Unique $R$-linear homomorphism $R\to M$

Question

Let $R$ be a ring with unit and $M$ a $R$-module. I would like to prove that for each $x$ exists unique an $R$-linear homomorphism $f_x : R \to M$ such that $f_x(1) = x $. In my opinion, a good definition would be $f_x(a) = a\cdot x$ and this is an actual homomorphism that satisfies all the requirements. However, I am not able to prove that it is unique. My idea was to assume that for each $x$ it exists another $R$-linear homomorphism $g_x$ s.t $g_x(1) = x$ and then prove that for each $r\in R, f_x(r) = g_x(r)$. If it is not, then exists $\bar{x}$ such that $f_{\bar{x}}(r) - g_{\bar x}(r) \neq 0$ and $ f_{\bar{x}}(1) = g_{\bar x}(1) = \bar x $. However I can't understand how to conclude.

Answer 1

If $g_x:R\to M$ is another choice, then for any $a\in R$ $$g_x(a)=g_x(a\cdot 1)=a\cdot g_x(1)=ax$$ since $g_x$ is $R$-linear.

So actually $g_x=f_x$.