Determinant of one stably free projective module

Question

A projective $A$-module $P$ is called one stably free if $P\oplus A\simeq A^n$ for some $n\in \mathbb{Z}$. It can be seen that such projective modules are in one-to-one correspondence with the orbits of the natural $GL_n(A)$ action on the set $Um_n(A)=\{[a_1,\ldots,a_n]\in A^n\mid Aa_1+Aa_2+\dots+Aa_n=A\}$. It can be seen that $\det(P)\simeq A$, if $P$ is one stably free. How to express a choice of isomorphism $\det(P)\simeq A$ interms of an element in $Um_n(A)$?

Answer 1

The projective module $P$ is generated by $n$ elements $e_1,\ldots,e_n$ with one relation given by the unimodular row, say $a_1e_1+\cdots+a_ne_n=0$ where one has $a_1b_1+\cdots+a_nb_n=1$ for some $b_i$. Rank of the projective module is $n-1$ and its determinant is generated by the $n$ elements $v_i=e_1\wedge e_2\wedge\cdots\wedge e_{i-1}\wedge e_{i+1}\wedge\cdots e_n$.

I claim that the determinant is generated by $b_1v_1-b_2v_2+\cdots+(-1)^{n-1}b_nv_n$. The algebra is a bit messy and long to write, so let me just do this for $n=3$.

So, we have $v_1=e_2\wedge e_3, v_2=e_1\wedge e_3, v_3=e_1\wedge e_2$ and we want to show that $\alpha=b_1v_1-b_2v_2+b_3v_3$ generates $\wedge^2 P$. Clearly suffices to show that $\alpha$ generates when one of the $a_i$ is inverted, since these cover the spectrum of the ring. Without loss of generality, let us assume $a_1$ is inverted. Then we will show that $a_1\alpha$ generates $\wedge^2P$. From $a_1e_1+a_2e_2+a_3e_3=0$, one gets $a_1v_3=a_3v_1$ and $a_1v_2=-a_2v_1$ and thus when $a_1$ is inverted, $v_1$ generates $\wedge^2 P$. But, $a_1\alpha=a_1b_1v_1-a_1b_2v_2+a_1b_3v_3$ and substituting for the terms involving $v_2,v_3$, one gets $a_1\alpha=(a_1b_1+a_2b_2+a_3b_3)v_1=v_1$.

I hope the argument is clear.