# $R$-linear injection [duplicate]

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?

• $R=0$ is a counterexample. ;) – 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 $\bar{f}=f \otimes_R R/\mathfrak{m}$.
• This doesn't work, since $\overline{f}$ is not injective in general. This proof only works for the corresponding statement about surjective homomorphisms. – Martin Brandenburg Feb 21 '13 at 15:40
• For example, $2 : \mathbb{Z} \to \mathbb{Z}$ gets the zero map when tensored with $\mathbb{F}_2$. – Martin Brandenburg Feb 21 '13 at 15:51
• @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? – Fredrik Meyer Feb 21 '13 at 16:18
• 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 $d=x$. – Martin Brandenburg Feb 21 '13 at 16:46