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Feb
19
comment Freyd's Geometric Finiteness : An Example Computation
No, continuous does not mean preserving arbitrary joins. It would be more precise to say that it preserves "jointly epimorphic families", in the sense that if $\mathfrak{U} = \{ U_i \to V \}$ is an open cover of $V$, then the image of $\mathfrak{U}$ under $F$ is jointly epimorphic. However it is true that a left exact functor $F : \mathcal{O}(\mathbb{R}) \to \mathcal{B}$ must have image in the subterminals of $\mathcal{B}$, so in this case it is just a matter of preserving joins.
Feb
18
comment Show that the ideal of $k[X_1, X_2, X_3]$ generated by $X_1^3-X_3$ and $X_2^2-X_3$is a prime ideal.
I am referring to the notion of a graded ring. But perhaps your supervisor has a better solution; I'm not a professional algebraic geometer.
Feb
18
comment Show that the ideal of $k[X_1, X_2, X_3]$ generated by $X_1^3-X_3$ and $X_2^2-X_3$is a prime ideal.
You can use a kind of degree argument where $X_1$ has degree $2$, $X_2$ has degree $3$, and $X_3$ has degree $6$. (Also, are you one of my students?)
Feb
18
comment An example of decomposing a projective variety
This one is somewhat difficult. You have probably seen the solution by now, but if not, I'll just quickly say that one component is the line $V(X_2, X_3)$, and the other is the twisted cubic. (Also, the variety is supposed to be considered in $\mathbb{P}^3$.)
Feb
18
answered Show that the ideal of $k[X_1, X_2, X_3]$ generated by $X_1^3-X_3$ and $X_2^2-X_3$is a prime ideal.
Feb
18
comment To what extent is a scheme morphism determined by its topological map?
In the same vein, it is not quite true that a scheme morphism $X \to Y$ is determined by its action on points even if $X$ and $Y$ are varieties over some algebraically closed field $k$. You also have to know that the morphism commutes with the structural morphisms down to $\operatorname{Spec} k$.
Feb
18
comment Why isn't this a valid argument to the “proof” of the Axiom of Countable Choice?
No, there is certainly no problem with creating finite choice functions. Everyone has explained this. The problem is in assembling them together to make an infinite choice function.
Feb
18
comment Why isn't this a valid argument to the “proof” of the Axiom of Countable Choice?
@AloizioMacedo More to the point, mathematics is not temporal. A variable is not undefined at one time and then defined at another, so that you can jump over infinitely many steps and have them all simultaneously defined. Even if you could perform infinitely many tasks and jump ahead to a time when they have all been completed, that does not mean that things will be in a consistent state at that point. For example, consider the process where I turn on a lamp at $t = 0$, turn it off at $t = \frac{1}{2}$, turn it on again at $t = \frac{3}{4}$, etc.; is the lamp on or off at $t = 1$?
Feb
18
comment Are the right derived functors of the inclusion sheaf cohomology?
Yes, that's what I said.
Feb
18
comment Are the right derived functors of the inclusion sheaf cohomology?
As I said, this is because $\check{H}{}^0(U, -)$ commutes with cohomology. Yes, $\check{H}{}^0(U, -) = \Gamma (U, -)$, but this $\Gamma$ is a functor on presheaves, not sheaves! (Recall, colimits of presheaves are computed differently. It turns out that taking sections of presheaves is exact.)
Feb
17
comment In what sense is the forgetful functor $Ab \to Grp$ forgetful?
If you incarnate categories as sets (or classes) of objects and morphisms, then the naïve notion of subcategory works well enough.
Feb
17
comment Projective objects in the category of rings
I know what a projective object is. So let me spell it out for you: If we have a ring homomorphism $0 \to B$, then $B$ must be the zero ring. Then $f : A \to B$ is automatically surjective (hence epic), but there is no chance of getting a ring homomorphism $0 \to A$ unless $A = 0$ as well. But there are plenty of non-zero rings. So $0$ is not projective.
Feb
17
comment How did the ancients view *infinitesimals*?
Surreals seem even less likely than nilpotent infinitesimals, to be honest.
Feb
17
comment In what sense is the forgetful functor $Ab \to Grp$ forgetful?
That's not such a good notion though. It's like trying to define homomorphisms of groups using only the presentation of the source and target groups, without referring to the elements of the actual groups.
Feb
17
comment How can we know arithmetical axioms are consistent?
Variables are syntactic constructs and can be considered outside the context of any particular model. This should be explained in any good logic textbook.
Feb
17
comment How can we know arithmetical axioms are consistent?
I said variables, not values.
Feb
17
comment How can we know arithmetical axioms are consistent?
@ChrisEagle For an algebraic theory (e.g. the theory of rings), it is more conventional to define "consistent" to mean that the equation $x = y$, where $x$ and $y$ are distinct variables, is not provable in that theory. This is because every algebraic theory is consistent in the sense of full first-order logic.
Feb
17
comment Height and minimal number of generators of an ideal
I believe the minimal number is 3, but I don't have an entirely convincing proof. My claim is based on the observation that $I$ is a homogeneous ideal when $x, y, z$ have degrees $3, 4, 5$, respectively, and so the smallest homogeneous relation must be of degree $8$, etc. However it is not obvious that the minimal number of homogeneous generators of a homogeneous ideal is the same as the minimal number of not-necessarily-homogeneous generators.
Feb
17
comment A pedantic question about defining new structures in a path-independent way.
There is no problem doing that on a pairwise basis: you just translate the theorems and proofs from one axiomatisation to another. But there is no universal way of doing so: if there were, all you'd be doing would be picking a distinguished axiomatisation.
Feb
17
comment A pedantic question about defining new structures in a path-independent way.
interface-independence, which any programmer will tell you is an insane proposition that could only be suggested by someone who does not understand what an interface is.