0

Let $R=\mathbb Z[\sqrt{-5}]$ and $I=(3,2+\sqrt {-5})$ be the ideal generated by $3$ and $2+\sqrt{-5}$. I'm trying to prove that $I$ is a projective $R$-module.

I'm using the lifting property which is the most used definition of projective modules. I'm having troubles to prove that $I$ is projective using this definition, maybe it's easy and I'm forgetting something or there is another definition which could help more?

I need help.

Thanks

Community
  • 1
user42912
  • 24,130
  • This question was asked and solved many times on M.SE: see http://math.stackexchange.com/questions/144177/why-is-this-ideal-projective-but-not-free and http://math.stackexchange.com/questions/321255/what-is-a-projective-ideal. – user26857 Mar 10 '14 at 16:33

1 Answers1

3

The general strategy here is to take the exact sequence $$ 0\to \ker \pi\to R^2\overset{\pi}\to I\to 0 $$ and find a splitting for $\pi$. Then we'd have $R^2\cong \ker\pi\oplus I$, so $I$ is projective as a direct summand of a free module. You have to implicitly use fractional ideals of $R$, so if you know what that is do some Googling, but if you don't then just try to split the sequence.

Ian Coley
  • 6,060
  • could you help me to find the right inverse for $\pi$? I couldn't find one. – user42912 Mar 06 '14 at 22:19
  • 1
    Finding a right inverse amounts finding an expression $$a_1i_1+a_2i_2=1$$ such that $i_1,i_2\in I$, and $a_1,a_2\in I^{-1}$. Here $I^{-1}$ is the fractional ideal in the field of quotients of $R$ consisting of such numbers $a$ that $ai\in R$ for all $i\in I$. Then you can define $\pi$ by declaring $\pi(r_1,r_2)=r_1i_1+r_2i_2$, and the splitting homomorphism $s:I\to R^2$ by declaring $s(i)=(a_1i,a_2i)$. For examples of this technique follow the links given by user121097. – Jyrki Lahtonen Mar 07 '14 at 11:47
  • 1
    Addendum: IIRC with this particular $R$ we do get an expression as above. In general we may need to use more terms, like $\sum_{j=1}^n a_ji_j=1$, and then we can write $I$ as a summand of $R^n$. – Jyrki Lahtonen Mar 07 '14 at 11:57
  • Great comments, dear @Jyrki Lahtonen! – Georges Elencwajg Dec 17 '19 at 19:07