Example 52.27.8. Proposition 52.27.6 cannot be extended to quotients

$A = B/(f_1, \ldots , f_ r)$

where $B$ is regular and $\dim (B) - r \geq 4$. In other words, the condition that $f_1, \ldots , f_ r$ be a regular sequence is (in general) needed for vanishing of the Picard group of the punctured spectrum of $A$. Namely, let $k$ be a field and set

$A = k[[a, b, x, y, z, u, v, w]]/(a^3, b^3, xa^2 + yab + zb^2, w^2)$

Observe that $A = A_0[w]/(w^2)$ with $A_0 = k[[a, b, x, y, z, u, v]]/(a^3, b^3, xa^2 + yab + zb^2)$. We will show below that $A_0$ has depth $2$. Denote $U$ the punctured spectrum of $A$ and $U_0$ the punctured spectrum of $A_0$. Observe there is a short exact sequence $0 \to A_0 \to A \to A_0 \to 0$ where the first arrow is given by multiplication by $w$. By More on Morphisms, Lemma 37.4.1 we find that there is an exact sequence

$H^0(U, \mathcal{O}_ U^*) \to H^0(U_0, \mathcal{O}_{U_0}^*) \to H^1(U_0, \mathcal{O}_{U_0}) \to \mathop{\mathrm{Pic}}\nolimits (U)$

Since the depth of $A_0$ and hence $A$ is $2$ we see that $H^0(U_0, \mathcal{O}_{U_0}) = A_0$ and $H^0(U, \mathcal{O}_ U) = A$ and that $H^1(U_0, \mathcal{O}_{U_0})$ is nonzero, see Dualizing Complexes, Lemma 47.11.1 and Local Cohomology, Lemma 51.2.2. Thus the last arrow displayed above is nonzero and we conclude that $\mathop{\mathrm{Pic}}\nolimits (U)$ is nonzero.

To show that $A_0$ has depth $2$ it suffices to show that $A_1 = k[[a, b, x, y, z]]/(a^3, b^3, xa^2 + yab + zb^2)$ has depth $0$. This is true because $a^2b^2$ maps to a nonzero element of $A_1$ which is annihilated by each of the variables $a, b, x, y, z$. For example $ya^2b^2 = (yab)(ab) = - (xa^2 + zb^2)(ab) = -xa^3b - yab^3 = 0$ in $A_1$. The other cases are similar.

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