Ah no that's stronger

Well wait, what does a split exact sequence of groups mean to you? Groups are not an abelian category

I have not, but may as well ask

I'm not sure; that's what I am asking about.

@Balarka do you know a non-cancellative f.g. module over a commutative ring

@Balarka Yesterday me and Thorgott where thinking about the cancellation property for modules. An $R$-module $B$ has the cancellation property if for all $R$-modules $C$ and $D$, $B\oplus C\cong B\oplus D$ implies $C\cong D$. What is an example of a f.g. $B$ without the cancellation property?

Apparently $R$ must be pretty ugly. What was the result you found @Thorgott? $R$ can't be a Dedekind domain, right?

yeah, that was it

(if we drop the f.g. of course it's easy because $R^\omega$ absorbs finite powers of $R$)

$1 \to H \to G \to K \to 1$ is an exact sequence if the map from $H$ to $G$ is injective, the map from $G$ to $K$ is surjective, and the image of the first map is the kernel of the second map.

and if we drop commutativity, any ring without IBN gives an example

From what I recall, the sequence splits if there is a map going back from $K$ to $G$ satisfying some condition, but I don't recall the condition.

The composition is the identity on $K$ is the condition

Ah, okay.

Thanks!

You need rings over which there are non-free projective modules first of all

Nevermind

Wait but there are such examples

$TS^2 \oplus \Bbb R \cong \Bbb R^2 \oplus \Bbb R$

Pass to modules over $C^\infty(S^2)$ using Serre-Swan

Wait I'm confused half of those things are not bundles over $S^2$

But I like going for a Serre-Swan trick

When I say $\Bbb R^n$ I mean the trivial bundle of rank $n$ over $S^2$

Oh ok sorry

Yeah that should work then

Nuked

No I like that approach

Very neat

Is something interesting here

Murdering algebraists, yes

I do feel murdered

Balarka is turning into a noncommutative geometer

should do anabelian geometry instead

There should be a purely algebraic translation. Look at the ring $R = \Bbb R[x, y, z]/(x^2 + y^2 + z^2 - 1)$, consider the kernel $K$ of $R^3 \to R$, $(a, b, c) \to ax + by + cz$

Then $K \oplus R \cong R^2 \oplus R$

This is because you have a short exact sequence $0 \to K \to R^3 \to R \to 0$, and the last term is free

SES where the last term is free splits

The only difficult is to prove $K$ is not free, but this is manageable.

This is that thing

What do you call it

Fucking hell I knew this very well

I think the geometric picture was nicer

That needs to go on the starboard. Alessandro liked something geometric.

Ah yeah

Finding a unimodular row

Oh but I will often prefer the geometric argument over doing the algebraic version of the same argument

Look up Bass stable range

My main complaint with the AG course I took was that is was 99% algebra and 1% geometry

That often happens with algebraic algebraic geometers :(

Here's the algebraic problem. You're given a vector $(v_1, \cdots, v_n) \in R^n$ such that there exists $u_1, \cdots, u_n \in R$ such that $\sum u_i v_i = 1$

Is $(v_1, \cdots, v_n)$ the first row of a matrix in $M_n(R)$ with determinant $1$?

yeah, I'd also be annoyed at that 1% geometry

The above example is a negative answer; $(x, y, z)$ satisfies the criterion, but suppose it was part of a matrix with determinant $1$. The other two rows can be chosen to be orthogonal to $(x, y, z)$ (because say $(a, b, c)$ is another row, then $ax + by + cz = k \in R^\times$, then $(a - kx)x + (b - ky)y + (c - kz)z = 0$ and this new vector $(a - kx, b - ky, c - kz)$ is nonzero by det = 1 condition)

But that would give us a basis of $K$

Note how this is true if $R$ is a PID or whatever

Just reverse this argument; suppose $K \cong R^2$. Choose a basis $(a, b, c)$, $(u, v, w)$ for $K$ and consider the determinant of $(x, y, z|a, b, c|u, v, w)$ which lands in $R^\times$. Elements of $R$ are functions on the sphere $S^2$, so evaluating this determinant at points on $S^2$ says the determinant is nonvanishing and in particular $(a, b, c)$ is a nowhere vanishing tangent vector field on $S^2$

Contradicting hairy ball

A sequence of elements in a commutative ring $R$ which generates the unit ideal is called a unimodular row, and the question of when it is the first row of some matrix of unit determinant is called completability

The classical phrasing of Serre's conjecture (proved by Quillen and Suslin) was if every unimodular row in a polynomial ring is completable

Fields medal work

what on earth is this

Very classical algebraic K-theory

In a ring where every unimodular row is completable, the first K-group vanishes