Lemma 53.15.5. Let $k$ be an algebraically closed field. If $f : X' \to X$ is the glueing of two points $a, b$ as in Example 53.15.2, then there is an exact sequence

$k^* \to \mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (X') \to 0$

The first map is zero if $a$ and $b$ are on different connected components of $X'$ and injective if $X'$ is proper and $a$ and $b$ are on the same connected component of $X'$.

Proof. The map $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (X')$ is surjective by Varieties, Lemma 33.37.7. Using the short exact sequence

$0 \to \mathcal{O}_ X^* \to f_*\mathcal{O}_{X'}^* \xrightarrow {ab^{-1}} x_*k^* \to 0$

we obtain

$H^0(X', \mathcal{O}_{X'}^*) \xrightarrow {ab^{-1}} k^* \to H^1(X, \mathcal{O}_ X^*) \to H^1(X, f_*\mathcal{O}_{X'}^*)$

We have $H^1(X, f_*\mathcal{O}_{X'}^*) \subset H^1(X', \mathcal{O}_{X'}^*)$ (for example by the Leray spectral sequence, see Cohomology, Lemma 20.13.4). Hence the kernel of $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (X')$ is the cokernel of $ab^{-1} : H^0(X', \mathcal{O}_{X'}^*) \to k^*$. If $a$ and $b$ are on different connected components of $X'$, then $ab^{-1}$ is surjective. Because $k$ is algebraically closed any regular function on a reduced connected proper scheme over $k$ comes from an element of $k$, see Varieties, Lemma 33.9.3. Thus $ab^{-1}$ is zero if $X'$ is proper and $a$ and $b$ are on the same connected component. $\square$

Comment #3004 by James Waldron on

In the statement of Lemma 49.15.5, I think that $f$ should be a map $f:X'\to X$ instead of $f:X\to X'$.

Comment #4869 by Grisha Konovalov on

This doesn't seem to be true that if $X'$ is affine than map $ab^{-1} : H^0(X', \mathcal{O}_{X'}^*) \to k^\ast$ is surjective. For example, $ab^{-1}$ is constant for glueing of two $k$-points on $\mathbf{A}^1_k$.

Comment #4870 by on

Oops, yes you are right. Luckily this parenthetical statement isn't needed for the proof. I will fix this soonish.

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