# Why is this ideal projective but not free?

Let $R=\mathbb{Z}[\sqrt{-5}]$ and $I=(2,1+\sqrt{-5})$. How can I prove that $I$ is projective but not free?

-
If it were free, it would have to be principal, right? What is its norm? – Hurkyl May 12 '12 at 13:16
Compute the fractional ideal that is its inverse. Using the ideal product then allows you to write it as a split summand of a free module of rank two. This implies that it is projective. – Jyrki Lahtonen May 12 '12 at 17:51

Please see $\S 3.5.4$ -- "Projective verus free" -- in my commutative algebra notes. In particular, Proposition 27 and the exercise follow it step you through showing that the ideal $\langle 3, 1+ \sqrt{-5} \rangle$ is projective but not free.

The same techniques apply to $I = \langle 2, 1+ \sqrt{-5} \rangle$. (In fact, I view it as a happy accident that your question is similar but not identical to what is treated in my notes. You'll learn more by figuring out what slight modifications you need to make.)

As I say in the notes, much of the presentation owes a debt to this blurb of Keith Conrad.


Added: Here is the relevant portion from the notes.

Proposition: Let $I$ and $J$ be comaximal ideals in a domain $R$, and consider the $R$-module map $q: I \oplus J \ra R$ given by $(x,y) \mapsto x+y$. Then:
a) The map $q$ is surjective.
b) $\Ker(q) = \{(x,-x) \ | \ x \in I \cap J\}$, hence is isomorphic as an $R$-module to $I \cap J$.
c) We have an isomorphism of $R$-modules
$I \oplus J \cong IJ \oplus R$.
d) Thus if $IJ$ is a principal ideal, $I$ and $J$ are projective modules.

It is clear that for any ideals $I$ and $J$, the image of the map $q$ is the ideal $I+J$, and we are assuming $I+J = R$, whence part a).
Part b) is essentially immediate: details are left to the reader. Combining parts a) and b) we get a short exact sequence $0 \ra I \cap J \ra I \oplus J \ra R \ra 0$.
But $R$ is free, hence projective, and thus the sequence splits, giving part c). Finally, a nonzero principal ideal $(x)$ in a domain $R$ is isomorphic as an $R$-module to $R$ itself: indeed, multiplication by $x$ gives the isomorphism $R \ra (x)$. So if $IJ$ is principal, $I\oplus J \cong R^2$ and $I$ and $J$ are both direct summands of a free module.

In particular, if we can find in a domain $R$ two comaximal nonprincipal ideals $I$ and $J$ with $IJ$ principal, then $I$ and $J$ are finitely generated projective nonfree $R$-modules. The following exercise asks you to work through an explicit example.

Exercise 3.30: Let $R = \Z[\sqrt{-5}]$, and put

$\pp_1 = \langle 3, 1+ \sqrt{-5} \rangle, \ \pp_2 = \langle 3, 1 - \sqrt{-5} \rangle.$
a) Show that $R/\pp_1 \cong R/\pp_2 \cong \Z/3\Z$, so $\pp_1$ and $\pp_2$ are maximal ideals of $R$.
b) Show that $\pp_1 + \pp_2 = R$ (or equivalently, that $\pp_1 \neq \pp_2$).
c) Show that $\pp_1 \pp_2 = (3)$.
d) Show that neither $\pp_1$ nor $\pp_2$ is principal. (Suggestion: show that if $\pp_1 = (x+\sqrt{-5}y)$ then $\pp_2 = (x-\sqrt{-5}y)$ and thus there are integers $x,y$ such that $x^2+5y^2 = \pm 3$.)
e) Conclude that $\pp_1$ and $\pp_2$ are (in fact isomorphic) nonfree finitely generated projective modules over the domain $R$.
f) Show that $\pp^2$ is principal, and thus that the class of $\pp$ in $\widetilde{K_0(R)}$ is $2$-torsion.

This construction looks very specific, and the number-theoretically inclined reader is warmly invited to play around with other quadratic rings and more general rings of integers of number fields to try to figure out what is really going on. From our perspective, we will (much later on) gain a deeper understanding of this in terms of the concepts of invertible ideals, the Picard group and Dedekind domains.

-
@ThomasAndrews and Pete: Please don't use the comment section to engage in extended meta discussions. I am purging the comments in a couple minutes. If you wish I (or another mod) can copy all the comments to a Meta post and you can continue your discussion there. Please flag if you would like that done. – Willie Wong Mar 7 '14 at 9:11