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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
up vote 2 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
Nice (+1). The same argument seems to show that product of prime ideals is radical? (Well at least if those prime ideals are principal ideals) – Prism Dec 3 '14 at 23:59
@Prism In general, the product of maximal ideals is radical. For prime ideals consider $(X)(X,Y)$ whose radical is ...? – user26857 Dec 29 '14 at 9:10
$(4) = (2)(2)$ is a product of prime ideals in $\mathbb{Z}$ which is not radical. $(x^2, xy) = (x)(x,y)$ is a product of prime ideals in $\mathbb{C}[x,y]$ which is not radical. A product of distinct prime principal ideals in a commutative domain is indeed radical. – Konstantin Ardakov Dec 29 '14 at 9:10
@KonstantinArdakov: I did not know you are at here! – Bombyx mori Dec 29 '14 at 9:39

Why don't you use a simple example ... Suppose the ring to be $R=\mathbf{Z}$ and take the ideal $6\mathbf{Z}$. Since the $\sqrt{m\mathbf{Z}}=r\mathbf{Z}$ where $r=\Pi_{p|m} p$ and $p$ is a prime number. Here $\sqrt{6\mathbf{Z}}=6\mathbf{Z}$ but $6\mathbf{Z}$ is not a prime ideal since 6 is not prime.

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A source of radical but not prime:
Let $K$ be a field, and let $S = K[x_1, . . . , x_n]$ be the polynomial ring in $n$ variables over $K$. The monomial prime ideals are all of the form $(x_{i_1}, . . . , x_{i_k} )$. But A monomial ideal $I$ is a radical ideal, iff $I$ is a squarefree monomial ideal (This is because: $\{\sqrt u : u ∈ G(I)\}$ is a set of generators of $\sqrt I$. [denote the unique minimal set of monomial generators of the monomial ideal I by G(I).] ) .
So for example: $I=\langle{x_1 x_3}\rangle$ is radical but not prime.
See also here.

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