Lemma 37.4.1. Let $X \subset X'$ be a first order thickening with ideal sheaf $\mathcal{I}$. Then there is a canonical exact sequence

of abelian groups.

Some material on Picard groups of thickenings.

Lemma 37.4.1. Let $X \subset X'$ be a first order thickening with ideal sheaf $\mathcal{I}$. Then there is a canonical exact sequence

\[ \xymatrix{ 0 \ar[r] & H^0(X, \mathcal{I}) \ar[r] & H^0(X', \mathcal{O}_{X'}^*) \ar[r] & H^0(X, \mathcal{O}^*_ X) \ar `r[d] `d[l] `l[llld] `d[dll] [dll] \\ & H^1(X, \mathcal{I}) \ar[r] & \mathop{\mathrm{Pic}}\nolimits (X') \ar[r] & \mathop{\mathrm{Pic}}\nolimits (X) \ar `r[d] `d[l] `l[llld] `d[dll] [dll] \\ & H^2(X, \mathcal{I}) \ar[r] & \ldots \ar[r] & \ldots } \]

of abelian groups.

**Proof.**
This is the long exact cohomology sequence associated to the short exact sequence of sheaves of abelian groups

\[ 0 \to \mathcal{I} \to \mathcal{O}_{X'}^* \to \mathcal{O}_ X^* \to 0 \]

where the first map sends a local section $f$ of $\mathcal{I}$ to the invertible section $1 + f$ of $\mathcal{O}_{X'}$. We also use the identification of the Picard group of a ringed space with the first cohomology group of the sheaf of invertible functions, see Cohomology, Lemma 20.6.1. $\square$

Lemma 37.4.2. Let $X \subset X'$ be a thickening. Let $n$ be an integer invertible in $\mathcal{O}_ X$. Then the map $\mathop{\mathrm{Pic}}\nolimits (X')[n] \to \mathop{\mathrm{Pic}}\nolimits (X)[n]$ is bijective.

**Proof for a finite order thickening.**
By the general principle explained following Definition 37.2.1 this reduces to the case of a first order thickening. Then may use Lemma 37.4.1 to see that it suffices to show that $H^1(X, \mathcal{I})[n]$, $H^1(X, \mathcal{I})/n$, and $H^2(X, \mathcal{I})[n]$ are zero. This follows as multiplication by $n$ on $\mathcal{I}$ is an isomorphism as it is an $\mathcal{O}_ X$-module.
$\square$

**Proof in general.**
Let $\mathcal{I} \subset \mathcal{O}_{X'}$ be the quasi-coherent ideal sheaf cutting out $X$. Then we have a short exact sequence of abelian groups

\[ 0 \to (1 + \mathcal{I})^* \to \mathcal{O}_{X'}^* \to \mathcal{O}_ X^* \to 0 \]

We obtain a long exact cohomology sequence as in the statement of Lemma 37.4.1 with $H^ i(X, \mathcal{I})$ replaced by $H^ i(X, (1 + \mathcal{I})^*)$. Thus it suffices to show that raising to the $n$th power is an isomorphism $(1 + \mathcal{I})^* \to (1 + \mathcal{I})^*$. Taking sections over affine opens this follows from Algebra, Lemma 10.32.8. $\square$

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## Comments (1)

Comment #9740 by Fiasco on