I've to solve the following problem.
$R=\mathbb Q [x]$ and for $S \subset \mathbb C$ define $I(S):=\{f \in \mathbb Q [x]: f(x)=0 \forall x \in S \}$.
I have already shown, that this is an ideal in $\mathbb Q [x]$.
The question is now:
For which $S_i$ is $I(S_i)$ a prime ideal ?
$S_1 := \{3,7\}$, $S_2 := \{7+3i,7-3i\}$, $S_3 := \{\sqrt[3]2,-\sqrt[3]2\}$,$ S_4 := \{\sqrt2,-\sqrt2\}$
For $S_1$: it is a prime ideal.
$I(S_1) \neq R$, because of the polynomial $f=x$ with $f(3)=3 \neq 0$ and $f(7)=7 \neq 0$.
Let $a,b \in R$ such that $ab \in I(S_1)$, so $(ab)(x)=0 \forall x \in S_1$. Now I know that $0=(ab)(x)=a(x)b(x) \forall x \in S_1$. This leads to $a(x)=0$ or $b(x)=0$ , because there are no zero divisors in $\mathbb Q[x]$. So $a \in I(S_1)$ or $b \in I(S_1)$, which is the definition of prime ideal.
So now I need the other $S_i$. I hope someone of you can help me.
Edit: This "proof" is wrong, because of the polynomials $a(x)=x-3$ and $b(x)=x-7$.
Hints: Argue that \begin{align*} I(S_1)&=((x-3)(x-7))\\[4pt] I(S_2)&=(x^2-14x+58)\\[4pt] I(S_3)&=((x^3-2)(x^3+2))\\[4pt] I(S_4)&=(x^2-2)\\[4pt] \end{align*} and recall that an ideal $(f)$ of $\mathbb{Q}[x]$ is a prime ideal if and only if $\deg(f) \ge 1$ and $f$ is irreducible in $\mathbb{Q}[x]$.