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1d
asked Do these non-homotopic maps induce the same map in reduced homology?
Nov
30
asked Why Do We Associate Quadratic Forms To Symmetric Matrices Rather Than Upper Rriangular Matrices?
Oct
26
comment Showing explicitly that a cubic surface is unirational.
which term? Originally, a variety $X$ was called birational (over $k$) if there was an invertible map $\mathbb{P}^n_k\dashrightarrow X$ for some $n$ and unirational if the map $\mathbb{P}^n_k\dashrightarrow X$ just goes one way. Then, at some stage, the prefix `bi' got dropped from birational.
Oct
25
revised Showing explicitly that a cubic surface is unirational.
added 13 characters in body
Oct
25
asked Showing explicitly that a cubic surface is unirational.
Oct
17
asked Set union in the category of sets
Oct
11
accepted $(1-e^{\frac{2\pi i}{n} })(1-e^{\frac{4\pi i}{n} })…(1-e^{\frac{(n-1)2\pi i}{n} })=n$ for each natural $n\geq 2$.
Oct
10
accepted Showing that a subset of the complex plane is open.
Oct
10
asked Showing that a subset of the complex plane is open.
Oct
9
revised $(1-e^{\frac{2\pi i}{n} })(1-e^{\frac{4\pi i}{n} })…(1-e^{\frac{(n-1)2\pi i}{n} })=n$ for each natural $n\geq 2$.
edited body
Oct
9
reviewed Approve $(1-e^{\frac{2\pi i}{n} })(1-e^{\frac{4\pi i}{n} })…(1-e^{\frac{(n-1)2\pi i}{n} })=n$ for each natural $n\geq 2$.
Oct
9
awarded  Custodian
Oct
9
revised $(1-e^{\frac{2\pi i}{n} })(1-e^{\frac{4\pi i}{n} })…(1-e^{\frac{(n-1)2\pi i}{n} })=n$ for each natural $n\geq 2$.
edited body; edited title
Oct
9
revised $(1-e^{\frac{2\pi i}{n} })(1-e^{\frac{4\pi i}{n} })…(1-e^{\frac{(n-1)2\pi i}{n} })=n$ for each natural $n\geq 2$.
added 1 character in body; edited title
Oct
9
comment $(1-e^{\frac{2\pi i}{n} })(1-e^{\frac{4\pi i}{n} })…(1-e^{\frac{(n-1)2\pi i}{n} })=n$ for each natural $n\geq 2$.
the statement for $n$ cannot be obtained by simply multiplying the statement for $n-1$ by an additional factor, because all the factors for $n$ and $n-1$ would be different.
Oct
9
asked $(1-e^{\frac{2\pi i}{n} })(1-e^{\frac{4\pi i}{n} })…(1-e^{\frac{(n-1)2\pi i}{n} })=n$ for each natural $n\geq 2$.
Sep
29
awarded  Yearling
Sep
21
accepted The existence of a bound for degrees of subsheaves of a coherent sheaf.
Sep
21
accepted Does $\operatorname{D}(X)\cong \operatorname{D}(Y)$ imply $X\cong Y$?
Sep
21
comment Does $\operatorname{D}(X)\cong \operatorname{D}(Y)$ imply $X\cong Y$?
it's very unnatural to talk about isomorphisms of categories even when they are not derived