Proof:
Pick a $n \in \Bbb{N}$ and $a_n \in \Bbb{Q}$. Then
$$p(x)=\sum_{i=0}^n a_nx^n$$

Now by assumption $\forall k \in \Bbb{A}$

$$p(k)=\sum_{i=0}^n a_n k^n=0$$

The zero polynomial can be written as $0(x)=\sum_{i=0}^n 0 x^n$. Therefore

$$p(k)=\sum_{i=0}^n a_n k^n=\sum_{i=0}^n 0 x^n=\sum_{i=0}^n 0 k^n$$

Since $x^i$ and $x^j$ are linearly independent for $i \neq j$, we can compare coefficients and get

$a_n=0$

Now since $n$ is arbitrary, the above applies to all $n\ in \Bbb{N}$ hence $\Bbb{C}[T]$. Therefore $p=0$ as required