Jesko Hüttenhain
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5,952
76/100 score
 Aug 30 awarded Self-Learner Aug 12 revised Why would we a priori expect $V(I)$ to satisfy axioms to define the closed sets for a topology on $\text{Proj}(S)$? added 873 characters in body Aug 5 comment Continuous maps are morphisms of varieties? Why are $g\circ\phi$ and $h\circ\phi$ polynomials? Aug 4 revised $A+B+C=2149$, Find $A$ deleted 51 characters in body Aug 4 answered $A+B+C=2149$, Find $A$ Aug 4 comment Why would we a priori expect $V(I)$ to satisfy axioms to define the closed sets for a topology on $\text{Proj}(S)$? Whoops, if you liked the reply then here it is again. I thought I might have misunderstood your question =). Aug 4 answered Why would we a priori expect $V(I)$ to satisfy axioms to define the closed sets for a topology on $\text{Proj}(S)$? Jul 28 comment Finding a path in a graph by its hash value At this point you'd need to make the question more formal, I think. What are your exact requirements and what properties does the hash function have? Jul 28 answered Finding a path in a graph by its hash value Jul 27 comment Finding a path in a graph by its hash value This is trivially possible if there are finitely many vertices, so I assume there is an infinite number of vertices - but in this case, there could be infinitely many paths of length $n$, how are you "given" all these hash values? Jul 17 comment Algorithmic question about algebraic varieties and affinely independence What precisely is your input and what is your algorithm allowed to do? If you are allowed to use randomization for example, just pick the $u^{(i)}$ randomly and you'll be fine. Jul 16 answered Reference request for a theorem on maps to normal varieties with equidimensional fibers being open Jul 13 comment Fiber dimension theorem for locally closed sets @guest_09072015: Yes, this is true, but simply because a nonempty, open subset of an irreducible algebraic variety is again an irreducible algebraic variety. It might not be affine, but it is a variety. Jul 12 comment Fiber dimension theorem for locally closed sets It is open in $\mathbb C^n$, but it can be given the structure of an affine variety. Inside $\mathbb C^{n}\times\mathbb C$, it correponds to the points $(z,t)$ subject to the polynomial condition $f(z)\cdot t = 1$. Jul 9 answered Fiber dimension theorem for locally closed sets Jul 5 revised Projection is an open map cosmetics Jul 5 comment Projection is an open map @GFR: True & done. Thanks for the comment btw, I realize now that tomasz already pointed it out before but I completely overlooked that in his comment. Jul 5 revised Projection is an open map deleted 6 characters in body Jul 5 comment Projection is an open map @GFR: Indeed, that was a serious blunder. Should be fixed now. Jul 5 revised Projection is an open map Fixed a serious mistake