I have stumbled across the following exercise on radicals of ideals of rings. I shall show that: $\operatorname{rad}(x+y^2,x^2+2xy^2)$ is a maximal ideal of $\mathbb{C}[x,y]$, but $(x+y^2,x^2+2xy^2)$ is not. How can I show this?

Also, is the radical of a prime ideal equal to the ideal itself? so does $\operatorname{rad}(p)=p$ hold?

And a last question, I shall find a ring $R$ with exactly 17 ideals. My first idea was to take something similar to $\mathbb{Z}/2^{17}\mathbb{Z}$, where the ideals are $(0), (\mathbb{Z}/2^i\mathbb{Z})$ for any $i\le17$. But are those then the only ideals? Is not the union of one ideal with the zero ideal a new ideal? Then it would be difficult to find a Ring with the required number of ideals, wouldn't it?

Thanks in advance for any help!

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The last question has little or nothing to do with the rest; it should really be posted separately. – Arturo Magidin Nov 15 '11 at 18:14
The union of one ideal with the zero ideal is an ideal all right, but not exactly a new one. – Henning Makholm Nov 15 '11 at 18:19
The ideals if $\mathbb{Z}/2^{17}\mathbb{Z}$ are precisely the ideals of $\mathbb{Z}$ that contain $2^{17}\mathbb{Z}$, which are precisely the ideals generated by divisors of $2^{17}$. Unfortunately, there's 18 of them, not 17. Fortunately, it should be easy to fix. – Arturo Magidin Nov 15 '11 at 18:24

Well, the radical of an ideal is the intersection of all the prime ideals containing it (and proving that is a fairly canonical exercise). Thus, if $P\subseteq R$ is a prime ideal, then $rad(P)=P$.

So, let $I=(x+y^2,x^2+2xy^2)\subseteq \mathbb{C}[x,y]=R$. Note that

$x^2+2xy^2-x(x+y^2)=xy^2\in I$, thus $2xy^2\in I$, and hence $x^2\in I$. Therefore, $x\in rad(I)$. Also, $x+y^2-x=y^2\in rad(I)$, implying that $y\in rad(I)$. Therefore, $(x,y)\subseteq rad(I)\subsetneq R$, and since $(x,y)$ is clearly maximal in $R$, $rad(I)=(x,y)$.

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