0
$\begingroup$

On page 4 of Categories for the Working Mathematician Mac Lane writes that "the definition of a monoid is more general because the cartesian product may be replaced by another operation". He then explains that a monoid in the category of abelian groups (with the tensor product operation) is a ring.

Can we not extend the definition of groups in a similar fashion, ridding ourselves of the cartesian product? For example we could think of a skew field as a group in the category abelian groups (with the tensor product operation).

Since it seems that we can what does Mac Lane mean by his quote?

Sorry, you will likely need a copy of the text to answer this question.

$\endgroup$
  • 1
    $\begingroup$ This doesn't work out -- you wind up forcing $0$ to be invertible, so the zero ring is the only one you can construct in this way. $\endgroup$ – Hurkyl Oct 1 '13 at 3:05
5
$\begingroup$

Well, it's not quite so straightforward to define groups $G$ under a monoidal product (other than cartesian product) as you might think.

Consider the axioms for inverses: if $i$ is the operation taking an element $x$ to its inverse ($x^{-1}$), and $m$ is the multiplication, then the axiom for inverses is

$$m(x, i(x)) = e = m(i(x), x).$$

This axiom can be interpreted in a category with cartesian products; it would say that the composite

$$G \stackrel{\delta}{\to} G \times G \stackrel{i \times 1_G}{\to} G \times G \stackrel{m}{\to} G$$

equals the composite

$$G \stackrel{!}{\to} 1 \stackrel{e}{\to} G$$

(where $\delta$ denotes a diagonal map, $1_G$ an identity morphism, $1$ a terminal object, and $!$ the unique morphism to the terminal object).

For another monoidal product, the monoidal unit $I$ would replace $1$, but: there isn't a morphism $G \to G \otimes G$ to play the role of a diagonal map, and there isn't a morphism $G \to I$ to play the role of a projection. Indeed, a cartesian product can be characterized as a monoidal product that carries such a (natural) diagonal map and such a (natural) projection map, satisfying suitable properties. So the point is that although you can define a monoid object in a general monoidal category, you really can't define the notion of a group object in a general monoidal category.

Also, it is not true that a skew field could be defined this way as a group object. The notion of skew field cannot be interpreted even in a cartesian monoidal category, because the multiplicative inversion operation is not globally defined. Close, but no cigar, and in fact from a structural categorical point of view, fields do not behave as algebraic categories, in that the category of fields (skew or not) doesn't have most limits or colimits, there are no "free fields", and so on.

$\endgroup$
  • $\begingroup$ Right, now I see that inversion is not globally defined and that there is no homomorphism to play the role of the diagonal map (forgot this had to be a homomorphism). I think there is a projection map though. If the integers are the monoidal unit then we can just map everything to 1 I believe? $\endgroup$ – Jonathan Aronson Oct 1 '13 at 3:21
  • $\begingroup$ For the category of abelian groups $A$, there is no natural transformation from the identity functor to the constant functor with value $\mathbb{Z}$. For example, there is no map $\mathbb{Z}/(n) \to \mathbb{Z}$, no map $\mathbb{Q} \to \mathbb{Z}$, etc. But even in the case of the category of vector spaces over a field $k$, there is no natural transformation from the identity functor to the constant functor at the unit (a 1-dimensional space). $\endgroup$ – user43208 Oct 1 '13 at 3:59
  • $\begingroup$ (Make that no natural transformation except for the zero transformation.) $\endgroup$ – user43208 Oct 1 '13 at 10:46
  • $\begingroup$ Alright, this all very new to me. I will think about what you are saying. $\endgroup$ – Jonathan Aronson Oct 10 '13 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.