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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.

Martin Sleziak
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Alexander Sibelius
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    Is $6\mathbb Z$ not a radical ideal? – Thomas Andrews Dec 07 '12 at 04:26
  • Oh! Well, that's certainly a better example then the one I had. Thanks. – Alexander Sibelius Dec 07 '12 at 04:33
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    Generally, any squarefree integer will generate one. – rondo9 Dec 07 '12 at 04:33
  • If $R$ is a PIR, then any non-zero prime is maximal, so that if $P \neq Q$ are distinct primes, there are comaximal and thus $PQ = P \cap Q$, which is radical, since $\sqrt{P \cap Q} = \sqrt P \cap \sqrt Q$, but not prime for it would be maximal, and we have $P \cap Q \subsetneq P$. – Watson Jan 25 '17 at 19:02

4 Answers4

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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.

user26857
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Arvin
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$x(x-1)$ 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.

user26857
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Bombyx mori
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  • Great. Very clear. Thank you. – Alexander Sibelius Dec 07 '12 at 04:35
  • 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 03 '14 at 23:59
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    @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 09:10
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    $(4) = (2)(2)$ 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 09:10
  • @KonstantinArdakov: I did not know you are at here! – Bombyx mori Dec 29 '14 at 09:39
  • @user26857, Konstantin Ardakov: Thanks a lot for the examples. – Prism Dec 30 '14 at 19:15
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Not sure how much algebraic geometry you have at your disposal, but since your attempt used polynomials maybe this will work well for you. It's certainly not as simple as the $6\mathbb{Z}$ answer, but how about an intentional algebro-geometric construction? Let's take $\mathbb{C}[x,y]$ which suffices to be an integral domain, and will help us out for being algebraically closed. We will use the following results:

  1. $\textbf{I}(V)$ is a radical ideal.
  2. $\textbf{I}(V)$ is prime if and only if $V$ is irreducible.

So the game plan now is to construct a reducible variety $V$, then take $\textbf{I}(V)$, which cannot be prime but must be radical.

The variety $V_1 = \textbf{V}(y-x^2)$ is the variety who's real part corresponds to a classic upward opening parabola, and $V_2 = \textbf{V}(y+x^2)$ is the variety who's real part is a downward opening parabola. It is easy enough to form the variety $V_3 = V_1 \cup V_2$ by taking $V_3 = \textbf{V}((y-x^2)(y+x^2))$. Thus $\textbf{I}(V_3)$ is radical, but cannot be prime.

Prince M
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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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