# An easy half of Quillen-Suslin theorem

Michael Artin's Algebra, exercise 14.9.4, mixed with the context

Suppose $V$ is a finitely generated free module over the ring $R=\mathbb C[x_1,\dots,x_k]$. Let $A$ be a presentation matrix for $V$. Denote by $A(c)$ the evaluation of $A$ at a point of $\mathbb C^k$, viz. $x=(x_1,\dots,x_k)\mapsto c\in\mathbb C^k$. Then the rank of $A(c)$ is constant for any $c\in\mathbb C^k$.

What I all know is that $R$ is Noetherian. Since apparently, $R$ isn't PID when $n>1$, the structure theorem of finitely generated module doesn't apply. Moreover, after some searching work, I know that there's some commutative ring $R$ such that there's some maximal independent subset of cardinality $m\neq n$ in the free module $R^n$, therefore the properties of modules are quite different from those of vector spaces. After some messy tries, I have no result or no idea to proceed.

Any idea? Thanks!

Edit.

1. An $m\times n$ matrix $A$ is called a presentation matrix of an $R$-module $M$ iff $M\simeq R^m/AR^n$, where $AR^n=\{\,AX\colon X\in R^n\,\}$ is a submodule of $R^m$. Note that even a free module could have a collection of dependent generators with relations. It seems strange, but obvious after a second glance.

2. I misunderstood the proposition about the module. The original thread is here. It's true that the cardinality of the independent set is no greater than the rank of the free module. It could be easily proved by Cayley-Hamilton theorem. See here.

• For a commutative ring A with 1, there is no injective A-module map from A^m to A^n if m>n. – wxu Aug 1 '13 at 2:06
• The ring A we require is not the zero ring in the statement. – wxu Aug 1 '13 at 2:39

Let $A_1,A_2,\ldots, A_m$ be the column vectors of matrix $A$. Suppose $V$ is isomorphic to $R^k$. Then we have an exact sequence $R^m \to R^n\to R^k \to 0$. Let matrix $B$ denote the second map and let $C_1,\ldots, C_k$ such that $B(C_1,\ldots,C_k)=I_k$. On the other hand, show that $R^n$ is the direct sum of the module $\langle A_1,\ldots,A_m\rangle$ and the module generated by $C_i$'s. In other words, we can find a matrix $M$ such that $I_n = (A\mid C)M$. Since under any evaluation $v$ the later module is always of rank $k$, then matrix $C(v)$ is of rank $k$ hence the first module is of rank $n-k$, $A(v)$ is of rank $n-k$.
• Seems true. The key point might be: once $V$ is free, the exact sequence $R^n\xrightarrow AR^m\to V\to0$ decomposes $R^m\approx(AR^n)\oplus V$, therefore there's some invariant of $AR^n$. – Yai0Phah Aug 3 '13 at 12:29