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assume we have a curve X of any genus with one non separating node. Call it $x$. Which is the explicit form of the boundary map $ \mathbb{G}_m\rightarrow H^1(X,\mathcal{O}_X^{*}) $ It seems to me that I have to do the following: given a covering of the curve $\{U_{\alpha}\}$ and $\lambda\in\mathbb{G}_m$ I have to extend locally on the normalization to $\{f_{\alpha}\}$ with $\frac{f_{\alpha}(p)}{f_{\alpha}(q)}=\lambda$, where $p$ and $q$ are the points mapped to the node and then I have to consider the line bundle on $X$ with transition functions $g_{\alpha\beta}=\frac{f_{\alpha}}{f_{\beta}}$. This is correct? How is it described geometrically the action of $\mathbb{G}_m$ on $H^1(X,\mathcal{O}_X^*)$? Why if I have a separating node I get an isomorphism between the Picard groups of $X$ and the one of its normalization?

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Let $\widetilde{X}$ be the normalization of $X$, and let $x_0$ and $x_1$ be the two points of $\widetilde{X}$ lying above the node $x$.

If $\widetilde{\mathcal L}$ is a line bundle on $\widetilde{X}$ and $\phi$ is an element of Isom$(\widetilde{\mathcal L}_{x_0},\widetilde{\mathcal L}_{x_1})$, then you can make a line bundle $\mathcal L$ on $X$ by identifying the fibres $\widetilde{\mathcal L}_{x_0}$ and $\widetilde{\mathcal L}_{x_1}$ via $\phi$.

Note that $\widetilde{\mathcal L}_{x_0}$ and $\widetilde{\mathcal L}_{x_1}$ are both one-dimensional vector spaces, and so the Isom space between them is non-canonically identified with $\mathbb G_m$. More canonically, it is a $\mathbb G_m$-torsor, i.e. we can multiply any isomorphism $\phi$ by an element of $\mathbb G_m$, and this makes the Isom space a principal homogeneous space for $\mathbb G_m$.

Conversely, if $\mathcal L$ is a line bundle on $X$, then its pull-back to $\widetilde{X}$ is a line bundle $\widetilde{L}$, equipped with an isomorphism $\phi$ between $\widetilde{\mathcal L}_{x_0}$ and $\widetilde{\mathcal L}_{x_1}$. (Because both are canonically identified with the fibre $\mathcal L_x$.)

Thus line bundles on $X$ are the same as pairs $(\widetilde{L},\phi)$ where $\widetilde{L}$ is a line bundle on $\widetilde{X}$ and $\phi$ is an isomorphism between $\widetilde{\mathcal L}_{x_0}$ and $\widetilde{\mathcal L}_{x_1}$.

The $\mathbb G_m$-action is just the action on $\phi$ given by scaling.

The boundary map is given by taking $\widetilde{L} = \mathcal O_{\widetilde{X}}$, the trivial line bundle on $\widetilde{X}$. In this case, the fibres at $x_0$ and $x_1$ are both canonically equal to $k$ (the ground field), and so the Isom space between them is canonically $\mathbb G_m$.

Thus we get $\mathbb G_m$ being canonically identified with those $\mathcal L$ for which $\widetilde{\mathcal L}$ is trivial. (Actually, this identification, and also the $\mathbb G_m$-action, depend on the labelling of $x_0$ and $x_1$. If we switch those two points, it is the same as applying $\lambda \mapsto \lambda^{-1}$ to $\mathbb G_m$.)

If $x$ is a separating node, then the situation is different: the normalization is not connected, it has two pieces, say $X_0$ and $X_1$, each with a unique point $x_0$ and $x_1$ lying over $x$. To get a line bundle on $X$ we have to choose $\mathcal L_0$ and $\mathcal L_1$ on $X_0$ and $X_1$ and then identify the fibre of $\mathcal L_0$ at $x_0$ with the fibre of $\mathcal L_1$ at $x_1$. Again, this identification is only determined up to scaling by an element of $\mathbb G_m$, but if we choose two different identifications, which say differ by a scaling factor of $\lambda$, we nevertheless get the same line bundle on $X$, the reason being that we can apply the automorphism $1 \times \lambda$ to the line bundle $\mathcal L_0 \coprod \mathcal L_1$ on $X_0 \coprod X_1$ (i.e. the automorphism which is $1$ on the first factor and multiplication by $\lambda$ on the second factor), and this will take the first identification to the second.

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