$$xy\in\mathfrak q\:\Rightarrow\:\text{either $x\in\mathfrak q$ or $y^n\in\mathfrak q$ for some $n\gt0$}.$$

Primary ideals can be regard as the generalization of prime ideals and radical.

But why it's defined like that? It's not symmetry. Why not define like that:

$$xy\in\mathfrak q\:\Rightarrow\:\text{either $x^n\in\mathfrak q$ or $y^n\in\mathfrak q$ for some $n\gt0$}.$$

  • 1
    $\begingroup$ The second definition is not as strong as the first. $\endgroup$
    – rschwieb
    Mar 27, 2014 at 2:54
  • 3
    $\begingroup$ The definition of primary ideals is symmetric: see here. $\endgroup$
    – user26857
    Apr 1, 2014 at 10:23
  • $\begingroup$ The question in the title hasn't been answered yet. $\endgroup$ Oct 12, 2014 at 14:06
  • 2
    $\begingroup$ It seems a more accurate version of the title would be "Why is the definition of primary ideals the way it is?" $\endgroup$
    – rschwieb
    Sep 13, 2019 at 16:09

1 Answer 1


I've seen this question many times. The problem is that the second definition is strictly weaker than the first.

Consider $(x^2,xy)$ in the ring $F[x,y]$ where $F$ is a field. According to the normal definition, it is not primary since it doesn't contain any powers of $y$ and doesn't contain $x$.

However, it does satisfy the second definition. If $ab$ is in $(x^2, xy)$, then $x$ divides one of $a$ or $b$, and then that element's square is in this ideal.

The ordinary definition links zero divisors to nilpotent elements. In the quotient of a ring by a primary ideal, the elements are divided into regular elements and nilpotent elements. Said another way, zero divisors are nilpotent in such a ring.

Another way to think about primary ideals is that the condition is one half of primeness. An ideal is prime iff it is radical and primary. (But actually, you could replace primary with your definition and that would still hold!)


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