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On a commutative ring K, an ideal I is prime when: $$ I \text{ is a subgroup of }K,\text{ } I \neq K, \text{and if } x,y \in K, \text{the relation } xy \in I \text{ implies } x \in I \text{ or }y\in I. $$

But is this not always the case in a commutative ring K? Because by definition, if I is an ideal of K, we have $ax \in I$ for any $a \in K$ and $x \in I$.

Where my understanding went wrong?

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    What you wrote is **not** the definition of "primary ideal" but of **prime ideal** ...and anyway the answer is no. – 2017-02-21
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    sorry it was a translation problem. – 2017-02-21
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    Consider $\mathbb{Z}$ and the ideal $6\mathbb{Z}$. – 2017-02-21

2 Answers 2

1

Take $\;2\cdot 3\in6\Bbb Z\;$ . Yet $\;2,\,3\notin 6\Bbb Z\;$

I think the understanding problem here is: a prime ideal certainly fulfills the condition$\;xy\in I\implies (x\in I\;\;or\;\;y\in I\,$) , and this is not the same as the very definition of ideal, as the example above shows: observe that

$$a\in6\Bbb Z\implies a=6m,\,m\in\Bbb Z\implies\;\;\forall\,y\in\Bbb Z\;,\;\;ay=6my\in\Bbb Z$$

2

You wrote the definition of a prime ideal, and no, it is not the same as the later thing you wrote. Here is a side-by side comparison of the two:

  1. $\forall x,y\in K$, if $xy\in I$, then $x\in I\text{ or } y\in I$
  2. $\forall x\in I\, \forall y\in K$, we have $xy\in I$

The first one is a conclusion about the factors of a product, and the second one is a conclusion about a product.

In fact, it's a basic theorem that all proper ideals of a commutative ring are prime iff the ring is a field. See for example Kaplansky's Commutative rings chapter 1, exercise 1, page 7, or most other introductory commutative algebra texts.