Verification of an R-Module isomorphism between $R^n$ and its dual With one step at a time, I am getting slightly more used to $R$-Modules.

Let $R$ denote a commutative Ring with $\mathbb{1}$ and $n$ a natural number. For the tuple $a:= (a_i)_{i=1}^n \in R^n$ we have the mapping: $$\varphi_a: R^n \to R ,\  (x_i)_{i=1}^n \mapsto \sum_{i=1}^n a_i x_i $$

It is easy to see that the above mapping is $R$ linear, which means that for scalars $r,s \in R$ and 'vectors' $x,y$ we have $\varphi_a(rx + sy)= r\varphi_a(x) + s\varphi_a(y)$
I now want to show that the following mapping $$\psi : R^n \to \text{dual}(R^n), \ a \mapsto \varphi_a $$
is an $R$-Module isomorphism. Here dual$(R^n)$ just denotes linear mappings from $R^n \to R$

My approach:  First I want to show that $\psi$ is a homomorphism. I did that as follows.
Let $r,s \in R$ and $a,b \in R^n$ be arbitrary. Then it follows that $$\psi(ra + sb) = \varphi_{ra + sb} \tag{*} $$
By definition. So lets take $x \in R^n$ arbitrary and continue from there $$\varphi_{ra + sb}(x) =\sum_{i=1}^n (ra_i + sb_i)x_i = r\sum_{i=1}^n a_ix_i + s\sum_{i=1}^nb_i x_i = r \varphi_a(x) + s\varphi_b(x) = (r\varphi_a + s\varphi_b)(x) $$
Since $x \in R^n$ was arbitrary it follows that $\varphi_{ra + sb} = r \varphi_a + s \varphi_b$ which means for the above calculation at (*) that  $$\psi(ra + sb) = \varphi_{ra + sb}= r \varphi_a + s \varphi_b = r \psi(a) + s \psi(b) $$
Which means that $\psi$ is R-linear.

Injectivity: I want to show that the kernel of $\psi$ is trivial. Choose $a \in \ker \psi$ arbitrary. Thus I have $\psi(a) = \varphi_a = 0$. I want to show that my 'vector' $a$ is the zero vector.
I thought choosing an arbitrary $x \in R^n$ would be the next thing to do but then I ran into troubles. So instead I choose $x \in R^n \setminus \ker \psi$ and continue my calculations from there.
$\varphi_a(x)=x\varphi_a(1)= x(a_1 + \dots + a_n ) =0 \implies a_1 + \dots + a_n =0, \text{ because } x \neq 0$
And here I am stuck again. I somehow want to argue that it must follow that $a_1 = a_2 = \dots = a_n =0$ but I don't see a rigorous statement for that yet, so I assume my entire approach was flawed.
Surjectivity: I didn't come up with a good idea here yet, I might must give it some more thought or maybe you could provide me a hint here as well.

Given $\varphi\colon R^n\to R$ you want to find $a=(a_1,a_2,\dots,a_n)\in R^n$ such that $\varphi=\varphi_a$. In particular, you want that
$$
\varphi(e_1)=\varphi_a(e_1)=a_1
$$
where $e_1=(1,0,\dots,0)$.
Can you go on from here?
A similar idea can be used for injectivity.
The rest is maybe a bit too verbose, but correct.
