$n+1$ elements in the free module $R^n$ must be linear dependent?

Question

Suppose that $R$ is a nonzero commutative ring with unit, and $R^n$ is the free module of dimension $n$ over $R$. Is it right that $n+1$ elements in $R^n$ must be linearly dependent?

Answer 1

Suppose that there are $n+1$ linearly independent elements $x_1, x_2, …, x_{n+1}$ in $M=R^{n}$. Consider the submodule $F$ generated by these $n+1$ elements. Then clearly $F\cong R^{n+1}$. Since $F$ is a submodule of $M=R^{n}$, it follows that there is an injective $R$-linear map $R^{n+1}\to R^{n}$. But this is impossible!

That there is no injective $R$-linear map from $R^{n+1}\to R^{n}$ follows from this (famous?) exercise in Atiyah & Macdonald's commutative algebra textbook (Exercise 2.11). See this Mathoverflow thread. Let me copy the statement of the exercise and a beautiful solution (given by Balasz Strenner) for convenience.

Proposition. If $A$ is a nonzero commutative ring with $1$, and there is an injective $A$-linear map $A^{m}\to A^{n}$, then $m\leq n$.

Proof. Assume by contradiction that there is an injective map $\phi: A^m \to A^n$ with $m>n$. The first idea is that we regard $A^n$ as a submodule of $A^m$, say the submodule generated by the first $n$ coordinates. Then, by the Cayley-Hamilton Theorem (Proposition 2.4 in Atiyah & Macdonald), $\phi$ satisfies some polynomial equation \begin{equation} \phi^k + a_{k-1} \phi^{k-1} + \cdots + a_1 \phi + a_0 = 0. \end{equation} Using the injectivity of $\phi$ it is easy to see that if this polynomial has the minimal possible degree, then $a_0 \ne 0$. But then, applying this polynomial of $\phi$ to $(0,\ldots,0, 1)$, the last coordinate will be $a_0$ which is a contradiction as it should be zero.