# spurious torsion under compositions of linear maps

Say we have a PID $R$, integers $1 \leq a \leq b$, and $R$-homomorphisms $R^a \stackrel f\to R^b \stackrel g\to R^a$ with $g \circ f$ of full rank.

For $h = f, g, g \circ f$, let $c(h)$ be the characteristic ideal of $\mathrm{coker}(h)_\mathrm{tors}$, i.e. the index of the image of $h$ in its saturation. (In particular $c(g \circ f) = \det(g \circ f)R$, though this interpretation isn't available for $c(f),c(g)$ when $a < b$.)

Keeping $c(f),c(g)$ fixed, can I arrange $c(g \circ f)$ to be divisible by whatever primes of $R$ I want, beyond to those forced to appear by $c(f),c(g)$? If so, I would like a obvious way of constructing examples.

-
Dear Jay, is there a reference for characteristic ideal? I do not understand the definition (though I have a suspect in commutative algebra language). Where can I read more about it? – Hailong Dao Jun 28 '11 at 2:18
More to the point, is it related to the Fitting ideal? – Hailong Dao Jun 28 '11 at 2:21
If $R$ is a PID and $M$ a finitely generated torsion module, hence isomorphic to $\bigoplus_{i=1}^n R/f_i$, then the characteristic ideal of $M$ is generated by $\prod_{i=1}^n f_i$. In this case, I believe it agrees with the Fitting ideal. The terminology comes from the case where $V$ is a finite-dimensional vector space over a field $k$, equipped with a $k$-linear endomorphism $A$. Repackaging $V$ as a finitely generated torsion module over the PID $R=k[X]$ (with $X$ acting via $A$), the the characteristic ideal is simply the one generated by the characteristic polynomial of $A$. – Jay Jun 28 '11 at 3:04
If I understand the question, the answer is yes if $(a,b)=(1,2)$ and therefore also more generally when $a<b$. – Tom Goodwillie Jun 28 '11 at 3:33

Here is an obvious way to construct examples with $a=1,b=2$. By elementary divisor theory, with these numerics one always has $f=uf_0,g=vg_0$ with $c(f_0),c(g_0)=1$ for some $u,v \in R$, so it suffices to study the situation with $f,g$ replaced by $f_0,g_0$, i.e. to assume that $c(f),c(g)=1$.
Take $w \in R$ arbitrary. Then $f(x)=(x,0)$ up to choice of basis, and $g(y,z) := wy+z$ satisfies all the conditions, but has $c(g \circ f) = w$.