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I have a particular graded ring whose Picard group I am trying to compute, namely $$\mathbb{Z}[x,y,z]/(2x, x^3, xy, z^2-4y),$$ where $|x|=1$, $|y|=4$, and $|z|=8$. How does one go about computing this? I know that if I can do some algebra to write this ring as, say, a polynomial ring over $\mathbb{Z}$ ,or any other UFD, then I know that its Picard group is trivial, but I am at a lost for how to compute the Picard group here or in general. Any advice is appreciated.

Ben Steffan
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categorically_stupid
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1 Answers1

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One very useful property of Picard groups is that they are unaffected by modding out by nilpotents. This is an immediate consequence of the basic fact that finite projective modules lift modulo nil ideals. See, for example, Stacks [Lemma 0D47]. In particular, for any ring $A$, $\operatorname{Pic}(A) = \operatorname{Pic}(A_{red})$.

In your ring, both $x$ and $z$ are nilpotents, so the computation becomes trivial.

Badam Baplan
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  • Thanks! I had a typo in my original ring: the relation should be $z^2-4y$, not $z^2-4x$, but knowing that the Picard is unaffected by modding out by nilpotents solves the problem. Since $x$ is nilpotent, we are reduced to computing the Picard group of $\mathbb{Z}[x,y]/(z^2-4y)$. This is isomorphic via a change of variables to $\mathbb{Z}[t]$, hence the Picard group is $\mathbb{Z}$. – categorically_stupid Mar 13 '25 at 19:27
  • @categorically_stupid when you say the Picard group is $\mathbb{Z}$, I think you mean $0$? – Badam Baplan Mar 13 '25 at 19:59
  • I meant $\mathbb{Z}$, but now I am confused because it should certainly be 0. In Lemma 4.9 of this paper https://arxiv.org/pdf/1705.02810 it is claimed that $\text{Pic}(ku_) = \mathbb{Z}$, where $ku_ = \mathbb{Z}_2^\wedge[u]$ is the homotopy ring of connective complex $K$-theory with $|u|=2$. This contradicts what you and others have said now, which leaves me confused. – categorically_stupid Mar 13 '25 at 22:00
  • @categorically_stupid what's your definition of the picard group? is it from the derived module category? If so, that could account for the difference. – Badam Baplan Mar 13 '25 at 23:26