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1h
answered Converse of the implication $V(S)\subseteq V(T)\iff T\subseteq\sqrt{\langle S\rangle}$.
5h
comment Why aren't those Cartier Divisors equivalent?
Dear adrido: I have deleted my first comment which was unfairly harsh. Indeed you had an excellent idea and just replacing divisors by divisor classes makes it technically quite correct.
5h
answered Geometric meaning of intersection multiplicities?
6h
answered Localization module is nonzero
7h
comment Why aren't those Cartier Divisors equivalent?
Dear adrido: as I explained in another comment, restricting a line bundle is not the same as restricting a divisor but is the same as restricting a divisorclass. So, although the morphism $CaCl(X)\to Pic(X)$ is indeed bijective here (and practically everywhere else) we cannot replace the study of line bundles by divisors but only by divisor classes. I have written a little edit on the subject.
7h
revised Why aren't those Cartier Divisors equivalent?
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8h
comment Why aren't those Cartier Divisors equivalent?
Dear adrido, you are perfectly right : the subtle point is that while one may not (a priori) pull back divisors one may pull back divisor classes or line bundles. This is an old trick in intersection theory: if you want to self-intersect a line in $\mathbb P^2$, you move one copy of it in its divisor class, obtain a distinct line and conclude that the self-intersection was $1$.
15h
comment Noetherian ring and prime ideal contained in an invertible maximal ideal.
Dear@user26857: no I don't mind, especially since I don't remember why I used the unfamiliar notation in the first place!
16h
answered Why aren't those Cartier Divisors equivalent?
1d
comment Etymology of “flabby” or “flasque” sheaf
No, flabby is a totally accurate translation of flasque.
1d
revised Etymology of “flabby” or “flasque” sheaf
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1d
answered Etymology of “flabby” or “flasque” sheaf
1d
comment Is the ring of entire functions coherent?
Dear @rschwieb, be assured that I meant nothing disturbing in my comment (but I'm happy to hear that it mostly amused you!). I understand and even sympathize with the wish to have a completely elementary solution but since nobody (yet?) has given such a solution I am happy to have an opportunity to advertize the power of sheaves , which I consider a beautiful and natural tool in the context of this question. And you are quite right that one can disagree without being disagreeable :)
1d
comment Contraction of non-zero prime ideals in the ring of algebraic integers
Dear @Pavel, everything in your comment is absolutely correct: +1 for your answer and +1 for your comment! By the way (for commutative algebraists): an analogous result is still true for any integrally closed (=normal) domain and any extension field of its field of fractions.
1d
comment Contraction of non-zero prime ideals in the ring of algebraic integers
What is "the minimal polynomial" of an integral element ?
1d
comment Is the ring of entire functions coherent?
Dear @rschwieb, allow me to disagree with your "Alas": I would be very happy if there were no elementary proof of coherence! This would show Martin Brandenburg and people of similar disposition that techniques of complex analytic geometry involving coherent sheaves and the like (which were my first love in mathematics) are really necessary, even for apparently elementary problems on holomorphic functions :-)
1d
comment Is the ring of entire functions coherent?
Dear @rschwieb: ok, here are the answers. The ring $R=\mathcal O(\mathbb C^n)$ is not Prüfer for $n\geq2$ because the ideal $I=(z_1,...,z_n)$ is not invertible, since $I^{-1}=R$ and thus $I\cdot I^{-1}=I\neq R$ . Since Bézout implies Prüfer, that ring is not Bézout either. For $n=1$ the ring $\mathcal O(\mathbb C)$ is well known (en.wikipedia.org/wiki/B%C3%A9zout_domain#Examples) to be Bézout and is thus Prüfer too.
1d
comment Is the ring of entire functions coherent?
Dear @rschwieb: why not ask these questions on a separate thread, for clarity's sake?
1d
revised Is the ring of entire functions coherent?
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1d
comment Is the ring of entire functions coherent?
@Martin: Yes, I am quite happy to use that the sheaf $\mathcal O_X$ is coherent. However I do not claim that one needs to use it: if you (or someone else) can provide a more elementary proof , I'd be delighted to read it. As for references, I have only used standard facts which you can look up in Grauert-Remmert's Theory of Stein Spaces.