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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeSep 20th 2010
   • (edited Sep 20th 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexTake a look over at [[module]]. Let $Ab$ be the category of abelian groups. Then a commutative ring $R$ is a commutative monoid in this category. A module over $R$ seems that it should not be an abelian group object in the category of commutative rings over $R$, but should be a commutative monoid in the category of abelian groups over $R$ (more generally, a commutative monoid in the symmetric monoidal category over a commutative monoid $M$) (even more generally, a monoid in the symmetric monoidal category over a monoid $M$, but to define the multiplication map, we first compose with the symmetry on one of the $R\otimes N$). To see this in generality, let $N$ be an $R$-module in the classical sense. We define the underlying object of $R\oplus N$ to be the direct sum of abelian groups. We define the product operation on this object as follows: $\ast:(R\oplus N)\otimes (R\oplus N)\cong (R\otimes R)\oplus (R\otimes N)\oplus (R\otimes N) \oplus (N\otimes N)\overset{p}{\to} R\oplus N$ Where $p$ is the map induced on the coproduct by the following three maps: * $R\otimes R\to R\overset{\iota_0}{\to} R\oplus N$, the ordinary product map on $R$ composed with the canonical injection $\iota_0$. * $R\otimes N\to N\overset{\iota_1}{\to} R\oplus N$, the $R$-action on $N$ composed with the canonical injection $\iota_1$. * $N\otimes N \overset{0}{\to} N\overset{\iota_1}{\to} R\oplus N$. We let the projection back to $R$ be the induced map from the coproduct defined by the identity on the first factor, and the zero map on the second. This gives us on elements $(r,n)\ast (r',n')=(rr',rn'+r'n+nn')$ naively, but by definition, $nn'=0$, so we have that $(r,n)\ast (r',n')=(rr',rn'+r'n)$. Further, a section of the projection map is a map $f$ mapping to the identity on the first factor and to $\eta_f$ on the second, such that $$f(rr')=f(r)f(r')=(r,\eta_f(r))(r',\eta_f(r'))=(rr',r'\eta_f(r)+r\eta_f(r'))=r'(r,\eta_f(r))+r(r',\eta_f(r'))=r'f(r)+rf(r'),$$ a derivation. Further, it is equipped with a canonical section representing the zero derivation.

   Take a look over at module.

   Let $Ab$ be the category of abelian groups. Then a commutative ring $R$ is a commutative monoid in this category. A module over $R$ seems that it should not be an abelian group object in the category of commutative rings over $R$, but should be a commutative monoid in the category of abelian groups over $R$ (more generally, a commutative monoid in the symmetric monoidal category over a commutative monoid $M$) (even more generally, a monoid in the symmetric monoidal category over a monoid $M$, but to define the multiplication map, we first compose with the symmetry on one of the $R\otimes N$).

   To see this in generality, let $N$ be an $R$-module in the classical sense. We define the underlying object of $R\oplus N$ to be the direct sum of abelian groups. We define the product operation on this object as follows:

   $\ast:(R\oplus N)\otimes (R\oplus N)\cong (R\otimes R)\oplus (R\otimes N)\oplus (R\otimes N) \oplus (N\otimes N)\overset{p}{\to} R\oplus N$

   Where $p$ is the map induced on the coproduct by the following three maps:

   • $R\otimes R\to R\overset{\iota_0}{\to} R\oplus N$, the ordinary product map on $R$ composed with the canonical injection $\iota_0$.
   • $R\otimes N\to N\overset{\iota_1}{\to} R\oplus N$, the $R$-action on $N$ composed with the canonical injection $\iota_1$.
   • $N\otimes N \overset{0}{\to} N\overset{\iota_1}{\to} R\oplus N$.

   We let the projection back to $R$ be the induced map from the coproduct defined by the identity on the first factor, and the zero map on the second.

   This gives us on elements $(r,n)\ast (r',n')=(rr',rn'+r'n+nn')$ naively, but by definition, $nn'=0$, so we have that $(r,n)\ast (r',n')=(rr',rn'+r'n)$. Further, a section of the projection map is a map $f$ mapping to the identity on the first factor and to $\eta_f$ on the second, such that

   $$f(rr')=f(r)f(r')=(r,\eta_f(r))(r',\eta_f(r'))=(rr',r'\eta_f(r)+r\eta_f(r'))=r'(r,\eta_f(r))+r(r',\eta_f(r'))=r'f(r)+rf(r'),$$

   a derivation.

   Further, it is equipped with a canonical section representing the zero derivation.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeSep 20th 2010
   • (edited Sep 20th 2010)
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   Author: Urs
   Format: MarkdownItexHi Harry, I need to rush off now to get to a talk. Can't check your computations right now. Let me just remark that the statement as in the nLab entry is classical, due to Quillen in the 60s or soemthing. Don't we even give the link to the original source? But I'll look at what you wrote later.

   Hi Harry,

   I need to rush off now to get to a talk. Can’t check your computations right now. Let me just remark that the statement as in the nLab entry is classical, due to Quillen in the 60s or soemthing. Don’t we even give the link to the original source?

   But I’ll look at what you wrote later.

3. • CommentRowNumber3.
   • CommentAuthorTim_Porter
   • CommentTimeSep 20th 2010
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   Author: Tim_Porter
   Format: MarkdownItex@Harry and Urs I think it is attributable to Beck in his thesis which is slightly earlier than Quillen. I have not checked Harry's argument but would note that in the category of groups and internal category is an internal groupoid, so I suspect that in the case in question here, an analogous result may hold... but as I said I have not checked it.

   @Harry and Urs I think it is attributable to Beck in his thesis which is slightly earlier than Quillen. I have not checked Harry’s argument but would note that in the category of groups and internal category is an internal groupoid, so I suspect that in the case in question here, an analogous result may hold… but as I said I have not checked it.

4. • CommentRowNumber4.
   • CommentAuthorHarry Gindi
   • CommentTimeSep 20th 2010
   • (edited Sep 20th 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexYeah, what Quillen said is kind of right, but it doesn't generalize well (for instance, "derivations" of a commutative monoid into a pointed-set module (by which I mean $Hom_{A/Comm/M}(X,M\vee S)$ for $M,X$ a commutative monoids under $A$, and $S$ a pointed-set module over $M$) only forms a commutative monoid under multiplication). Then we can at least say that the correct objects are the commutative monoids in $(C,\otimes)$ that are also commutative monoids in $(C/Y,\times_Y)$. Edit: I think we must require that the original monoidal category has more structure for this notion to be nontrivial, in that we need at the very least kernels, pullbacks, and biproducts to recover the action definition. The definition over at [[module]] isn't so bad, but there's some stuff that seems pretty untrue over at [[Kähler differential]]

   Yeah, what Quillen said is kind of right, but it doesn’t generalize well (for instance, “derivations” of a commutative monoid into a pointed-set module (by which I mean $Hom_{A/Comm/M}(X,M\vee S)$ for $M,X$ a commutative monoids under $A$, and $S$ a pointed-set module over $M$) only forms a commutative monoid under multiplication). Then we can at least say that the correct objects are the commutative monoids in $(C,\otimes)$ that are also commutative monoids in $(C/Y,\times_Y)$.

   Edit: I think we must require that the original monoidal category has more structure for this notion to be nontrivial, in that we need at the very least kernels, pullbacks, and biproducts to recover the action definition.

   The definition over at module isn’t so bad, but there’s some stuff that seems pretty untrue over at Kähler differential

5. • CommentRowNumber5.
   • CommentAuthorTim_Porter
   • CommentTimeSep 20th 2010
   • PermaLink
   Author: Tim_Porter
   Format: MarkdownItexVarious people have referred to semi-modules. A module is traditionally an Abelian group with an action etc., and for working with monoid (co)homology you may need something more 'semi'. I do not like the use of module to mean things that are not linked to Abelian groups as I think the undercurrents of that theory can end up pushing the enquiries in too Abelian direction and I like non-Abelian group objects as well! I have been looking at Baues et al on cohomology of categories and there are some good ideas there that may help.

   Various people have referred to semi-modules. A module is traditionally an Abelian group with an action etc., and for working with monoid (co)homology you may need something more ’semi’. I do not like the use of module to mean things that are not linked to Abelian groups as I think the undercurrents of that theory can end up pushing the enquiries in too Abelian direction and I like non-Abelian group objects as well!

   I have been looking at Baues et al on cohomology of categories and there are some good ideas there that may help.

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeSep 20th 2010
   • (edited Sep 20th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexOkay, Harry, here I am in a ten minutes break between two talks. I enjoyed this one here: > what Quillen said is kind of right Kind, of eh? > but it doesn't generalize well Yeah, it just gives the right notion of modules over all notions of $\infty$-rings in huge generality and all of derived geometry is based on this -- but apart from that, it does not seem to generalize well. _What have the Romans ever done for us?_ > The definition over at module isn't so bad, Phew, thanks. > but there's some stuff that seems pretty untrue over at Kähler differential Now that makes me worried.

   Okay, Harry, here I am in a ten minutes break between two talks.

   I enjoyed this one here:

   what Quillen said is kind of right

   Kind, of eh?

   but it doesn’t generalize well

   Yeah, it just gives the right notion of modules over all notions of $\infty$-rings in huge generality and all of derived geometry is based on this – but apart from that, it does not seem to generalize well. What have the Romans ever done for us?

   The definition over at module isn’t so bad,

   Phew, thanks.

   but there’s some stuff that seems pretty untrue over at Kähler differential

   Now that makes me worried.

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeSep 20th 2010
   • (edited Sep 20th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexSeriously, Harry, let's try to do this in a more constructive fashion. I appreciate it a lot to see you countercheck nLab entries. But maybe there is a different way of talking about this. Possibly there is a mistake somewhere in the page on Kähler differentials. If you spot it, point it out, and maybe even correct it, I'll be very grateful. But if you just say the entry looks _pretty untrue_ right after having falsely claimed some other entry to be untrue and having trouble to acknowledge that Quillen might have had a right idea, I somehow tend to lose a bit of the joy of it all. After all we also want to have a bit of a good time here. Right?

   Seriously, Harry, let’s try to do this in a more constructive fashion.

   I appreciate it a lot to see you countercheck nLab entries. But maybe there is a different way of talking about this. Possibly there is a mistake somewhere in the page on Kähler differentials. If you spot it, point it out, and maybe even correct it, I’ll be very grateful.

   But if you just say the entry looks pretty untrue right after having falsely claimed some other entry to be untrue and having trouble to acknowledge that Quillen might have had a right idea, I somehow tend to lose a bit of the joy of it all. After all we also want to have a bit of a good time here. Right?

8. • CommentRowNumber8.
   • CommentAuthorHarry Gindi
   • CommentTimeSep 20th 2010
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItex@Urs: I was reading [[Kähler differential]] and [[module]] at the same time and mixed up which one looked strange. The statement at [[Kähler differential]] that doesn't seem to work out right is near the bottom when you switch to arbitrary (commutative) monoids in an arbitrary (presumably symmetric monoidal) category. If you do so, you lose the standard definition of a module. You have to assume that there is an abelian category (or at least something like it, like, for example, assumptions on the base category of a "homotopical algebraic context" in the sense of Toen and Vezzosi (HAG II)). I agree that perhaps I was rude, and if you could find it in your heart to forgive me, I'd appreciate it.

   @Urs: I was reading Kähler differential and module at the same time and mixed up which one looked strange. The statement at Kähler differential that doesn’t seem to work out right is near the bottom when you switch to arbitrary (commutative) monoids in an arbitrary (presumably symmetric monoidal) category. If you do so, you lose the standard definition of a module. You have to assume that there is an abelian category (or at least something like it, like, for example, assumptions on the base category of a “homotopical algebraic context” in the sense of Toen and Vezzosi (HAG II)).

   I agree that perhaps I was rude, and if you could find it in your heart to forgive me, I’d appreciate it.