For a ring $R,$ there is no injective map from $R^m \hookrightarrow R^n$ for $m > n.$ (verification)

Question

Suppose $R$ is a ring and we have an $R$-module homomorphism of free modules $R^{n + 1} \hookrightarrow R^n.$ Then take a submodule of $R^{n + 1}$ that is isomorphic to $R.$ Say for any $r \in R,$ we can associate it to $(0, \ldots, 0, r)$ or even $r$ to $(r, \ldots, r).$ Now the image of this submodule under this injective map is also isomorphic to $R.$ Then taking the quotient of this submodule yields $R^{n + 1}/R = R^n \hookrightarrow R^n/R = R^{n - 1}.$ Proceeding in this manner yields $R \hookrightarrow 0$ which implies $R = 0.$ Does this work? Any comments are appreciated. (I guess this says the injectivity only holds when $R$ is the trivial ring but this is trivial...)

Answer 1

Your argument is wrong because (with your notation) $R/R$ may not be zero. For an example, first note:

If $R$ is an integral domain and $0\neq r\in R$, then $(r)\cong R$ as $R$-modules.

The isomorphism is given by multiplication with $r$. Once we have this, consider the polynomial ring (say, with complex coefficients) on infinitely many variables:

$$R=\mathbb{C}[x_1,x_2,\dots] $$

Then the ideal $(x_1)$ is isomorphic to $R$. But the quotient $R/(x_1)$ is isomorphic to $R$ (as a ring, because they satisfy the same universal properties), and in particular it is non-zero.