I'm trying to go through the details of an example in J. Milne's algebraic geometry notes (p.42, 2.22).
He gives us the general fact that for two algebraic subsets $W,W'\subseteq K^n$, we have $I(W\cap W')=rad(I(W)+I(W'))$$\;^{(1)}$, which I believe I understand, but then he goes on to illustrate this with a concrete example.
In his example, I presume we are working in $\mathbb{C}[x,y]$. He sets $W=V(x^2-y)$ and $W'=(x^2+y)$, and then claims that $$I(W\cap W')=rad(x^2, y)=(x,y)\,.\;^{(2)}$$ This is what I would like to understand.
First, I have shown that $(x^2-y)$ and $(x^2+y)$ are prime ideals, and thus radical. This means that $I(W)=IV(x^2-y)=(x^2-y)$ and similarly $I(W')=IV(x^2+y)=(x^2+y)$. So if we believe (1), then we get $$I(W\cap W')=rad\bigl((x^2-y)+(x^2+y)\bigr)\,.$$
So, the first equality of (2) suggest that $(x^2-y)+(x^2+y)=(x^2, y)$. Is this because I can write $x^2$ and $y$ as linear combinations of the generators $x^2-y$ and $x^2+y$? Should this be "obvious"?
Also, since $(x,y)$ is a maximal ideal it is radical, so we have $(x,y)=rad(x,y)$, and thus the second equality of (2) can be written as $rad(x,y)=rad(x^2,y)$, and this equality I would also like to understand. I have convinced myself that $rad(x^2,y)\subseteq rad(x,y)$, but I'm having trouble with the other containment.
Thank you for your help!
It "should" be obvious that $(x^2-y)+(x^2+y)=(x^2,y)$ because the left hand side is equal to $(x^2-y,x^2+y)$ by definition, and two ideals whose generators are contained in each other must be equal. But if it's not obvious, then prove the following carefully as an exercise: if $I=(i_1,...,i_n)$ and $J=(j_1,...,j_m)$ are two ideals (of any commutative ring, or left ideals of an arbitrary ring, or (left) modules over any ring, or...) such that $i_1,...,i_n\in J$ and $j_1,...,j_m\in I$ then $I=J$. The radical of $(x^2,y)$ contains $(x,y)$ because a power of $x$ is in $(x^2,y)$, and since $(x,y)$ is maximal, the radical of $(x^2,y)$ can't be any bigger! Note that none of this needed us to work over $\mathbb{C}$: any field not of characteristic $2$ will do.