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I'm preparing for an exam and, as part of this preparation, I'm looking for an ideal $I$ in an integral domain $R$ that is radical but not prime. Here is an example I'm fooling around with:

Let $R=\mathbb{R}[x]$ and let $I=(x(x-1))$. I'm having trouble showing that this ideal is in fact radical. My intuition is to consider the quotient ring $\mathbb{R}[x]/(x(x-1))$ and determine if it is reduced, that is, whether or not it has trivial nilradical. However, this has only led me in circles so far. $(x(x-1))$ is clearly not a prime ideal, so it suffices to show it's radical.

Any help would be appreciated.

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Is $6\mathbb Z$ not a radical ideal? –  Thomas Andrews Dec 7 '12 at 4:26
Oh! Well, that's certainly a better example then the one I had. Thanks. –  Alexander Sibelius Dec 7 '12 at 4:33
Generally, any squarefree integer will generate one. –  rondo9 Dec 7 '12 at 4:33

1 Answer 1

up vote 1 down vote accepted

$x(x-1)$ is certainly not prime, suppose $f^{n}\in \langle x(x-1) \rangle$. Since $f^{n}\in \langle x \rangle$, and $\langle x \rangle$ is prime, for some $m<n,f^{m}\in \langle x \rangle$. Proceed this way we can prove $f$ must be divisible by $x$, and similarly for $x-1$. This gives $f$ is divisible by $x(x-1)$. So this ideal is its own radical.

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Great. Very clear. Thank you. –  Alexander Sibelius Dec 7 '12 at 4:35

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