Let $I_{N}=\{f\in\Omega(\Bbb R,\Bbb R)|\forall x \in N\subset \Bbb R: f(x)=0\}$, where $\Omega(\Bbb R,\Bbb R)=\{f:\Bbb R \to \Bbb R|f\space is \space a\space function\}$, then the following statements are equivalent:
1) $I_{N}$ is a maximal Ideal
2) $I_{N}$ is a prime Ideal
3) $|N| = 1$
I could do 1) -> 2) by contradiction, but now I don't know how to continue, can somebody help me please?
For $2) \implies 3)$, assume that $I_N$ is prime, but $N$ has two distinct points $x, y$. Construct functions $f, g$ that don't belong to $I_N$, but their product does. (Hint: let $f$ be zero on $x$, but not on $y$).
For $3) \implies 1)$, note that if $A \subset B$, then $I_B \subset I_A$, and the latter inclusion is strict whenever the former is.