Define a function $\alpha$ :Hom$_R$($R^n,M$) $\rightarrow M^n$ which is a $R$-module isomorphism.

Question

Suppose $R$ is a commutative ring, $M$ is an $R$-module, and $n \ge 1$. Define a function $$\alpha : \hom_R(R^n,M) \to M^n$$ which is a $R$-module isomorphism.

Answer 1

Take $f \in \hom _R (R^n, M)$ - that is, $f : R^n \to M$ is a $R$-morphism of $R$-modules. You would like to associate to this $f$ an element of $M^n$. Let $e_i = (0, \dots, 0, 1, 0, \dots, 0)$ with $1$ on the $i$th position, $1 \le i \le n$. Define $\alpha(f) = (f(e_1), \dots, f(e_n))$.

Since

$$\alpha (f+g) = \big( (f+g) (e_1), \dots, (f+g) (e_n) \big) = (f(e_1) + g(e_1), \dots, f(e_n) + g(e_n)) = (f(e_1), \dots, f(e_n)) + (g(e_1), \dots, g(e_n)) = \alpha (f) + \alpha (g)$$

and

$$\alpha(rf) = (rf(e_1), \dots, rf(e_n)) = r (f(e_1), \dots, f(e_n)) = r \alpha(f)$$

it follows that $\alpha$ is a $R$-morphism of $R$-modules.

To show that it is surjective, let $(m_1, \dots, m_n) \in M^n$ and define $f : R^n \to M$ by

$$f(r_1, \dots, r_n) = r_1 m_1 + \dots + r_n m_n .$$

Check for yourself that this $f$ is a $R$-morphism and notice that $\alpha(f) = (m_1, \dots, m_n)$ - which proves that $\alpha$ is surjective.

To show that $\alpha$ is injective, notice that

$$\alpha(f) = 0 \iff (f(e_1), \dots, f(e_n)) = 0 \iff f(e_i) = 0 \ \forall i = 1, \dots, n$$

and since $\{e_i\} _{1 \le i \le n}$ is a basis in $R^n$ it follows that $f=0$, so $\ker \alpha = 0$, i.e. $\alpha$ is injective.

Being an injective and surjective morphism, $\alpha$ is an isomorphism.