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two very minor typose (but I found one of them confusing)
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user9072
user9072

Let $X$ be a either a projective scheme or a compact complex space. Then one has an exact sequence $$ (1) \quad 0 \to \textrm{Pic}(X) \to \textrm{Cl}(X) \to \bigoplus_{x \in \textrm{Sing}(X)} \textrm{Cl}(\mathcal{O}_{X,x}),$$ where $\textrm{Pic}(X)$ and $\textrm{Cl}(X)$ denote the groups of Cartier (resp. Weil) divisors on $X$ up to linear equivalence.

The last arrow is in general not surjective, as can be shown by simple examples.

Now, in the paper by J.Birgener and U. Storch

Zur Berechnung der Divisorenklassengruppen Completterkompletter lokaler Ringe, Nova Acta leopoldinaLeopoldina 52 Nr. 240, 7-63 (1981)

page 11, it is claimed that in the algebraic case one actually has the sequence $$ (2) \quad 0 \to \textrm{Pic}(X) \to \textrm{Cl}(X) \to \bigoplus_{x \in \textrm{Sing}(X)} \textrm{Cl}(\mathcal{O}_{X,x}) \to H^2(X, \mathcal{O}_X^{\ast}) \to 0.$$

My problem is that $(2)$ seems to me not true. For instance, take a smooth cubic threefold $X \subset \mathbb{P}^4$. Then the inclusion $\textrm{Pic}(X) \to \textrm{Cl}(X)$ is an isomorphism, so $(2)$ would imply $ H^2(X, \mathcal{O}_X^{\ast})=0$. However, by the exponential sequence one finds $H^2(X, \mathcal{O}_X^{\ast})=H^3(X, \mathbb{Z})= \mathbb{Z}^{10}$, and this is a contradiction.

So my question is:

am I missing something? If not, can one correct $(2)$ in some way?

Any answer or reference to the existing literature will be appreciated. Thank you!

Let $X$ be a either a projective scheme or a compact complex space. Then one has an exact sequence $$ (1) \quad 0 \to \textrm{Pic}(X) \to \textrm{Cl}(X) \to \bigoplus_{x \in \textrm{Sing}(X)} \textrm{Cl}(\mathcal{O}_{X,x}),$$ where $\textrm{Pic}(X)$ and $\textrm{Cl}(X)$ denote the groups of Cartier (resp. Weil) divisors on $X$ up to linear equivalence.

The last arrow is in general not surjective, as can be shown by simple examples.

Now, in the paper by J.Birgener and U. Storch

Zur Berechnung der Divisorenklassengruppen Completter lokaler Ringe, Nova Acta leopoldina 52 Nr. 240, 7-63 (1981)

page 11, it is claimed that in the algebraic case one actually has the sequence $$ (2) \quad 0 \to \textrm{Pic}(X) \to \textrm{Cl}(X) \to \bigoplus_{x \in \textrm{Sing}(X)} \textrm{Cl}(\mathcal{O}_{X,x}) \to H^2(X, \mathcal{O}_X^{\ast}) \to 0.$$

My problem is that $(2)$ seems to me not true. For instance, take a smooth cubic threefold $X \subset \mathbb{P}^4$. Then the inclusion $\textrm{Pic}(X) \to \textrm{Cl}(X)$ is an isomorphism, so $(2)$ would imply $ H^2(X, \mathcal{O}_X^{\ast})=0$. However, by the exponential sequence one finds $H^2(X, \mathcal{O}_X^{\ast})=H^3(X, \mathbb{Z})= \mathbb{Z}^{10}$, and this is a contradiction.

So my question is:

am I missing something? If not, can one correct $(2)$ in some way?

Any answer or reference to the existing literature will be appreciated. Thank you!

Let $X$ be a either a projective scheme or a compact complex space. Then one has an exact sequence $$ (1) \quad 0 \to \textrm{Pic}(X) \to \textrm{Cl}(X) \to \bigoplus_{x \in \textrm{Sing}(X)} \textrm{Cl}(\mathcal{O}_{X,x}),$$ where $\textrm{Pic}(X)$ and $\textrm{Cl}(X)$ denote the groups of Cartier (resp. Weil) divisors on $X$ up to linear equivalence.

The last arrow is in general not surjective, as can be shown by simple examples.

Now, in the paper by J.Birgener and U. Storch

Zur Berechnung der Divisorenklassengruppen kompletter lokaler Ringe, Nova Acta Leopoldina 52 Nr. 240, 7-63 (1981)

page 11, it is claimed that in the algebraic case one actually has the sequence $$ (2) \quad 0 \to \textrm{Pic}(X) \to \textrm{Cl}(X) \to \bigoplus_{x \in \textrm{Sing}(X)} \textrm{Cl}(\mathcal{O}_{X,x}) \to H^2(X, \mathcal{O}_X^{\ast}) \to 0.$$

My problem is that $(2)$ seems to me not true. For instance, take a smooth cubic threefold $X \subset \mathbb{P}^4$. Then the inclusion $\textrm{Pic}(X) \to \textrm{Cl}(X)$ is an isomorphism, so $(2)$ would imply $ H^2(X, \mathcal{O}_X^{\ast})=0$. However, by the exponential sequence one finds $H^2(X, \mathcal{O}_X^{\ast})=H^3(X, \mathbb{Z})= \mathbb{Z}^{10}$, and this is a contradiction.

So my question is:

am I missing something? If not, can one correct $(2)$ in some way?

Any answer or reference to the existing literature will be appreciated. Thank you!

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Francesco Polizzi
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Exact sequence for divisor class groups

Let $X$ be a either a projective scheme or a compact complex space. Then one has an exact sequence $$ (1) \quad 0 \to \textrm{Pic}(X) \to \textrm{Cl}(X) \to \bigoplus_{x \in \textrm{Sing}(X)} \textrm{Cl}(\mathcal{O}_{X,x}),$$ where $\textrm{Pic}(X)$ and $\textrm{Cl}(X)$ denote the groups of Cartier (resp. Weil) divisors on $X$ up to linear equivalence.

The last arrow is in general not surjective, as can be shown by simple examples.

Now, in the paper by J.Birgener and U. Storch

Zur Berechnung der Divisorenklassengruppen Completter lokaler Ringe, Nova Acta leopoldina 52 Nr. 240, 7-63 (1981)

page 11, it is claimed that in the algebraic case one actually has the sequence $$ (2) \quad 0 \to \textrm{Pic}(X) \to \textrm{Cl}(X) \to \bigoplus_{x \in \textrm{Sing}(X)} \textrm{Cl}(\mathcal{O}_{X,x}) \to H^2(X, \mathcal{O}_X^{\ast}) \to 0.$$

My problem is that $(2)$ seems to me not true. For instance, take a smooth cubic threefold $X \subset \mathbb{P}^4$. Then the inclusion $\textrm{Pic}(X) \to \textrm{Cl}(X)$ is an isomorphism, so $(2)$ would imply $ H^2(X, \mathcal{O}_X^{\ast})=0$. However, by the exponential sequence one finds $H^2(X, \mathcal{O}_X^{\ast})=H^3(X, \mathbb{Z})= \mathbb{Z}^{10}$, and this is a contradiction.

So my question is:

am I missing something? If not, can one correct $(2)$ in some way?

Any answer or reference to the existing literature will be appreciated. Thank you!