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

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
    Please elaborate on your question. – user122283 Mar 26 '14 at 23:47
  • 10
    What is there to elaborate upon? The question is precise in what it's asking, and it seems like a very reasonable (albeit perhaps naïve) question to ask when confronted with the definition at first. – ah11950 Mar 26 '14 at 23:55
  • 1
    The second definition is not as strong as the first. – rschwieb Mar 27 '14 at 2:54
  • The definition of primary ideals is symmetric: see here. – user26857 Apr 1 '14 at 10:23
  • The question in the title hasn't been answered yet. – Martin Brandenburg Oct 12 '14 at 14:06
up vote 10 down vote accepted

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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