spurious torsion under compositions of linear maps - MathOverflow most recent 30 from http://mathoverflow.net2013-05-18T22:11:06Zhttp://mathoverflow.net/feeds/question/68994http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/68994/spurious-torsion-under-compositions-of-linear-mapsspurious torsion under compositions of linear mapsJay Pottharst2011-06-28T02:01:47Z2011-06-28T12:17:39Z
<p>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.</p>
<p>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 $a < b$.)</p>
<p>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.</p>
http://mathoverflow.net/questions/68994/spurious-torsion-under-compositions-of-linear-maps/69017#69017Answer by Jay Pottharst for spurious torsion under compositions of linear mapsJay Pottharst2011-06-28T12:17:39Z2011-06-28T12:17:39Z<p>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 $c(f),c(g)=1$.</p>
<p>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 $c(g \circ f) = w$.</p>