Let $\mathbb{C}[x,y]$ be a ring of complex polynomials in two variables.
Let $A = V(y^2-x, y^2-x^2) := \{(s,t) \in \mathbb{C}^2 |t^2 - s = 0 = t^2 - s^2\}$ (i.e $V(p(x,y))$ is the set of all zeros of $p(x,y)$)
Then $\mathscr{A} = \mathbb{I}(A)$ is an ideal of $\mathbb{C}[x,y].$ I would like to show that $\mathscr{A}$ is prime, where $$\mathbb{I}(S) = \{q(x,y) \in \mathbb{C}[x,y] | q(s,t) = 0 \ \mbox{for all} \ (s,t) \in S\}$$,$S \subset \mathbb{C}^2.$ (i.e. $\mathbb{I}(S)$ ia a set of all polynomails vanishing at all points in $S$)
Precisely, I think $$A = \{(0,0), (1,1), (-1,1)\}.$$ So $\mathscr{A} = \mathbb{I}(A)$ is the set of all polynomails in $\mathbb{C}[x,y]$ such that it vanishes at $(0,0), (1,1), (-1,1).$
--------------------------- $\textbf{Update}$ ----------------------------
Thank you very much for helps from various people.
So the problem does not go as I first expected (since $\mathscr{A}$ is not prime), and now the problem change to a different one.
Precisely, I need to decompose $A$ into smaller pieces, let say $B, C$ and $A = B \cup C$, so that $\mathscr{B} = \mathbb{I}(B)$ and $\mathscr{C} = \mathbb{I}(C)$ are prime ideals in $\mathbb{C}[x,y].$
So $A = V(y^2-x, (y-x)(y+x)) = V(\{y^2-x\} \cup \{(y-x)(y+x)\}).$
I know that $V(f \cup g) = V(f) \cap V(g)$. So $A = V(y^2-x) \cap V((y-x)(y+x)).$
Also, $V(fg) = V(f) \cup V(g).$ So $V((y-x)(y+x)) = V(y-x) \cup V(y+x)$.
Therefore, $$A = (V(y^2-x) \cap V(y-x)) \cup (V(y^2-x) \cap V(y+x)).$$ I think I should set $B = V(y^2-x) \cap V(y-x)$ and $C = V(y^2-x) \cap V(y+x).$
I think $B = \{(0,0), (1,1)\}$ and $C = \{(0,0), (1,-1)\}$ (if my calculation is correct). So what left is to find $\mathscr{B} = \mathbb{I}(B), \mathscr{C} = \mathbb{I}(C)$ and show that they are prime ideals of $\mathbb{C}[x,y]$.
Any suggestion on this ?