# Intuitive explanation of Serre's Conjecture

I came to know about Serre's conjecture in a lecture. The conjecture states the following.

If $$K$$ is principal ideal domain then all finitely generated projective modules over the polynomial ring $$K[x_1,\dots,x_n]$$ are free.

I am wondering why is this conjecture important and whether there is an intuitive way to explain this conjecture for someone who has only master's level knowledge in abstract algebra.

• Can you be a bit more clear about what you don't understand? A masters level knowledge of abstract algebra varies wildly from both place to place and based on your particular focus. For example, do you understand all the words in this definition? If not, which words do you not understand? – jgon Nov 21 '18 at 0:39
• Also, could you address what about the Wikipedia article doesn't address your question? en.wikipedia.org/wiki/Quillen–Suslin_theorem And finally, it would be more accurate to call this statement the Quillen-Suslin theorem, since it was proven. – jgon Nov 21 '18 at 0:44
• Warning, that link didn't parse correctly in the MSE system, try this link en.wikipedia.org/wiki/Quillen-Suslin_theorem – jgon Nov 21 '18 at 0:46
• @jgon: Thanks. I know the definitions of a module, PID, polynomial ring only. I was just wondering if it was possible to explain that result to a common (math) man using geometry etc. I don't get the intuitive explanation from the wikipedia page either. – Ashok Nov 21 '18 at 0:54
• For the "why important" question, here's an application of the theorem: mathoverflow.net/a/88446 – darij grinberg Nov 21 '18 at 3:16

Well, I'll answer anyway, since Wikipedia isn't MSE, but to be clear, while the math doesn't come from Wikipedia, I'm more or less going to directly copy the story of the conjecture from Wikipedia.

Next, I'd like to point out that the statement is now known as the Quillen-Suslin theorem after both gave independent proofs of the conjecture.

Projective Modules

Based on your comments, it seems like the main words that you don't understand are projective and free (also maybe finitely-generated, but that's less complicated). So let's talk about projective modules. A module $$P$$ over a ring $$R$$ is projective if any one of the following equivalent statements is satisfied.

1. The functor $$\newcommand\Hom{\operatorname{Hom}}\Hom_R(P,-)$$ is exact,
2. $$\Hom_R(P,-)$$ is right exact,
3. If $$f: M \to N$$ is surjective, and $$g:P\to N$$ is any map, then there exists a lift $$\tilde{g}:P\to M$$ such that $$g = f\circ\tilde{g}$$,
4. $$P$$ is a direct summand of a free module. (Free modules are defined below)

Drawing a picture of definition 3, we have that when we have $$g$$ and $$f$$, such that we have the following diagram with bottom row exact, then there exists $$\tilde{g}$$ such that the second diagram below commutes (apologies for the poor diagrams, but MSE doesn't support commutative diagrams very well). $$\require{AMScd} \begin{CD} @. P @.\\ @. @VgVV\\ M @>f>> N @>>> 0\\ \end{CD} \qquad\implies\exists_{\tilde{g}}\text{ such that}\qquad \begin{CD} P @= P @.\\ @V\tilde{g}VV @VgVV\\ M @>f>> N @>>> 0\\ \end{CD}$$

The reason I've put so much effort into the third definition is because it will be the most relevant for us (at least at this stage of the explanation). And now let's introduce free modules.

Free Modules

A module over a ring $$A$$ is free if it is isomorphic to the module $$A^{\oplus \alpha\in S}$$ (the possibly infinite direct sum of copies of $$A$$) for some index set $$S$$. Free modules are in many ways analagous to vector spaces. Indeed, a module $$M$$ is free if and only if it has a basis, i.e. a collection of elements $$e_s$$ for $$s\in S$$ such that the $$e_s$$ are linearly independent over $$A$$ and also span $$M$$. Thus the fact that every vector space has a basis tells us that every module over a field is free.

Now we can prove that free modules are projective (using definition 3). Let $$e_s$$ be a basis for a free module $$F$$, then a map from $$F$$ to any module $$M$$ is determined by its values on the basis (since a basis spans $$F$$), and any choice of values on the basis determines a map (since the basis is linearly independent), so if $$f:M\to N$$ is surjective and we have a map $$g:F\to N$$, then for each $$s$$, since $$f$$ is surjective, we can choose $$m_s\in M$$ such that $$f(m_s)=g(e_s)$$. Then if we define $$\tilde{g}(e_s)=m_s$$, we have $$f(\tilde{g}(e_s))=f(m_s)=g(e_s)$$, so $$g=f\circ \tilde{g}$$. Thus free modules are projective.

Side note, finitely generated

You've indicated you're not familiar with what a finitely generated module means, but the name sort of gives it away. A finitely generated module $$M$$ is one where there is a finite subset of $$M$$ such that its span over $$A$$ is all of $$M$$ (i.e. there is a finite subset of $$M$$ that generates $$M$$). It's analagous to a vector space being finite dimensional.

The relationship between projective and free modules

Since every free module is projective, it's natural to wonder if every projective module is free. Definition 4 of projective modules sort of answers this. We can prove that any direct summand of a free module is a projective module (and vice versa). Thus we can find projective modules that aren't free if we can find a summand of a free module that isn't itself free. See here for an example.

However, if our ring is sufficiently nice, nothing like that happens. In particular if our ring is a PID, all finitely generated projective modules are free.

Thus it is natural to wonder, well, what if our ring isn't a PID, but instead a polynomial ring over a PID. That should still be a fairly "nice" ring, so its modules should also be "nice". And this leads us to the actual story of the conjecture. For more on the relationship between projective and free, see this section of the Wiki page on projective modules.

The story

According to Wiki, Serre remarked that for certain rings $$K$$, like let's say as examples $$\Bbb{R}$$ or $$\Bbb{C}$$, facts about smooth and holomorphic manifolds tell us that every finitely generated projective module over $$K[x_1,\ldots,x_n]$$ is free.

Thus it shouldn't be too surprising if for any PID, all finitely generated projective modules over $$K[x_1,\ldots,x_n]$$ are free.

Indeed, this was proved by both Quillen and Suslin independently in 1976 (21 years after the initial conjecture).

Finally, the geometric interpretation

Algebraic geometry allows us to interpret algebraic statements geometrically. Under this correspondence, $$K[x_1,\ldots,x_n]$$ corresponds to affine space over $$K$$. If $$K$$ is an algebraically closed field, this is essentially $$K^n$$, although it can be much more complicated for a general PID.

Similarly under this correspondence projective modules over a commutative ring correspond to vector bundles on the geometric space the ring corresponds to. It would be fairly difficult for me to explain precisely why this is if you don't know any algebraic geometry, but for those who know some algebraic geometry, this lemma at the stacks project essentially answers this question.

Now if vector bundles correspond to projective modules, it's natural to ask which projective modules correspond to trivial vector bundles, and (perhaps unsurprisingly) the answer is free modules correspond to trivial vector bundles. Thus the fact that there are no nonfree projective modules corresponds to the geometric fact that any vector bundle on affine space over a PID is trivial.

Hopefully this was interesting and helpful.

• For the purpose of intuition, replace $K$ being a PID with the more restrictive condition that $K$ is a field (a trivial kind of PID). When $K$ is a field, algebraic geometry suggests that the ring $K[x_1,\ldots,x_n]$ should be considered as an algebraic incarnation of the space $K^n$ (more precisely, affine $n$-space over $K$ in the sense of algebraic geometry). This is why geometric ideas on $K^n$ may be converted into algebraic statements about the polynomial ring $K[x_1,\ldots,x_n]$ (e.g., the geometric concept of a vector bundle is like the algebraic idea of a projective module). – KCd Nov 21 '18 at 1:44
• That a result for $K[x_1,\ldots,x_n]$ when $K$ is a field might also work when $K$ is a PID should be thought of as purely a technical achievement and not part of the desired intuition. – KCd Nov 21 '18 at 1:45
• @KCd, excellent point – jgon Nov 21 '18 at 1:51
• @jgon Thank you for the efforts taken to explain. I will read closely and get back to you soon. – Ashok Nov 21 '18 at 2:16