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Stream: theory: mathematics

Topic: Quillen-Suslin with infinitely many variables

Jean-Baptiste Vienney (Nov 06 2023 at 09:07):

The Quillen-Suslin theorem asserts that every finitely-generated projective module over a polynomial ring $R=k[x_1,…,x_n]$ for $k$ a field (or a PID), is free.

Is the theorem false if we replace $R$ by $k[x_i, i \in \mathbb{N}]$? I’m sure it should be the case but I don’t know a lot about these matters…

John Baez (Nov 06 2023 at 11:12):

I have no intuition for that. So you're looking for a summand of a finitely generated free module of $k[x_1, x_2, \dots]$ that's not itself free.

Jean-Baptiste Vienney (Nov 06 2023 at 12:55):

Let me think two minutes. I'm looking for a finitely-generated projective module $M$ over $k[x_1,x_2,...]$ which is not free ie.:

• $M$ is a $k[x_1,x_2,...]$-module
• $M$ is not free
• Finitely-generated: $M= k[x1,x_2,...].m_1 + ... + k[x1,x_2,...].m_p$ for some elements $m_1,...,m_p \in M$
• Projective: $M \oplus Q$ is free for some $k[x_1,x_2,...]$-module $Q$

Jean-Baptiste Vienney (Nov 06 2023 at 13:01):

Yes, I'm looking for one which is finitely-generated

Jean-Baptiste Vienney (Nov 06 2023 at 13:03):

I think that every finitely-generated submodule of $k[x_1,x_2,...]$ is a finitely-generated submodule of some $k[x_I]$ for a finite set of variables $I$. It might help if it is true.

Jean-Baptiste Vienney (Nov 06 2023 at 13:05):

(Maybe the theorem is true as an easy consequence of Quillen-Suslin in some way like this.)

Jean-Baptiste Vienney (Nov 06 2023 at 13:09):

Jean-Baptiste Vienney said:

Yes, I'm looking for one which is finitely-generated

Maybe it is equivalent to look for any summand and this is why you didn't mention finitely-generated?

John Baez (Nov 06 2023 at 13:10):

No, I just left out that condition by accident. I stuck it back in. For all I know there might be non-finitely-generated modules of $k[x_1, \dots, x_n]$ that are projective but not free! I imagine that the assumption was put in for a good reason.

Jean-Baptiste Vienney (Nov 06 2023 at 13:10):

Ahah ok. This is why I think it's probably false indeed.

Jean-Baptiste Vienney (Nov 06 2023 at 13:13):

Hmm, I was thinking about the assumption "polynomial ring with finitely many variables": the finite number of variables is probably here for a reason

Jean-Baptiste Vienney (Nov 06 2023 at 13:18):

John Baez said:

So you're looking for a summand of a finitely generated free module of $k[x_1, x_2, \dots]$ that's not itself free.

I think it's still not that, no :sweat_smile: ?

Jean-Baptiste Vienney (Nov 06 2023 at 13:19):

I'm confused, sorry

Jean-Baptiste Vienney (Nov 06 2023 at 13:21):

Yes, the summand must be finitely generated but not necessarily the free module

Tobias Fritz (Nov 06 2023 at 13:41):

Jean-Baptiste Vienney said:

(Maybe the theorem is true as an easy consequence of Quillen-Suslin in some way like this.)

Yes, I think this is what happens. To see this, for a fg projective module $P$, consider the composite

$$R^n \cong P \oplus Q \to P \to P \oplus Q \cong R^n ,$$

where all maps are the obvious ones. Then this map can be represented by a matrix $e \in M_{n \times n}(R)$ with $e^2 = e$. The module is free if and only if there is an isomorphism $P \cong R^m$ for some $m$, and using this we can also represent the projection $P \oplus Q \to P$ as a matrix $p \in M_{n \times m}(R)$, and the inclusion $P \to P \oplus Q$ as a matrix $i \in M_{m \times n}(R)$, such that $e = ip$ and $pi = 1$. Like this, I believe that we get the following:

Theorem: A ring $R$ has the property that every fg projective module is free if and only if every idempotent matrix with entries from $R$ splits.

Now the punchline: over $R = k[x_1,x_2,\dots]$, every matrix can only contain finitely many variables! Like this, the problem is reduced to the Quillen-Suslin theorem, and it should indeed be true that every fg projective module is free. I'm saying "should" because I haven't carefully worked out the proof of the above theorem. But assuming that it's true, we can think of this geometrically as saying that every algebraic vector bundle in infinite dimensions can vary in only finitely many of them, and it's trivially trivial in the other directions.

Jean-Baptiste Vienney (Nov 06 2023 at 14:27):

Thank you! I will try to understand your theorem better a bit later. But for now, I think if it works, it works also for $R=k[x_i,~i \in I]$ for any set $I$. In fact I'm interested by these rings also. The reason is that I wanted to understand if we can replace $R$ by the symmetric algebra of any vector space over field (which is isomorphic to a $k[x_i,~i \in I]$).

Tobias Fritz (Nov 06 2023 at 14:31):

Yes, it should work for any $I$ regardless of cardinality.

Jens Hemelaer (Nov 06 2023 at 17:49):

In general, if $\mathbf{P}(R)$ is the monoid of finitely generated projective $R$-modules, up to isomorphism,
then $\mathbf{P}$ preserves filtered colimits, see here, 2.1.6.
In our situation, $k[x_i, i \in I]$ is the filtered colimit of the $k[x_i, i \in J]$ for $J$ going over the finite subsets of $I$.
And $\mathbf{P}(k[x_i, i \in J])$ is always the monoid $(\mathbb{N},+)$, so it is still this monoid of natural numbers in the limit.