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Dec
28
comment Hom functor and left exactness
However, if we know that the hypothesis holds for all $M$, then the claim is true.
Dec
28
answered filtered colimit of $Hom_{A_i}(M_0\otimes_{A_0} A_i, N_0\otimes_{A_0} A_i)$
Dec
28
comment Partition of Unity for the Divisor Sheaf
Couldn't you do it arbitrarily? Unlike constructing smooth partitions of unity, it seems to me here that a discontinuous integer-valued one will work just as well...
Dec
28
comment Hom functor and left exactness
Use the Yoneda lemma.
Dec
28
comment On colim $Hom_{A-alg}(B, C_i)$
A direct proof is ugly. You of all people should know that. Anyway, an $R$-module $M$ is of finite presentation if and only if there is an exact sequence of the form $R^n \to R^m \to M \to 0$. My coequaliser diagram is the non-additive version of this fact.
Dec
28
comment On colim $Hom_{A-alg}(B, C_i)$
That's a somewhat high-tech definition if you think about it. Fundamentally, an algebra of finite presentation is something that is generated by finitely many elements and finitely many equations. This is what the coequaliser diagram expresses.
Dec
27
comment Does fiber product preserve limits?
This functor is used whereever Grothendieck's relative point of view is fruitful: algebraic geometry is just one of them. There is literally no algebraic geometry content to this question or its answer.
Dec
27
comment Does fiber product preserve limits?
I don't really know what you're asking. This argument can be found in various category theory textbooks – certainly in Awodey (Ch. 9).
Dec
27
revised Does fiber product preserve limits?
deleted 1 characters in body
Dec
27
answered On colim $Hom_{A-alg}(B, C_i)$
Dec
27
revised On colim $Hom_{A-alg}(B, C_i)$
edited tags
Dec
27
comment On colim $Hom_{A-alg}(B, C_i)$
Yes. This is actually an "if and only if" and is true for all kinds of finitary algebraic structures.
Dec
27
comment Does fiber product preserve limits?
The notation we use is completely standard in algebraic geometry, for example, and fairly common elsewhere. It is an accepted abuse of notation.
Dec
27
comment “Too simple to be true”
The fake proof is easy to fix though: change the base field from $\mathbb{C}$ to $\overline{\mathbb{C}(x_{1,1}, \ldots, x_{n,n})}$...
Dec
27
comment Quickie on Boolean valued models
A boolean algebra is in particular a poset, so you can embed it in a complete boolean algebra as described. (Alternatively, you could take its completion, using Stone duality.) You can construct boolean-valued models of any first-order theory, but there are extra complications when working with set theory in particular.
Dec
27
comment Does fiber product preserve limits?
How could it possibly be a functor on $\mathcal{C}$? There is no chance of confusion.
Dec
27
comment Does fiber product preserve limits?
True, but this functor is not a limit. The functor $(\mathcal{C} \downarrow S) \times (\mathcal{C} \downarrow S) \to (\mathcal{C} \downarrow S)$ that sends $(X \to S, Y \to S)$ to $X \times_S Y$ is, and it is right adjoint to the diagonal functor $\Delta : (\mathcal{C} \downarrow S) \to (\mathcal{C} \downarrow S) \times (\mathcal{C} \downarrow S)$.
Dec
27
answered Does fiber product preserve limits?
Dec
25
comment Integral domain of prime characteristic
Binomial theorem.
Dec
25
comment If every assignment that models $F$ also models $G$, does it mean that $F=G$?
It depends on what you mean by $=$, really...