The n-Category Café

I guess this means I fail the test, but “ring under a field” is not a term I’m familiar with. Googling doesn’t bring up anything relevant…

The only obvious meaning I could think of was a ring that is a subring of the field in question, for example Z is a subring of Q. But Z is definitely not an algebra over Q, so that can’t be what you mean.

I’m with Noah – “under” seems a very odd word here. I think I understand what the question is about if it’s replaced by “containing”.

I suppose it is well known that a commutative unital ring $R$ is an algebra “over” a field $F$ if there is a ring homomorphism $$F \hookrightarrow R \,.$$ Moreover, a morphism of two algebras over a field a is a commuting triangle of ring morphisms $$\array{ && F \\ & \swarrow && \searrow \\ R &&\to&& R' }$$

In this sense algebras over $F$ are objects in the under-category of rings under $F$.

I think I can see it!

Okay, so we’re getting some comments from people who are unfamiliar with the general concept of an object ‘under’ another object. Urs explained the general concept of an under category — a category of all objects under a given one. But let me go through it much more gently:

In any category, an object over an object $B$ is an object $E$ equipped with a morphism to $B$. The terminology makes the most sense if you draw the morphism vertically like this:

$$\array{ E \\ \downarrow p \\ B }$$

For example, $E$ could be a bundle ‘over’ a space $B$: then we say $p$ ‘projects down’ from $E$ to the ‘base’ $B$. Visualize $E$ as a puffy white cloud up in the sky and $B$ as its dark shadow: the map $p$ represents the rays of sunshine, sending each point up in the cloud to its shadow on the ground.

Dually, an object under an object $E$ is an object $B$ equipped with a morphism from $E$. Again the terminology makes the most sense if you draw the morphism vertically like this:

$$\array{ E \\ \downarrow p \\ B }$$

Okay, now we know what ‘under’ means in category theory. Urs went further and defined the category of all objects under a given object in some category. This is necessary to make my puzzle 100% precise, but I’ll leave that to the highly motivated reader.

Instead, I’ll just ask you: how does a ‘commutative algebra over a field $E$’ — in the usual algebra textbook sense of ‘over’ — compare with ‘commutative ring under the field $E$’ — in the above sense of ‘under’?

If you think about it, you should see they’re the same!

But how did ‘over’ turn into ‘under’? One could say it’s just a silly coincidence that we use language in two conflicting ways here. But in fact there’s a deeper explanation, which is well-known in algebraic geometry! So, I’ll pose another puzzle:

3. How is a commutative ring under a fixed commutative ring analogous to a space over a given base space?

And I’m still waiting for an answer to puzzle 2!

re your part 2, the definition of an algebra over a field according to wikipedia requires a compatibility condition

(ax)(by) = (ab)(xy)

for a,b in the field. I’m a bit over my head here but I’ll guess that your definition of a ring under the field won’t impose this condition if the ring is not commutative. Am I close?

(BTW n-category cafe is still not working in Vista+IE7. Have you formed a conspiracy to undermine Microsoft now?)

Yael correctly answered puzzle 3 — but I feel the need to sermonize a bit, since this puzzle explains why my first puzzle was worth bothering with. The experts already know this stuff, but still…

Algebraic geometry studies commutative rings, but thinks of the elements of these spaces as functions on cleverly devised spaces called ‘affine schemes’. And this change of viewpoint is contravariant: a map between commutative rings corresponds to a map between affine schemes going the other way! Algebra is geometry, only backwards!

So, a commutative ring under a fixed commutative ring is the same as an affine scheme over a fixed affine scheme — the ‘base’.

So, when people talk about commutative algebras ‘over’ a field, and call this field the ‘ground field’ or ‘base’, they’re talking about algebra — but they’re using metaphors appropriate for algebraic geometry.

I think it was Grothendieck who really pushed the idea of ‘relative’ algebraic geometry: always studying schemes over a given ‘base’, which can be something much more general than a field: any scheme, in fact.

(Above I focused attention on affine schemes to minimize the prerequisites for understanding my point, but of course the whole story works more generally.)

One can push the relative algebraic geometry
to the noncommutative case as well.

Gabriel-Rosenberg theorem says that
a commutative scheme can be reconstructed up to isomorphism from the abelian category
of quasicoherent sheaves over the scheme.
A.L. Rosenberg suggest considering
first a category Spc whose objects are abelian categories and morphisms are
(equivalence classes of) pairs of adjoint functors (direct and inverse
image functor). It is also useful to
think of adjoint pairs of
functors as bimodules in the case
the categories are categories of modules.

Now take a fixed object V, that is “ground” category and consider over-category Spc/V. An object in
Spc/V is called a noncommutative scheme
if it has a cover (conservative family
of flat functors) by localizations
satisfying some natural exactness properties, including that the localized
categories (members of the cover) are
affine over V. Flat localizations should be considered analogues of open sets.
The details can be found in

A. L. Rosenberg, Noncommutative schemes, Compositio Math. 112 (1998), 93–125

or in chapter 11 of my localization survey

Z. Skoda, Noncommutative localization in noncommutative geometry,
in London Math. Soc. LNS 330, pp.219-309 (preprint version arXiv:math.QA/0403276)