We defined the Generalized Reed-Solomon codes the following way: $\alpha=(\alpha_0,\alpha_1,\ldots,\alpha_{n-1})\in \mathbb{F}_q$, distinct elements of the finite field $\mathbb{F}_q$, $n\leq q$, $v=(v_0,v_1,\ldots,v_{n-1})$. For $k\leq n$, $GRS_{n,k}(\alpha ,v)=\left \{ v_0 f(\alpha_0), v_1 f(\alpha_1), \ldots, v_{n-1} f(\alpha_{n-1})\mid \deg\, f< k \right \}$. Given is $h=\prod_{i=0}^{n-1}(x-\alpha_i)$.
I have to show that the map $\varphi :\mathbb{F}_q[x]/(h)\rightarrow \mathbb{F}_q^n, f+(h) \mapsto(f(\alpha_0),f(\alpha_1),\ldots,f(\alpha_{n-1}))$ is an isomorphism.
Since the polynomial $h$ is reducible, i've been thinking that here we can apply the Chinese Remainder theorem, which says that $f+(h) \mapsto (f+(x-\alpha_0), f+ (x-\alpha_1),\ldots,f+(x-\alpha _{n-1}))$ but i don't see how this should be equal to $(f(\alpha_0), f(\alpha_1),\ldots,f(\alpha_{n-1}))$, if the idea is correct at all. To show the isomorphism, i need to point that the there is a homomorphism, which is clear, because the map preserves the operation, but how to show the existence of an inverse map (the bijectivity of this homomorphism)?
Can anybody help me with this question, please? Thank you very much in advance!
As you showed in your question,
$$\begin{align} \mathbb{F}_q[x]/(h(x)) &\cong \prod_{i=0}^{n-1} \mathbb{F}_q[x]/(x-\alpha_i) \\ f(x) + (h(x)) &\mapsto \left( f(x) + (x-\alpha_0), \ldots, f(x) + (x-\alpha_{n-1}) \right). \end{align}$$
Furthermore, for each $i$,
$$\begin{align} \mathbb{F}_q[x]/(x-\alpha_i) &\cong \mathbb{F}_q[\alpha_i] \\ f(x) + (x-\alpha_i) &\mapsto f(\alpha_i). \end{align}$$
Note that $\mathbb{F}_q[\alpha_i] = \mathbb{F}_q$. Therefore
$$\begin{align} \mathbb{F}_q[x]/(h(x)) &\cong \prod_{i=0}^{n-1} \mathbb{F}_q = \mathbb{F}_q^n \\ f(x) + (h(x)) &\mapsto \left( f(\alpha_0), \ldots, f(\alpha_{n-1}) \right). \end{align}$$
This is your isomorphism $\varphi$.