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An ideal $P\subset R$ is strongly prime, if for any $x$ and $y$ in the quotient field of $R$, $xy\in P$ implies $x\in P$ or $y\in P$.

What is the difference between strongly prime ideal of $R$ and a prime ideal of $R$?

It seems that they are same .Thank you for your helping ..

Davide Giraudo
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izaag
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  • What’s a strongly prime ideal? I googled and found this paper where the definition is: An ideal $P ⊂ R$ is strongly prime, if for any $x$ and $y$ in the quotient field of $R$, $xy ∈ P$ implies $x ∈ P$ or $y ∈ P$. (So these are only defined for integral domains, I assume.) Is that your definition, too? – k.stm Mar 09 '14 at 09:25
  • yes, this is definition of strongly prime ideal . – izaag Mar 09 '14 at 10:00
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    they are not the same i think: regard $(2)$ as an ideal of ${Z}$ and $\frac{2}{3}$ and $3$, for an example of prime but not strongly prime – jorst Mar 09 '14 at 10:19
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    It seems to me $\mathbb Z$ has no strongly prime ideals. – alex Mar 09 '14 at 10:36
  • @user104847 Looks like R being an integral domain was implicitly implied? – rschwieb Mar 09 '14 at 17:11

1 Answers1

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As should be abundantly clear from the comments yesterday, they are far from similar. Jorst pointed out that $\frac23\cdot 3\in (2)\subseteq\Bbb Z$, and yet neither one of $\frac23$ or $3$ lies in $(2)$. I'll answer talking about a domain $R$ with field of quotients $Q$.

As for your question "what is the difference?" the first thing to point out is that the definitions quantify over different sets. "For all $x,y\in R$" is a wholly different statement from "For all $x,y\in Q$".

By adapting the argument above, you can show that no nonzero ideal of $\Bbb Z$ is strongly prime. For any $(p)$ with $p$ a nonzero prime, you can just take a prime $q$ other than $p$ and note $q\cdot \frac{p}{q}\in (p)$, but $\frac{p}{q}, q\notin (p)$. The zero ideal, however, is always strongly prime.

What can be said? Of course, any strongly prime ideal not equal to $R$ is also a(n ordinary) prime ideal. I'm not sure how many sources include/exclude $R$ from this definition you gave.

Secondly, if $R$ is allowed into the definition above, then if $R$ is strongly prime, then the ideals of $R$ are linearly ordered. (This happens because $xx^{-1}=1\in R$ implies $x$ or $x^{-1}$ is in $R$, and this is a way to define a valuation domain.)

cansomeonehelpmeout
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rschwieb
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