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lefuneste
  • Member for 4 years, 7 months
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On the definition of a principal ideal sheaf
Thanks for your answer. In Mumford-Oda's book they explain that you can define a sheaf by giving its values on a basis of open subsets stable under finite intersections and checking the natural sheaf conditions for the basis. But on a non separated scheme the open affines are not stable under finite intersection. Does Qing Liu's recipe nevertheless yield a sheaf?
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How to write down an explicit equation of given degree yielding a smooth hypersurface in a projective space?
@alpoge: You are very kind but I think my examples were actually flawed. You really saved my day and I heartily congratulate you for finding such an advanced and pertinent reference. Thanks a lot!
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How to write down an explicit equation of given degree yielding a smooth hypersurface in a projective space?
@alpoge Ofer Gaber's example in your reference is much better than mine. Could you please explain in an answer, (which can then be upvoted) , since I have deleted my own answer .
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How to write down an explicit equation of given degree yielding a smooth hypersurface in a projective space?
In the case $p\vert d$ the equation mentioned by @alpoge is due to Ofer Gabber and is $x_0^d+\sum _{i=0}^{n-1}x_ix_{i+1}^{d-1}=0$. This is much better than my answer, which I'm thus deleting.
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Bézout ring with non-trivial Picard group?
My example above shows that the assertion in your following sentence according to which $\text{Cl}(R)$ and $\text{Pic}(R)$ are isomorphic for integral domains or noetherian rings is false too.
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Bézout ring with non-trivial Picard group?
You write "there is a canonical inclusion $\text{Cl}(R) \longrightarrow \text{Pic}(R)$...". This is not true: there is no canonical map $\text{Cl}(R) \longrightarrow \text{Pic}(R)$, let alone an injective one. Hartshorne shows on page 134 of his Algebraic Geometry that for the ring $R$ of regular functions on the the affine cone $xy-z^2=0$ in affine $3$-space $\mathbb A^3_k$ over a field $k$ we have $\text{Cl}(R)=\mathbb Z/2 \mathbb Z$ but $\text{Pic}(R)=0$.
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Spectrum of a ring (studied by Krull?) of rational functions
Thanks again, Piotr: your quite subtle Edit perfectly explains why I was uneasy with my calculations, which prompted me to post the question in the first place. (By the way, are you of Russian origin as your first name seems to indicate?)
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Spectrum of a ring (studied by Krull?) of rational functions
Dear @Piotr: you write "Maybe I'm mistaken, but what you wrote shows that the maximal ideal is NOT principal" You are not mistaken at all and your fantastic comment explains away all the problems I had with my erroneous belief that the maximal ideal was principal. I can't thank you enough for this brilliant explanation.
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Spectrum of a ring (studied by Krull?) of rational functions
I think that $R$ is a one-dimensional local domain with principal maximal ideal (and spectrum of cardinality $2$) . However it is not a DVR, because $x$ and $xy$ are in the maximal ideal but neither is divisible by the other. The only explanation for this strange result would be that $R$ is not noetherian.