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1d
comment Cutting a torus enough times disconnects it
@Dan, I'd be interested in that answer. I'm learning homology theory right now just to solve this problem.
2d
comment Cutting a torus enough times disconnects it
How can this be proven? Btw, I don't care about the value of $k$, just want to show it's finite. And I prefer an easier proof to a proof of an exact bound. Also you cannot relax it too much. You have to avoid the simple $M_1=M_2=\ldots=M_k$ for example.
2d
comment Cutting a torus enough times disconnects it
@SammyBlack, $\mathbb T^1$ is trivial but in $\mathbb T^2$ I already don't know how to show it.
2d
revised Cutting a torus enough times disconnects it
Changed the phrasing of the question, after realizing the way it was written was probably wrong.
2d
asked Cutting a torus enough times disconnects it
Oct
9
answered What is an example of function f: N to Z that is a bijection?
Sep
30
awarded  Explainer
Sep
23
comment If $r^{n-r}=k^{n-k}$, when is this true other than $r=k$?
You basically rephrased the problem with new variables $p = n - r$ and $q = n - k$...
Sep
16
accepted Discrete set of zeroes of polynomials must be finite?
Sep
14
comment Discrete set of zeroes of polynomials must be finite?
It does not seem to follow by logic alone. I understand that a point is irreducible. But what if some infinite sets of points were also irreducible? Maybe there is somehow a Zariski-closed discrete set such that removing 1 point makes it no longer Zariski-closed, and thus, it is irreducible? It would not logically contradict that Z has finitely many irreducible components if Z were such a set. I am missing another piece of the puzzle, which is that an infinite discrete set cannot be irreducible.
Sep
14
comment Discrete set of zeroes of polynomials must be finite?
I know about the Zariski topology, but I don't know much about it beyond the basic definitions, particularly why an infinite 0-dimensional set cannot be irreducible.
Sep
13
comment Discrete set of zeroes of polynomials must be finite?
Unfortunately I lack a lot of the algebraic background, and find it hard to think about my problem in terms of ideals, spectrums, Noetherian and Artinian conditions... So for now I prefer jo wehler's answer. Thanks!
Sep
13
comment Discrete set of zeroes of polynomials must be finite?
Thanks! I suspected the answer might be something like this. But I don't really know algebraic geometry so there are two things I still don't understand (probably very trivial): 1. Why isn't an infinite subset of $Z$ irreducible? 2. Why is every algebraic variety a finite union of irreducible components?
Sep
13
revised Discrete set of zeroes of polynomials must be finite?
Small edit to make the question clearer
Sep
13
comment Discrete set of zeroes of polynomials must be finite?
@Krokop I think you misunderstood the question. I am asking whether if you suppose $Z$ is discrete then is it finite? I edited the question to make it a bit clearer.
Sep
13
comment Discrete set of zeroes of polynomials must be finite?
@Krokop this is irrelevant, my $F$ is a multivariate function.
Sep
13
comment Discrete set of zeroes of polynomials must be finite?
Yes, it doesn't really pertain to my question as I am considering intersections of several zero sets, which can be discrete...
Sep
13
comment Discrete set of zeroes of polynomials must be finite?
@Crostul, that's a good point, but each $F_i$ has a zero set that is, in general, not bounded. For instance $f(x,y) = x-y$ has not at all a bounded zero set. It is not clear to me why the intersection must be bounded, once it is discrete.
Sep
13
comment Discrete set of zeroes of polynomials must be finite?
@idm I don't understand. $\mathbb{C}^n$ is not a field, it's a vector space... By Bezout's theorem if $Z$ is finite then it has at most $\deg F_1 \cdot \ldots \cdot \deg F_n$ values, but I don't know it's finite, only that it's discrete.
Sep
13
asked Discrete set of zeroes of polynomials must be finite?