38,487 reputation
244116
bio website
location
age
visits member for 3 years, 10 months
seen 13 mins ago

Dec
29
comment On colim $Hom_{A-alg}(B, C_i)$
I think it is, when you think about how to construct coequalisers in these concrete categories. But if you insist...
Dec
29
revised On colim $Hom_{A-alg}(B, C_i)$
added 318 characters in body
Dec
29
comment On colim $Hom_{A-alg}(B, C_i)$
In that case, $F Y = A[Y_1, Y_2]$, $p_1(Y_1) = f_1$, $p_1(Y_2) = f_2$, $p_2(Y_1) = 0$, $p_2(Y_2) = 0$.
Dec
28
answered A number of dice rolls to see every number at least once
Dec
28
comment On colim $Hom_{A-alg}(B, C_i)$
$F Y$ is also a polynomial ring over $A$. It tabulates the relations that generate $B$. (More precisely, if we write $p_1, p_2 : F Y \to F X$ for the two homomorphisms, the ideal generated by the image of $p_1 - p_2$ is the kernel of $F X \to B$.)
Dec
28
comment filtered colimit of $Hom_{A_i}(M_0\otimes_{A_0} A_i, N_0\otimes_{A_0} A_i)$
The proof I posted here covers this.
Dec
28
comment filtered colimit of $Hom_{A_i}(M_0\otimes_{A_0} A_i, N_0\otimes_{A_0} A_i)$
$\textrm{Hom}_{A_0}(M_0, -)$ preserves filtered colimits if (and only if) $M_0$ is a finitely-presented $A_0$-module.
Dec
28
awarded  Custodian
Dec
28
reviewed No Action Needed A number of dice rolls to see every number at least once
Dec
28
reviewed No Action Needed How to prove that the absolute difference of two numbers
Dec
28
comment Is it possible to prove everything in mathematics by theorem provers such as Coq?
The set of theorems of a computably-axiomatised first-order theory is computably enumerable – this is indisputable. So perhaps the question should be, can all of mathematics be captured in a sufficiently powerful first-order system?
Dec
28
comment What exactly is $\mathcal{O}_X$ as featured in the exact sequence of a hypersurface?
See this question.
Dec
28
comment Prove that $\Diamond\Box p \rightarrow \Diamond (\Box (p\land q) \lor \Box(p\land\neg q))$ does not define a first-order condition on frames
@AsafKaragila The forcing relation is, in some sense, a special case of the $\Vdash$ relation for modal semantics...
Dec
28
answered Hom functor and left exactness
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.