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Jun
18
comment What do I call a covariant functor which is a filtered colimit of representable functors?
Well, the category of flat functors $\mathcal{C} \to \mathbf{Set}$ is actually equivalent to the opposite of the category of pro-objects in $\mathcal{C}$, so it's not entirely unreasonable.
Jun
18
answered What do I call a covariant functor which is a filtered colimit of representable functors?
Jun
18
accepted Quasicoherent ideal sheaves on open subschemes
Jun
17
comment Is there a universal property for the ultraproduct?
Actually, I was thinking about this one, but I wasn't able to find it just now.
Jun
17
comment Is there a universal property for the ultraproduct?
I recall a question on MO. It's a colimit over a certain diagram derived from the filter.
Jun
17
comment Why is there apparently no general notion of structure-homomorphism?
Sometimes there is no obvious notion of homomorphism. For instance, what's the appropriate notion of morphism for Banach spaces?
Jun
16
revised How to find a minimal polynomial (field theory)
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Jun
16
answered How to find a minimal polynomial (field theory)
Jun
16
comment Every $n$-dimensional variety is birationally equivalent to a hypersurface in $\mathbb{A}^{n+1}.$
If the conclusion holds, then $K (X)$ must admit a transcendence basis over which it is simply generated – separability is sufficient but not necessary. So for a counterexample, I suppose one has to find a field that is not simply generated over any transcendence basis. Perhaps something like the variety $\{ x^p - z, y^p - w \} \subset \mathbb{A}^4$ over $\overline{\mathbb{F}_p}$?
Jun
16
answered Number of generators of the maximal ideals in polynomial rings over a field
Jun
15
comment Alternate pullback bundle construction
@Lord_Farin The question has not been answered – my comment is a general one and does not address whether or not it can be done in this specific case.
Jun
15
revised How does one show that definition of Betti number and its “informal definition” are equal?
edited tags
Jun
14
awarded  Notable Question
Jun
13
answered Every $n$-dimensional variety is birationally equivalent to a hypersurface in $\mathbb{A}^{n+1}.$
Jun
13
comment A commutative ring $A$ is an algebra over a field $F$ if and only if $A$ contains (an isomorphic copy of) $F$ as a subring
The headline claim is false: the trivial ring is an $F$-algebra. But that is the only problem. For the second problem you don't have to take $R = M$, but what you suggested works too.
Jun
13
comment Every $n$-dimensional variety is birationally equivalent to a hypersurface in $\mathbb{A}^{n+1}.$
That's where the separability hypothesis comes in – you need to be able to know that $k (X)$ is generated by one element over your transcendence basis.
Jun
13
comment Every $n$-dimensional variety is birationally equivalent to a hypersurface in $\mathbb{A}^{n+1}.$
You probably should assume some separability hypothesis. But here's a hint: take a transcendence basis for the function field...
Jun
13
answered Monads on Set and their strength
Jun
13
revised which of the following is/are algebraic over rationals
added 4 characters in body
Jun
13
comment Definition of the Coproduct of Categories?
The coproduct of ordinary categories differs from the coproduct of $\mathbf{Ab}$-enriched categories, which also differs from the coproduct of additive categories.