Define $\mathbb{Q}[r] = \{q_0+q_1r+ ... +q_{n-1}r^{n-1}|\;q_i \in \mathbb{Q}\}$, where $r\in \mathbb{C}\backslash \mathbb{Q}$ is a root for $x^n + a_{n-1}x^{n-1}+...+a_0=0, a_i\in \mathbb{Q}$.
Define $T: \mathbb{Q}^n \rightarrow \mathbb{Q}[r]$, $T((q_0,q_1,...,q_n-1))=q_0+q_1r+ ... +q_{n-1}r^{n-1}$, is there way to show that $T$ is injective?
This isn't true in general. For example, $r=\sqrt{2}$ is a root of $(x^2-2)(x-1)=x^3-x^2-2x+2$, but $\mathbb Q[\sqrt{2}]$ has degree $2$ over $\mathbb Q$, not $3$.
You need that $r$ is the root of a irreducible polynomial of degree $p(x)$ of degree $n$ to show that $T$ is injective.
If $T$ is not injective, then $\sum_{i=0}^{n-1} q_ir^i=0$ for some $q_i\in\mathbb Q$. This is a polynomial of smaller degree than $p(x)$. Since $p(x)$ is irreducible, it is prime (why?) so relatively prime to $q(x)=\sum q_ix^i$, a polynomial of smaller degree. But then $p(x)m(x)+q(x)n(x)=1$ has a solution for $m(x),n(x)\in\mathbb Q[x]$. Evaluating at $x=r$, we'd have $0=1$, which is not allowed.