Exercise 3.6.4 ($ R^{\oplus n} / R^{\oplus (n-1)} \cong R $)

Question

Let $R$ be a ring and let $n > 1$. View $R^{\oplus (n-1)}$ as a submodule of $R^{\oplus n}$ via the 
injective homomorphism 
$$
\varphi: R^{\oplus (n-1)} \hookrightarrow R^{\oplus n}
$$
defined by 
$$
(r_1, \dots, r_{n-1}) \mapsto (r_1, \dots, r_{n-1}, 0).
$$
Prove that 
$$
R^{\oplus n} / R^{\oplus (n-1)} ~ \cong ~ R.
$$

Shinkenjoe · Accepted Answer

Consider the projection homomorphism 
$$
\psi: R^{\oplus n} \twoheadrightarrow R
$$
given by 
$$
(r_1, \dots, r_n) \mapsto r_n.
$$
This is surjective, as is obvious. Then, by the First Isomorphism Theorem, we have 
$$
R^{\oplus n} / \ker \psi ~ \cong ~ R.
$$

What is $\ker \psi$? It consists of all $(r_1, \dots, r_{n-1}, r_n) \in R^{\oplus n}$ such that $r_n = 0$. This equals $\operatorname{im} \varphi$. Since $\varphi$ is injective and surjective onto its image, we conclude that 
$$
R^{\oplus (n-1)} ~ \cong ~ \ker \psi.
$$
The claim follows.

Exercise 3.6.4 (\( R^{\oplus n} / R^{\oplus (n-1)} \cong R \))

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Exercise 3.6.4 (\( R^{\oplus n} / R^{\oplus (n-1)} \cong R \))

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