Radical İdeal-Nullstellensatz $f=y{^2}+2xy-1$ and $g=x^2+1$.Prove $&lt;f,g&gt;$ is not radical ideal.
Hint :What is $f+g$ $?$

I have an idea: neither $x$ nor $y$ belong the ideal.but $(x\cdot y)^{2}\in I$.Since 
$x^{2}\in I$ ; thus $x\cdot y\in\sqrt{I}$.So $x+y\in\sqrt{I}$
To show that the ideal is not radical, you can exhibit some polynomial $p(x)$ so that $p(x)^n \in I$, while $p(x)$ is not. 

hint: Can you show that $(x+y)^2 \in I$?

As for your work: how did you show that $x^2 \in I$?. If $x^2 \in I$, why would this imply that $xy \in I$? If $xy \in I$, how do you know that $x+y \in I$? Where is the contradiction?
you say $p(x)^n$ is radical but $p(x)$ is not.I say $x^2$.I have a theorem : An ideal $I$ is $radical$ if $f^m\in I$ for some integer $m&gt;=1$ implies that $f\in I$