December 9, 2008

A Quick Algebra Quiz

Posted by John Baez Here’s a quick algebra quiz. It’s really a test of your reflexes when it comes to algebra and category theory. It should take less than a minute if you have the right mental training. If you don’t, you may be doomed.

So, take a deep breath and give it a try.

1. Why is a commutative algebra over a field the same thing as a commutative ring under that field?
2. What happens if we leave out the word ‘commutative’ both places here?
Posted at December 9, 2008 2:20 AM UTC

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Re: A Quick Algebra Quiz

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.

Posted by: Noah Snyder on December 9, 2008 6:50 AM | Permalink | Reply to this

Re: A Quick Algebra Quiz

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”.

Posted by: Allen Knutson on December 9, 2008 6:57 AM | Permalink | Reply to this

Re: A Quick Algebra Quiz

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 $F$.

Posted by: Urs Schreiber on December 9, 2008 3:27 PM | Permalink | Reply to this

Re: A Quick Algebra Quiz

See also the diagram at the Wikipedia entry. Scroll down to “Algebras and rings”.

Posted by: Urs Schreiber on December 9, 2008 4:29 PM | Permalink | Reply to this

Re: A Quick Algebra Quiz

Posted by: Toby Bartels on December 9, 2008 8:05 PM | Permalink | Reply to this

Re: A Quick Algebra Quiz

Thanks. I tried …#Algebras-and-rings, which didn’t work… :-)

Posted by: Urs Schreiber on December 9, 2008 8:26 PM | Permalink | Reply to this

Re: A Quick Algebra Quiz

I think I can see it!

Posted by: Anonymous on December 9, 2008 3:58 PM | Permalink | Reply to this

Re: A Quick Algebra Quiz

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!

Posted by: John Baez on December 9, 2008 9:31 PM | Permalink | Reply to this

Re: A Quick Algebra Quiz

Thanks for the explanation. Now I can try to tackle question 2: the quaternions are a ring under the complex numbers, but not an algebra over them (multiplication isn’t C-linear in the second factor).

Posted by: Noah Snyder on December 9, 2008 10:20 PM | Permalink | Reply to this

Re: A Quick Algebra Quiz

Right — that illustrates the distinction!

The distinction dissolves when we stay in the world of commutative rings.

Posted by: John Baez on December 10, 2008 2:15 AM | Permalink | Reply to this

Re: A Quick Algebra Quiz

For a noncommutative algebra the story still goes through when we demand that $F \hookrightarrow R$ factors through the center of $R$ $\array{ F &&\hookrightarrow&& R \\ & \searrow && \nearrow \\ && Z(R) } \,.$

Posted by: Urs Schreiber on December 10, 2008 7:37 AM | Permalink | Reply to this

Re: A Quick Algebra Quiz

concerning 3), I guess you are referring to the idea that a map f:E->B gives birth dually to a map f*:Fun(B)->Fun(E).

Posted by: yael on December 10, 2008 6:48 AM | Permalink | Reply to this

Re: A Quick Algebra Quiz

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?)

Posted by: PhilG on December 10, 2008 8:30 AM | Permalink | Reply to this

Re: A Quick Algebra Quiz

PhilG wrote:

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?

Yes, that’s exactly right: if $k$ is a field and $R$ is commutative ring any homomorphism $k \to R$ is enough to make $R$ into an algebra over $R$, but if $R$ is noncommutative this only works if the image of $k$ lies in the center of $R$, since we need

$(a x)(b y) = (a b)(x y)$

for $x,y$ in $R$ and $a,b$ in the image of $k$.

This is what Noah was talking about: he gave the classic example where $k$ is the complex numbers and $R$ is the quaternions.

Have you formed a conspiracy to undermine Microsoft now?

No, but Microsoft has.

Posted by: John Baez on December 10, 2008 4:49 PM | Permalink | Reply to this

Re: A Quick Algebra Quiz

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.)

Posted by: John Baez on December 10, 2008 5:05 PM | Permalink | Reply to this

Re: A Quick Algebra Quiz

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
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)

Posted by: zoran skoda on December 11, 2008 1:26 AM | Permalink | Reply to this

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