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I've been looking for the definition of projective ideal but haven't found anything, all I've seen is the definition of projective module (but I don't know how these are related, if they are ¿?). Does anyone know a book where I can look up basic facts about projective ideals?

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2 Answers 2

I don't know of any other meaning of a projective ideal other than the one suggested by Boris Novikov, i.e. an ideal of a ring $R$ that is also projective as an $R$-module. I want to emphasize that such an ideal $I$ need NOT be a direct summand of $R$ (Boris never implied that condition to be necessary - only sufficient!) as well as give more examples.

The obvious examples of that are principal ideals of a commutative domain - they are projective by virtue of being free of rank one.

Another set of projective ideals that comes to mind are the ideals of rings of integers of a number field. If $K$ is a finite extension of $\mathbb{Q}$, and $R={\cal O}_K$ its ring of integers, then every ideal of $R$ is projective. This is because the ring is a Dedekind domain. When the ideal is not principal, it is not isomorphic to $R$ as a module. For example, if $R=\mathbb{Z}[\sqrt{-5}]$ is the ring of integers of the field $\mathbb{Q}[\sqrt{-5}]$, then the prime ideal $P$ generated by the elements $2$ and $1+\sqrt{-5}$ is a direct summand of $R^2$. It is easy to verify that for all $x\in P$ we have $x(1-\sqrt{-5})/2\in R$. Therefore we have a well-defined monomorphism $s: P\to R^2, x\mapsto (-x,x(1-\sqrt{-5})/2)$. It is easy to check that this splits the obvious epimorphism $p:R^2\to P, (r_1,r_2)\mapsto 2r_1+(1+\sqrt{-5}) r_2$ by another banal calculation.

Yet another set of examples consists of the Wedderburn components of a complex group algebra $R=\mathbb{C}[G]$ of a finite group $G$. These are the isotypic components of the (left) regular module, and serve as examples of Boris' more general scheme.

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,you said "I want to emphasize that such an ideal I need to be a direct summand of R" so since we have $IJ\subseteq I\cap J$ for every ideal J hence we dont have $R=I\oplus J$ because $I\cap J\neq\emptyset$.I assume that R is commutative. –  R Salimi Nov 15 '13 at 8:07
@RSalimi: It's a typo. Supposed to read ...need NOT be a summand of.... (but may be summands of $R^n$ for some $n$). Fixed. –  Jyrki Lahtonen Nov 15 '13 at 9:52

A (left) projective ideal is just a projective submodule of the regular module ${}_RR$ (i.e. the ring $R$ considered as a left $R$-module). So if ${}_RR$ can be decomposed in a direct sum, then every summand is a projective ideal.

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