Radical ideal of $(x,y^2)$ 
How does one show that the radical of $(x,y^2)$ is $(x,y)$ over $\mathbb{Q}[x,y]$?

I have no idea how to do, please help me.

Recall that the radical of an ideal $\mathfrak{a}$ is equal to the intersection of the primes containing $\mathfrak{a}$. Here, let $\mathfrak{p}$ be a prime containing $(x, y^2)$. Then $y^2 \in \mathfrak{p}$ implies $y \in \mathfrak{p}$. Can you finish it off from there?
