I'm giving a seminar talk on the ideal class group, and am looking for an example of a fractional ideal that is not invertible. Does know of a simple example in say $k[x,y]$ or $\mathbb{Z}[\sqrt{-3}]$. Such examples must exist but computation there is very hard.



2 Answers 2


A fractional ideal is invertible iff it is a projective module. To find an example of a non-invertible one it is therefore enough to exhibit an example of an ideal which is not projective.

A simple example is the ideal $(x,y)$ in $R=k[x,y]$.

Let us check this ideal is not projective «by hand» showing that the natural map $\phi:(a,b)\in R\oplus R\mapsto ax+by\in (x,y)$, which is surjective, does not admit a section $\psi:(x,y)\to R\oplus R$. We suppose otherwise, that is, that there is such a $\psi$, and we let $\psi(x)=(f,g)$ and $\psi(y)=(r,s)$. We have $0=\psi(xy-yx)=x\psi(y)-y\psi(x)$, so that $yf=xr$ and $yg=xs$. It follows that $r$ and $s$ are multiples of $y$, and $f$ and $g$ are multiples of $x$. Now, as $\phi\circ\psi$ is the identity map of $(x,y)$, we have $x=\phi(\psi(x))=\phi(f,g)=xf+gy$, but now we know that $xf+gy$ is in the ideal $(x^2,y)$, which does not contain $x$: this is impossible.

  • $\begingroup$ Mariano, is there a super easy way $(x,y)$ isn't projective? I know it's not even flat, by the classic computation of $\text{Tor}$. Even more ridiculous, is that it would have to be free by Quillen-Suslin, which is ridiculous since it's not principal. But, is there quick and easy way to see this directly? Thanks! :) $\endgroup$ Oct 22, 2013 at 6:21
  • $\begingroup$ This was the calculation with Tor I was referring to. I'm wondering if there is a way to see the non projectivity without having to restort to a Tor calculation. Thanks! $\endgroup$ Oct 22, 2013 at 6:25
  • $\begingroup$ One should be able to prove directly that the obvious map $k[x,y]^2\to I$ is not split. $\endgroup$ Oct 22, 2013 at 6:28
  • $\begingroup$ @AlexYoucis, I added a simle argument. $\endgroup$ Oct 22, 2013 at 8:32
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    $\begingroup$ Invoking Quillen-Suslin for this is absurd... $\endgroup$ Oct 23, 2013 at 5:21

Hint: Consider the ring $\mathbb{Z}[\sqrt{5}]$. Since this is a non-maximal order in the Dedekind domain $\mathbb{Z}[\varphi]$, you know it must be not integrally closed, and so must have a non-invertible fractional ideal.

To actually find one, a good place to start might be at $I=(2)$ (why?). Indeed, let $\mathfrak{P}=(2,1-\sqrt{5})$. Show that $\mathfrak{P}$ is maximal (just compute the quotient ring). Then show that $\mathfrak{P}\ne I$, but


  • $\begingroup$ What about the fractional ideal $J=2^{-1}R$ is that not a fractional ideal such that $(2)J=R$? $\endgroup$ Oct 23, 2013 at 5:11
  • $\begingroup$ Yes, the non-invertible ideal is $\mathfrak{P}$... $\endgroup$ Oct 23, 2013 at 7:51
  • $\begingroup$ Oh ok, sorry for the confusion. $\endgroup$ Oct 23, 2013 at 17:01

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