User paul vankoughnett - MathOverflow

Answer by Paul VanKoughnett for Composition in the category quotient

2013-05-18T04:57:24Z

First, you actually want $((N'' + N')/N')$ every time you have $(N'' + N')/N$. If this was the source of your confusion, sweet. If not, here goes. (Also, the double underscores in the NUMDAM url got messed up by MO's processing; here's a working link.)

There's a short exact sequence $0 \to M'/M'' \to M/M'' \to M/M' \to 0$, and $M'/M'' \cong N/(N'' + N')$, since $M''$ is the kernel of the composition $M' \to N/N' \to N/(N'' + N')$; finally, $N/(N'' + N')$ is a quotient of $N/N''$, which is in $\mathcal{C}$, so it's in $\mathcal{C}$ as well.
$\mathcal{C}$ is closed under taking subobjects and quotient objects. $N'$ is in $\mathcal{C}$, $N' \cap N''$ is a subobject, and $g(N' \cap N'')$ is a quotient of that.
A sum of two subobjects of another object is a quotient of their direct sum. Here, $P'$ and $g(N'' \cap N')$ are in $\mathcal{C}$, so their direct sum is as well, and so this internal sum is as well.
The point of this was to compose $\overline{f}$ and $\overline{g}$ in the quotient category, but to do this, we had to lift them to maps $f$ and $g$ in the original category. Gabriel's saying that we get the same composition in the quotient category even if we had picked different lifts of $\overline{f}$ and $\overline{g}$. Let's say we picked $\tilde{f}:\widetilde{M'} \to N/\widetilde{N'}$ and $\tilde{g}:\widetilde{N''} \to P/\widetilde{P'}$ instead. Since these have to become $\overline{f}$ and $\overline{g}$ in the direct limit, then, for example, $\widetilde{f}$ and $f$ must be the same as maps from some subobject of $M' \cap \widetilde{M'}$ to some quotient of $N/(N' + \widetilde{N'})$, and likewise for $g$ and $\widetilde{g}$.

So we reduce to the case where $\widetilde{f}$ is defined 'further along in the direct system' than $f$. That is, $\widetilde{f}$ is a map $\widetilde{M'} \to N/\widetilde{N'}$ where $\widetilde{M'}$ and $\widetilde{N'}$ are subobjects of $M'$ and $N'$ respectively, with the obvious restrictions on their quotients, and we have to show that $gf$ and $g\widetilde{f}$ become the same map in the quotient category (and then do the same but vary $g$). I'm not going to do this whole proof for you, but if you're confused about things like this, it's probably worth doing yourself. Good luck!

Answer by Paul VanKoughnett for Definition of subobject classifier in presheaves

2013-05-06T14:28:20Z

What's going on is that, since $E$ is a contravariant functor to $\mathbf{Sets}$, by definition $f:D \to C$ induces a map of sets $f^*:E(C) \to E(D)$. Then $u_C(e)$ consists of the arrows $f$ such that $f^*(e)$ actually lands in the subobject $U(D)$ of $E(D)$. You can then check that this is a sieve for each $e$, and so on.

It might be helpful to let $E$ be a representable functor, say $E = \mathbf{C}(-,X)$. Then $e$ is a map $C \to X$, $U(D)$ is a set of maps $D \to X$, and $u_C(e)$ is the set of maps $D \to C$ such that the composition $D \to C \stackrel{e}{\to} X$ is in the set $U(D)$.

Answer by Paul VanKoughnett for injectivity is a local property over noetherian rings

2013-03-04T20:56:02Z

Let $A \to B$ be an injection of $R$-modules. We want to show that $\mathrm{Hom}(B,M) \to \mathrm{Hom}(A,M)$ is surjective. It suffices to check this locally (since localization is an exact functor and the cokernel of the map is zero iff it's zero locally). So let $p$ be a prime, and consider the localized map $\mathrm{Hom}_R(B,M) \otimes R_p \to \mathrm{Hom}_R(A,M) \otimes R_p$. Now, for $B$ and $A$ finitely presented over $R$, this is $\mathrm{Hom}_{R_p}(B_p,M_p) \to \mathrm{Hom}_{R_p}(A_p,M_p)$. Again, localization is exact so $A_p \to B_p$ is injective, and $M_p$ is injective by hypothesis, so this map is surjective.

What's left is to show that we only need to check injectivity against maps of finitely presented $R$-modules; since $R$ is noetherian, 'finitely presented' is equivalent to 'finitely generated.' In fact, there's a thing called Baer's injectivity criterion that says you only have to check injectivity against inclusions $I \to R$ for $I$ an ideal! So we're done.

Algebraic closure as a fibrant replacement?

2012-11-09T01:51:25Z

Emil Artin's construction of the algebraic closure of a field $K$ is as follows. Let $K_0 = K$, and inductively let $\{x_f\}$ be a set of indeterminates indexed by the irreducible $f$ in one variable over $K_i$. Let $\mathfrak{m}_i$ be a maximal ideal of $K_i[x_f]$ containing the ideal generated by all $f(x_f)$ for $f$ irreducible (one can prove this ideal is proper). Define $K_{i+1} = K_i[x_f]/\mathfrak{m}_i$ . Clearly, each $K_{i+1}$ is algebraic over $K_i$ (and thus over $K_0$), and every polynomial with coefficients in $K_i$ splits in $K_{i+1}$. In particular, the union $\bar{K}$ of the $K_i$ is algebraic over $K$ and algebraically closed.

To me, a fan of homotopy theory, this seems an awful lot like a small object argument. We can express a solution of the polynomial $f$ over the field $K$ as a solution to a lifting problem: we want the map $0:K[x] \to K$ sending $x$ to $0$ to factor through the map $f:K[x] \to K[x]$ sending $x$ to $f(x)$, so we want a lift to a square which has $K \to 0$ on the right, $f:K[x] \to K[x]$ on the left, and $0$ on the top. (I'd draw this here, but I don't know how.) To construct $K_1$, we are doing something like taking the pushout of the diagram $$\bigotimes_f K[x_f] \stackrel{\otimes f}{\leftarrow} \bigotimes_f K[x_f] \stackrel{0}{\rightarrow} K$$ whose left arrow is a coproduct of the left sides of the aforementioned lifting problems. The algebraic closure itself is then the colimit of these pushouts, which is doing something like giving a fibrant replacement for $K$.

As you've probably noticed, this naïve outline has quite a few problems. The pushouts constructed above are not actually the fields $K_i$, but rather become them after a quotient by a maximal ideal. Also, the lifting problems we are solving differ between the $K_i$: the left sides of the squares that appear are of the form $f:K_i[x] \to K_i[x]$, where $f$ is an irreducible polynomial over $K_i$. On the other hand, the small object argument is done via a single set of generating cofibrations which appear on the left sides of squares at every stage.

I'd like to think that, after passing from the category of $K$-algebras to something presumably more complicated, one can construct a model structure in which the algebraic closure of a field appears as its fibrant replacement. Does anyone know if this is true?

Comment by Paul VanKoughnett

2013-05-19T23:08:38Z

$g$ is a map $N'' \to P/P'$; taking a further quotient induces $N'' \to P/P''$; and then by definition, $N' \cap N''$ is a subobject of the kernel of this map, so it passes to $N''/(N'\cap N'')\to P/P''$. There's a good reason why arguments about abelian categories look like arguments about modules: the Freyd-Mitchell embedding theorem, which says that any small abelian category embeds into some category of modules over a ring. So any statement you can phrase in terms of an arbitrary small set of objects in an abelian category, and prove when those objects are modules, is true in general!

Comment by Paul VanKoughnett

2013-05-16T17:27:16Z

This should answer your question: mathoverflow.net/questions/40499/…

Comment by Paul VanKoughnett

2013-05-03T15:24:32Z

Tensor products commute with direct sums, and if the sequence is split, $B \cong A \oplus C$.

Comment by Paul VanKoughnett

2013-04-28T23:35:04Z

This paper (math.uiuc.edu/K-theory/0462/combination2.pdf) by Benjamin Blander seems to imply that the answer to your first question is 'yes,' and you again get a Quillen equivalent model category, with sheafifying and forgetting being a Quillen equivalence.

Comment by Paul VanKoughnett

2013-03-22T02:34:23Z

That isn't really an answer to your questions, so for now let me just make an insinuation: what could you mean by a space 'naturally' having a multiplication on its homotopy groups, other than this being induced by a multiplication on the space itself?

Comment by Paul VanKoughnett

2013-03-22T02:32:28Z

In fact the $K$-groups of a ring are naturally modelled as the homotopy groups of a spectrum. (If you want to stick with spaces, you can in turn model this as an infinite loop space, i.e. a space which can be expressed as the form $\Omega^nX_n$ for some space $X_n$, for every $n$.) When you're starting with a ring, the spectrum you get is always a ring spectrum, which is the spectrum version of an $H$-space. I don't actually know why this is true, but I'd guess it pops out of one of the various constructions of this spectrum.

Comment by Paul VanKoughnett

2013-03-04T03:32:34Z

In order for this to be well-defined, you'd need to say precisely what it means for two cycles to '"deform" into each other continuously.' If both cycles are subcomplexes of $X$, you could ask for them to be simplicially homotopic, but then you'd want your simplicial set to be Kan, i. e. probably the singular complex of the space $X$ rather than a triangulation. Then you'd need to make sure that your definition extends well to the abelian group of cycles.