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I am confused with the definition of a primary ideal. The definition states that if $R$ is a commutative ring then $I$ is called a primary ideal of $R$ is the following condition holds. If $xy\in I$ then either $x\in I$ or $y^n\in I$ for some $n$. My confusion lies in the following case. Suppose we have that $xy\in I$ but both $x$ and $y$ are not in $I$ and further $y^n \notin R\ \forall n$. Then by definition we have that $I$ is not primary. However if there exists $k$ such that $x^k\in R$ then the element $yx$ does not lead us to a contradiction. But since our ring is commutative $yx=xy$. So what I don't understand is the following this definition might show me that an ideal turns out to be primary depending on how I write the elements (i.e $xy$ or $yx$) which simply doesn't make sense. So what am I saying wrong here?

Thanks in advance.

user26857
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TheGeometer
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1 Answers1

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You can look at the definition from the opposite side:

$I$ is primary when, for all $x,y\in R$, if $xy\in I$ and no power of one of the elements belongs to $I$, then the other element belongs to $I$.

You get a symmetric definition also looking at the quotient ring. The ideal $I$ is primary if and only if every zero divisor in $R/I$ is nilpotent.

In particular, the nilradical of $R/I$ contains all zero divisors, so if you kill all nilpotent elements, you get a domain, which is to say that the radical ideal of $I$ is a prime ideal.

egreg
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