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.

  • 10
    $\begingroup$ Is $6\mathbb Z$ not a radical ideal? $\endgroup$ Dec 7, 2012 at 4:26
  • $\begingroup$ Oh! Well, that's certainly a better example then the one I had. Thanks. $\endgroup$ Dec 7, 2012 at 4:33
  • 3
    $\begingroup$ Generally, any squarefree integer will generate one. $\endgroup$
    – rondo9
    Dec 7, 2012 at 4:33
  • $\begingroup$ 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 $P \cap Q \subsetneq P$. $\endgroup$
    – Watson
    Jan 25, 2017 at 19:02

4 Answers 4


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.

  • 1
    $\begingroup$ remark: and 6Z= (2) intersect (3) which are prime in 6Z $\endgroup$
    – Endre Moen
    Oct 28, 2020 at 20:05

$x(x-1)$ 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.

  • $\begingroup$ Great. Very clear. Thank you. $\endgroup$ Dec 7, 2012 at 4:35
  • $\begingroup$ 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) $\endgroup$
    – Prism
    Dec 3, 2014 at 23:59
  • 1
    $\begingroup$ @Prism In general, the product of maximal ideals is radical. For prime ideals consider $(X)(X,Y)$ whose radical is ...? $\endgroup$
    – user26857
    Dec 29, 2014 at 9:10
  • 4
    $\begingroup$ $(4) = (2)(2)$ 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. $\endgroup$ Dec 29, 2014 at 9:10
  • $\begingroup$ @KonstantinArdakov: I did not know you are at here! $\endgroup$ Dec 29, 2014 at 9:39

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.


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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