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Let $R$ be an integral domain with fraction field $K$, and let $I$ be an $R$-submodule of $K$. We say that $I$ is a fractional ideal of $R$ if $rI\subset R$ for some nonzero $r \in R$.

My question is: Is there any example of an $I$ which is not fractional? Please give as many examples as you can.

Your help (and hints) will be appreciated.

mdp
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2 Answers2

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Take $R=\mathbb Z$ so that $K=\mathbb Q$. Fix a finite set of primes $S$. Then $I_S=\{\frac{a}{b}\colon p\nmid b \,\,\forall p\in S\}$ is an $R$-module which is not a fractional ideal.

Ferra
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See this question: A question about the relationship between submodule and ideal. The goal of the introduction of fractional ideals is to extend the set of ideals of a ring by their "inverses". So every ideal is a fractional ideal but not inversely. As an example take $\theta=\sqrt{-5}$ and let the ideal $I_1=(3,1-\theta)$ and $I_2=(3,1+\theta)$ then the product of the two ideals $I_1I_2=(3)$, the principal ideal generated by $3$. Now we can say that $I2\over{3}$ is the inverse of the ideal $I_1$ but it is not an ideal since it contains elements that are not algebraic integers, but it is a fractional ideal because when multiplied with $3$ it becomes an ideal (whence the definition of fractional ideal).

Community
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Marc Bogaerts
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  • As an example, suppose we were talking about rational numbers, how would you answer the question: "Give me an example of a number that is not rational"? Or "Give me an example of a real number that is not complex"? Would you explain why this question is not motivated by the fact that adjectives like "rational" or "complex" are not properties of numbers but are likely to express extensions of the number system? What would be your answer to these questions? – Marc Bogaerts Oct 31 '14 at 23:54
  • Thanks @Nimda. Q contained in R so R-Q may be your answer. I know fractional ideals forms a group. My questions is :Is there any non fractional Ideal. –  Nov 02 '14 at 10:55
  • No, every ideal is fractional. But there are fractional ideals that are not ideals. – Marc Bogaerts Nov 02 '14 at 14:21
  • Yes it's possible. –  Nov 02 '14 at 14:48
  • Maybe the term "integral ideal" is the term you are looking for, as explained here: http://www.math.uiuc.edu/~r-ash/Ant/AntChapter3.pdf See section 3.2.4 where it is also stated that an "ordinary" ideal is fractional. – Marc Bogaerts Nov 02 '14 at 19:48
  • thanksl. all ideals of R are trivially fractional. I am following the same book whose link you have given. My question is :Is there any non fractional Ideal. means in the 3.2.4 of the link (http://www.math.uiuc.edu/~r-ash/Ant/AntChapter3.pdf) is there any I such that rI is not in R. –  Nov 03 '14 at 11:46
  • Indeed, $\Bbb{Q}$ is a $\Bbb{Z}$ module but there is no $r\in \Bbb{Z}$ such that $r\Bbb{Q}\subseteq \Bbb{Z}$. – Marc Bogaerts Nov 03 '14 at 12:28
  • Thanks @nimda. Q is field of fractions. so any I such that Z<I<Q. –  Nov 03 '14 at 15:55
  • This for example: http://math.stackexchange.com/questions/986277/is-this-ring-a-well-known-ring-and-if-so-how-is-it-called but also $\Bbb{Z}\times\Bbb{Z}$ with pointwise addition, multiplication and product with scalar. Or more generally $R\times R$. – Marc Bogaerts Nov 03 '14 at 16:41