Let $I\subseteq K[x_1,...,x_n]$ be an ideal. Let $\sqrt{I}$ be the radical of the ideal.
Assuming that $I$ is radical, i.e $I=\sqrt{I}$, show that $I$ is not prime only when there exist ideals $G,H$ s.t $G\neq I$ and $H\neq I$, where $G\cap H=I$.
I've had a lot of difficulty with this question, trying to figure out the best way to prove this claim. I've proved the converse of the claim as follows:
$I$ is radical, then $I$ is prime, only when there doesn't exist ideals $G\neq I$ and $H\neq I$ , where $G\cap H=I$.
Assume that the two ideals $G$ and $H$ exist
Let $a\in G$\ $H$ and $b\in H$\ $G$, such elements exist since $H$\ $G\neq\emptyset$ and $G$\ $H\neq\emptyset$. Then $ab\in G\cap H=I$, but $a,b\notin G\cap H=I$, so $I$ is not prime. However I din't use the fact that I is radical in my proof, is it still valid
First of all, a prime ideal is radical: if $P$ is prime and $a\in\sqrt{P}$, then $a^n\in P$ for some $n>0$, which means $a\in P$, by primality.
Now, suppose $I$ is radical.
Let's show that if there exist $G$ and $H$ properly containing $I$ such that $G\cap H=I$, then $I$ is not prime. Since $G\cap H\supseteq GH$, we have $GH\subseteq I$. If $I$ is prime, then either $G\subseteq I$ or $H\subseteq I$: contradiction.
Let $I$ be non prime. Then, since $K[x_1,\dots,x_n]$ is noetherian, we can write $I=Q_1\cap Q_2\cap\dots\cap Q_n$, where $Q_i$ are primary ideals and this decomposition is not redundant, that is, leaving off one of the ideals, the intersection is not $I$. (Lasker-Noether theorem.)
Now $n>1$. Indeed, if $I=Q_1$ is primary, its radical is a prime ideal. But $I$ is radical and non prime.
Therefore we have $I=Q_1\cap(Q_2\cap\dots\cap Q_n)$ as requested.