The Stacks project

A map of finite free modules cannot be injective if the source has rank bigger than the target.

Lemma 10.15.7. Let $R$ be a nonzero ring. Let $n \geq 1$. Let $M$ be an $R$-module generated by $< n$ elements. Then any $R$-module map $f : R^{\oplus n} \to M$ has a nonzero kernel.

Proof. Choose a surjection $R^{\oplus n - 1} \to M$. We may lift the map $f$ to a map $f' : R^{\oplus n} \to R^{\oplus n - 1}$ (Lemma 10.5.2). It suffices to prove $f'$ has a nonzero kernel. The map $f' : R^{\oplus n} \to R^{\oplus n - 1}$ is given by a matrix $A = (a_{ij})$. If one of the $a_{ij}$ is not nilpotent, say $a = a_{ij}$ is not, then we can replace $R$ by the localization $R_ a$ and we may assume $a_{ij}$ is a unit. Since if we find a nonzero kernel after localization then there was a nonzero kernel to start with as localization is exact, see Proposition 10.9.12. In this case we can do a base change on both $R^{\oplus n}$ and $R^{\oplus n - 1}$ and reduce to the case where

\[ A = \left( \begin{matrix} 1 & 0 & 0 & \ldots \\ 0 & a_{22} & a_{23} & \ldots \\ 0 & a_{32} & \ldots \\ \ldots & \ldots \end{matrix} \right) \]

Hence in this case we win by induction on $n$. If not then each $a_{ij}$ is nilpotent. Set $I = (a_{ij}) \subset R$. Note that $I^{m + 1} = 0$ for some $m \geq 0$. Let $m$ be the largest integer such that $I^ m \not= 0$. Then we see that $(I^ m)^{\oplus n}$ is contained in the kernel of the map and we win. $\square$

Comments (2)

Comment #1478 by jp on

The statement of the lemma is not quite right, since if the ring R is trivial (1=0), then the kernel is indeed zero. Perhaps it would be better to say that if the kernel is zero then 1=0 in R. This formulation allows one to make the argument constructive: one starts by proving that all of the a_ij are nilpotent. (Show 1=0 in R_{a_ij}--and hence that a_ij is nilpotent--by induction on n as in the current proof.) With notation as in the proof, the assumption that the kernel is trivial shows that if I^{m+1}=0, then I^m=0 and hence by induction that I^0=0, i.e. that 1=0.

Comment #1496 by on

Hi jp. OK, I fixed it here by assuming the ring is nonzero.

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  • 5 comment(s) on Section 10.15: Miscellany

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