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Problem with split exact sequences and free finitely generated modules

Mathematics Asked by aa_bb on September 22, 2020

Let the sequence $$0to Ato Bto Cto 0$$ be a split exact sequence of $R$-modules over a ring $R$. The ring $R$ is a commutative ring with identity. Let $A,B$ be free and finitely generated. Is $C$ also free and finitely generated? If it is not, then under which conditions on the ring $R$ we can say that $C$ is free and finitely generated?

I know that without splitting it is certainly not true, due to the common example: $0to Zto 2Zto Z/2Zto 0$. However, I have no idea on how to approach this…

2 Answers

Split exact implies that $B$ is isomorphic to the direct sum of $A$ and $C$. Thus, $C$ is a direct summand of the free and finitely generated module $B$ (and thus $C$ must be a projective module). So, for example, the structure theorem for finitely generated modules over principal ideal domains implies that a sufficient condition for this result to hold is that $R$ be a principal ideal domain.

Also, note that $C$ is always finitely generated, being a surjective image of a finitely generated module. The only issue is whether $C$ need be free.

Without the f.g., conditions, a module $C$ that can occur in this split-exact sequence is called stably free. (Note that not all projective modules are stably free, since $A$ is not required to be free in order for $C$ to be projective.) Thus, $C$ in your question is a finitely generated stably free module. There are commutative rings $R$ where not all such modules are free. For example, let $R=Bbb{R}[x,y,z]/(x^2+y^2+z^2-1)$ and let $T={(f,g,h)in R^3 | xf+yg+zh=0}$. Then $T$ is not free, but $Toplus Rcong R^3$. See here for a proof, and a discussion of stably free modules generally.

Also, note that not all projective modules are stably free. For example, if $R$ is a Dedekind domain that is not a principal ideal domain, then its non-principal ideals are projective modules that are not stably free. In fact, I think you could conclude from the structure theory for f.g. modules over Dedekind domains that $R$ being a Dedekind domain is sufficient to make $C$ free.

Correct answer by C Monsour on September 22, 2020

Consider $R=mathbb{Z}[x,y,z]/(x^2+y^2+z^2-1)$.

Then we have a split short exact sequence: $$0to Rstackrel ito R^3to Mto 0,$$ where $$icolon 1mapsto left(begin{array}{c} x\y\zend{array}right),$$ and $M$ is the quotient $R^3/i(R)$.

To see that this sequence is split, it suffices to note that we have a map $$pcolon R^3to R,$$ represented by the matrix $$ left(begin{array}{ccc} x&y&zend{array}right),$$and $pi=1_R$ (as $x^2+y^2+z^2=1$).

However $M$ is not free.

Proof: Suppose $M$ is free. Then it must have rank $2$ as $R$ is Noetherian and we cannot have $$R^icong R^ioplus R^jcong R^ioplus R^joplus R^jcongcdots,$$ with $j>0$.

Thus if $M$ free then $Mcong R^2$ and we have a projection $alphacolon R^3to Mto R$ onto the first summand and inclusion $beta$ back.

$$Rmathrel{mathop{rightleftharpoons}^{i}_{p}}R^3mathrel{mathop{rightleftharpoons}^{alpha}_{beta}} R$$

We have $$alpha i=0,quad pbeta=0, alphabeta=1_R.$$

Writing $$alpha= left(begin{array}{ccc} a&b&cend{array}right),quad beta=left(begin{array}{c} u\v\wend{array}right),$$ for some $a,b,c,u,v,win R$ we have:

$$left(begin{array}{c} a&b&c end{array}right) left(begin{array}{c} x\y\zend{array}right)=0,qquad left(begin{array}{c} x&y&zend{array}right)left(begin{array}{c} u\v\wend{array}right)=0 qquad left(begin{array}{c} a&b&cend{array}right)left(begin{array}{c} u\v\wend{array}right)=1$$

Thus $(a,b,c)$ and $(u,v,w)$ describe tangent vector fields to the unit sphere in $mathbb{R}^3$, whose inner product is everywhere equal to $1$. This cannot happen by the Hairy Ball theorem completing our proof.


As far as I know this by far the simplest proof of not only the non-freeness of this module $M$, but any finitely generated stably free module over a commutative ring. It is fascinating that the Hairy Ball theorem is used right at the end, given that prior to that point, it is not even necessary to define the real numbers, either in the argument or the construction of $M$, or indeed for the notion of a stably free module.

Answered by tkf on September 22, 2020

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