z})$ we can compute that we have in fact $0\rightarrow O\_{\mathbb{P}^1\_{\mathbb{F}\_q}}(d+1)\rightarrow E\otimes \mathbb{F}\_q \rightarrow O\_{\mathbb{P}^1\_{\mathbb{F}\_q}}(-1)\rightarrow 0$. Now we just have to compute some $Ext$ to see for which $d$ this already splits.



How do ideal sheaves behave on the special fibers of the projective line over the integers?
If you consider the ideal $I\_y=(x,p)$ and tensor with $\mathbb Z/p$, you get the module $(x,p)/( xp, p^2)$. On the other hand, the ideal of the point $x$ on $\mathbb A^1\_{\mathbb F\_p}$ is $x/(xp)$. The difference betweeen the two modules is that the first one contains $p/p^2$, where the second one doesn't. We can see that with your exact sequence, specialized to $k(x)=\mathbb F\_p$ and $\mathbb F\_q=\mathbb F\_p$:


$0 \to \mathcal Tor^1(\mathbb F\_p, \mathbb F\_p) \to I\_Y \otimes\_{\mathbb Z} \to \mathcal O\_{\mathbb P^1\_{\mathbb F\_P} } \to \mathbb F\_p \otimes\_Z \mathbb F\_p $


We have $\mathbb F\_P \otimes\_Z \mathbb F\_p = \mathcal Tor^1(\mathbb F\_p, \mathbb F\_p) = \mathbb F\_p$, so we have the exact sequence: 


$0 \to \mathbb F\_p \to I\_Y \otimes\_{\mathbb Z} \to \mathcal O\_{\mathbb P^1\_{\mathbb F\_P} } \to \mathbb F\_p $


The kernel of the third morphism is exactly the ideal sheaf of $x$ in $\mathbb P^1\_{\mathbb F\_p}$. So we see that the tensor product is the extension of this by an additional copy of $\mathbb F\_p$. Note that this is not an ideal sheaf.


The case of a different ideal is directly analogous.


