Why aren't those Cartier Divisors equivalent?

Question

Please refer to Gathmann's notes http://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2002/main.pdf at Example 9.3.6 for context.

It's trying to give an example that the map between $Div(X)$ and $Z_{0}(X)$ (see text) is not injective. $X$ is a 1-dimensional scheme (a curve) given by $X=X_1\cup X_2$ where $X_i$ is isomorphic to $\mathbb{P}^1$, $i=1,2$ and they are both glued together in a point $P$ that lies in $X_1\cap X_2\neq \emptyset$. We consider the covering $U=X-\{Q\}$, $Q\in X_1-X_2$, and $V=X_1-\{P\}$.

Now consider the Cartier Divisor on X given by the constant function 1 on $U$ and the linear function that has a simple zero on $Q$ on $V\cong \mathbb{A}^1$. So we have a Cartier divisor $D$ which Weil divisor class class is $[D]=1.[Q]$.

Symetrically, we can obtain another Cartier divisor $D'$ on $X$ which Weil divisor class is given by $[D']=1.[Q']$ for a point $Q'\in X_2-X_1$.

My question is: why is the divisor $D-D'$ not linearly equivalent to zero (that is, why is it not a divisor of a rational function on $X$)?

Answer 1

You can use the fact that rational equivalence is preserved under pullback. Consider the inclusion map $\pi:X_2\to X$. Then you can see that $\pi_i^*D$ is trivial while $\pi^*D'$ is class of a hyperplane in $\mathbb{P}^1$.

Edit: correction needed

This argument is technically incorrect, but I'd rather keep it in this form since when combined with Georges's comments it might explain some tricks (or even trickeries :)) in intersection theory. The correct argument should go sth like this: Let $\mathcal{D}$ and $\mathcal{D}'$ be the classes of $D$ and $D'$. Use the fact that pullback preserves linear equivalence of divisor classes (see Georges's comments on why we need to work with the class of a divisor here). Consider the inclusion map $\pi:X_2\to X$. Then you can see that $\pi_i^*\mathcal{D}$ is trivial while $\pi^*\mathcal{D}'$ is class of a hyperplane in $\mathbb{P}^1$.

Answer 2

Gathmann's example is bad first of all because there isn't enough notation and justification but much more seriously because you shouldn't consider Cartier divisors on non-integral schemes.
Fulton in the most advanced monograph on the subject only considers Cartier divisors on integral schemes and introduces pseudo divisors on general schemes.
Grothendieck defined Cartier divisors on general schemes but made a notorious mistake (discovered by Kleiman) already in his definition of rational function!

That said, here is an answer to the question:
To $D,D'$ we associate line bundles $\mathcal O(D), \mathcal O(D')$.
Since the association $CaCl(X)\to Pic(X)$ is injective (actually bijective), it is enough to prove that $\mathcal O(D), \mathcal O(D')$ are not isomorphic.
But this is clear, since $\mathcal O(D)|X_1\cong \mathcal O_{X_1}(1)$ on $X_1\cong \mathbb P^1$, whereas $\mathcal O(D')|X_1\cong \mathcal O_{X_1}$ is trivial.

Edit: Warning !
Although the morphism $CaCl(X)\to Pic(X)$ is practically always bijective we cannot replace the study of line bundles by divisors but only by divisor classes.
To take an example: if you consider a line $D\subset \mathbb P^2$ and the injection $j:D\hookrightarrow \mathbb P^2$, it does not make sense to talk of the restriction $j^\ast(D)$ of $D$ to $D$ but it does make sense to talk of the restriction $j^\ast([D])$ of the class $[D]$ of $D$ to $D$ and to thus obtain the divisor class of a point $p\in D$.
[Don't worry if you find this a bit confusing: these are very subtle points which have only be completely settled recently.
The ultimate reference on the subject is Fultons very difficult Intersection theory]