Exercise A0.III.6.04

Proposition

Let $R$ be a ring. Then

$$\frac{R^{\oplus n}}{R^{\oplus (n - 1)}} \cong R.$$

Proof

We are given an injective homomorphism $\iota : R^{\oplus (n-1)} \to R^{\oplus n}$, defined by

$$(r_1,r_2,\ldots,r_{n-1}) \mapsto (r_1,r_2,\ldots,r_{n-1}, 0).$$

Consider the short exact sequence

$$0 \to R^{\oplus (n - 1)} \overset{\iota}{\hookrightarrow} R^{\oplus n} \overset{\pi_n}{\twoheadrightarrow} R \to 0.$$

It induces an isomorphism $R \cong R^{\oplus n}/\im\iota,$ but $\im \iota$ is just $R^{\oplus (n - 1)}.$

$\blacksquare$