Assume $n>m$. Is it possible to have an injective $R$-module homomorphism from $R^n$ to $R^m$?
I feel this is possible, the homomorphism is determined by where we send each basis. So we just choose different element in $R^m$ and send different basis of $R^n$ to them. Is this right?
The following is the same as in linear algebra: an injective homomorphism $f$ preserves linear independence because if $\{ e_i \}$ are linearly independent and you have a dependence relation $$0 = \sum r_i f(e_i) = f\left(\sum r_i e_i\right)$$ then $\sum r_i e_i \in \ker f = \{ 0 \}$ which implies that all the $r_i$ are zero.
Thus, if your ring $R$ has a well-defined notion of rank, one cannot embed $R^n$ into $R^m$ when $n > m$. This holds in particular if $R$ is commutative.
However, this need not be true in general. The following counterexample can be worked out in exercise 27 of $\S$10.3 in Dummit & Foote: Consider the endomorphism ring $R = \operatorname{End}_\mathbb{Z}(\mathbb{Z} \times \mathbb{Z}\times \cdots)$. Then the elements $\phi_1,\phi_2 \in R$ defined by $$\phi_1(a_1,a_2,a_3,\ldots)=(a_1,a_3,a_5,\ldots), \quad \phi_2(a_1,a_2,a_3,\ldots)=(a_2,a_4,a_6,\ldots)$$ form a basis of $R$ as an $R$-module over itself. Thus $R \cong R^2$. (!)
