# How do ideal sheaves behave on the special fibers of the projective line over the integers?

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.

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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.

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Fantastic. In the other case we can write $O_Y=\oplus F_{p_i}$ for different primes $p_i$ and the same computation works. So we have an exact sequence $F_p\rightarrow I_Y\otimes F_p \rightarrow O(-1)$. Is it possible to choose $Y$ such that we get an exact sequence: $Tor(O_Y,F_{p_i})\rightarrow I_Y\otimes F_{p_i}\rightarrow O(-n_i)$ for finitely many primes $p_i$ with numbers $n_i$?. –  DonD Dec 6 '13 at 18:47
Yes. You just need to choose an ideal of degree $n_i$ in $\mathbb A^1_{\mathbb F_{p_i}}$. Then choose the ideal of elements that are in the $i$th ideal mod $p_i$. By the Chinese remainder theorem, adding conditions at $p_j$ for $j\neq i$ will not change the ideal modulo $p_i$, which was constructed to be $O(-n_i)$. –  Will Sawin Dec 7 '13 at 5:02
Very good. I accepted your answer, but unfortunately i don't have enough reputation to give a "+1". But this answer was really helpful. –  DonD Dec 7 '13 at 12:23