Radical of the ideal $(x^2,y^2)$

Question

Consider the ideal $(x^2,y^2)$ in $k[x,y]$ for algebraic closed field $k$ and the polynomials $f=x$, $g=y$. Then $f, g \in \sqrt{(x^2,y^2)}$ since $f^2, g^2 \in J$. But $f + g \notin \sqrt{(x^2,y^2)}$ since $(f+g)^p$ will never be of the form $r_1x^2+r_2y^2$ no matter what $p$ greater than 1 we choose. This to me is a contradiction to the fact that the radical of an ideal is an ideal…which leads me to my question: what am I misunderstanding?

Answer 1

No, it will be of the form $r_1x^2+r_2y^2$, if you follow the proof that $\sqrt{I}$ is an ideal. Indeed, already for $p=4$ we have $$ (x+y)^4=(x^2+4xy+6y^2)x^2+(4xy+y^2)y^2\in I=(x^2,y^2). $$ Actually, even $(x+y)^3=(x+3y)x^2+(3x+y)y^2\in I$.