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Aug
12
comment how to lift geometrically integralness using etale(+something else) morphisms
What is your definition of variety? If it's just "reduced and of finite type", then $X \amalg X \to X$ is a counterexample.
Aug
12
answered Morphism schemes after base extension
Aug
12
comment Every closed (not-necessarily symmetric) monoidal category is canonically self-enriched, right?
It is unusual to consider categories enriched in something other than a symmetric monoidal category in the first place – partly because a lot of the theory becomes awkward (e.g. opposite categories).
Aug
11
comment A question regarding a lemma in Perrin's Algebraic Geometry.
If $k$ is algebraically closed, it can't have any non-trivial algebraic extensions...
Aug
11
revised A question regarding a lemma in Perrin's Algebraic Geometry.
edited tags; edited title
Aug
11
revised Equality in de Rham cohomology
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Aug
11
revised De Rham cohomology group
added 1 character in body; edited tags; edited title
Aug
11
comment De Rham cohomology group
You should take $\Omega^{-1} (U) = 0$ for the purposes of this formula.
Aug
11
answered is tangent bundle of $S^n$ an algebraic variety?
Aug
10
comment Intuition behind $\sigma$-algebras: why union?
The inclusion-exclusion principle tells you what the measure of the union is, provided you know the measure of the intersection. Do you believe that the intersection of measurable sets should be measurable?
Aug
10
answered “Nice proof” that the unit of the left Kan extension of $F$ is an isomorphism, if $F$ is fully faithful
Aug
10
comment “Nice proof” that the unit of the left Kan extension of $F$ is an isomorphism, if $F$ is fully faithful
It's confusing to see $\epsilon$ for the unit. Normally, the unit is $\eta$.
Aug
9
comment Finding the ring of regular functions on $X-S$
I guess your varieties are integral. It looks correct.
Aug
9
comment Finding the ring of regular functions on $X-S$
Is $S$ supposed to be closed? If $X \setminus S$ admits an open cover then $X \setminus S$ must itself be open...
Aug
9
comment Map of monads and left adjoints
The left adjoint exists as soon as $X^{T'}$ has reflexive coequalisers, so first one has to find $X$ and $T'$ such that $X^{T'}$ does not have reflexive coequalisers...
Aug
8
comment Krull dimension of quotient by principal ideal
@zcn It appears to me that your example has a geometric analogue: take $k$ a field, $R = (k [s]_{(s)}) [t]$ and $x = s t - 1$. Then the picture one thinks of is the hyperbola $\{ s t = 1 \} \subset \mathbb{A}^2_k$ (which is 1-dimensional) and the germ of $\mathbb{A}^2_k$ around $\{ s = 0 \}$ (which is 2-dimensional), and the intersection is unsurprisingly 0-dimensional.
Aug
8
comment $\dim(R/x) = \dim(R)-1$ for Noetherian integral domains?
You should also require $x$ to be non-unit, otherwise $R / (x) \cong \{ 0 \}$.
Aug
8
comment Are the topology of a manifold and the topology induced by the metric of a manifold the same?
No, that wouldn't work. In a pseudo-Riemannian metric there can be distinct points that are distance zero apart, which cannot happen in a metric space.
Aug
8
comment Are the topology of a manifold and the topology induced by the metric of a manifold the same?
Of course, the metric must be Riemannian. You don't get a metric space otherwise.
Aug
8
revised Are the topology of a manifold and the topology induced by the metric of a manifold the same?
edited tags