If $f: R^n\rightarrow R^m$ is an injective map, which is also $R$-linear, where $R$ is a commutative ring with unity. Is it true that $n$ has to be less than or equal to $m$ always?

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    $\begingroup$ $R=0$ is a counterexample. ;) $\endgroup$ – Martin Brandenburg Feb 21 '13 at 15:49

When $R \neq 0$, this is true. This question has already appeared on math.SE and on mathoverflow. There you find lots of proofs. For example quite short one using the Cayley-Hamilton theorem.


Let $\mathfrak{m}$ be any maximal ideal of $R$. You can tensor the map with $R/\mathfrak{m}$, so that you have a map a $\bar{f}:(R/\mathfrak{m})^n \to (R/\mathfrak{m})^m$, where $\bar{f}=f \otimes_R R/\mathfrak{m}$.

Edit: This is wrong, see Martin's comment below.

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    $\begingroup$ This doesn't work, since $\overline{f}$ is not injective in general. This proof only works for the corresponding statement about surjective homomorphisms. $\endgroup$ – Martin Brandenburg Feb 21 '13 at 15:40
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    $\begingroup$ For example, $2 : \mathbb{Z} \to \mathbb{Z}$ gets the zero map when tensored with $\mathbb{F}_2$. $\endgroup$ – Martin Brandenburg Feb 21 '13 at 15:51
  • $\begingroup$ @MartinBrandenburg: Of course you're right. However, can we make it work if require $R$ to have infinitely many maximal ideals? (idea: $f$ is represented by some matrix $(m_{ij})$. If we choose a maximal ideal $\mathfrak{m}$ with $m_{ij} \not \in \mathfrak{m}$ for all $i,j$, then (maybe) injectivity survives? $\endgroup$ – Fredrik Meyer Feb 21 '13 at 16:18
  • $\begingroup$ A matrix with nonzero entries can have nontrivial kernel. $\endgroup$ – Martin Brandenburg Feb 21 '13 at 16:29
  • $\begingroup$ One can show that a homomorphism $R^n \to R^m$ is injective if and only if there is no element of $R \setminus \{0\}$ which annihilates all the $n \times n$-minors of the corresponding matrix (this provides another proof that we must have $n \leq m$). In particular, $R^n \to R^n$ is injective, but not injective when tensored with an arbitrary field over $R$, if and only if the determinant $d \in R$ is regular and in the Jacobson radical. There are examples, even for $n=1$, for example the local integral domain of germs of holomorphic functions at $0$, with $d=x$. $\endgroup$ – Martin Brandenburg Feb 21 '13 at 16:46

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