1. ## Invertible ideal

Hi there

I have the ring $\displaystyle \mathbb{Z}[\sqrt{-5}]$ with field of fractions $\displaystyle K$ and am to decide whether any of the ideals $\displaystyle (2), (1+\sqrt{-5}), (2, 1+\sqrt{-5})$ are invertible or not.

Well, if the ideal $\displaystyle I$ is invertible, we have that $\displaystyle II^*=R$ where $\displaystyle I^*=\{q\in K | qI \subset R\}$
so $\displaystyle q$ is on the form $\displaystyle \frac{a+b\sqrt{-5}}{c+d\sqrt{-5}}$, correct?

Then I would say that the ideal $\displaystyle (2, 1+\sqrt{-5})$ is invertible, because whatever the denominator is in the above fraction, it disappaers when multiplying with $\displaystyle (2, 1+\sqrt{-5})$.

Did I get that right?

2. Originally Posted by Carl
Hi there

I have the ring $\displaystyle \mathbb{Z}[\sqrt{-5}]$ with field of fractions $\displaystyle K$ and am to decide whether any of the ideals $\displaystyle (2), (1+\sqrt{-5}), (2, 1+\sqrt{-5})$ are invertible or not.

Well, if the ideal $\displaystyle I$ is invertible, we have that $\displaystyle II^*=R$ where $\displaystyle I^*=\{q\in K | qI \subset R\}$
so $\displaystyle q$ is on the form $\displaystyle \frac{a+b\sqrt{-5}}{c+d\sqrt{-5}}$, correct?

Then I would say that the ideal $\displaystyle (2, 1+\sqrt{-5})$ is invertible, because whatever the denominator is in the above fraction, it disappaers when multiplying with $\displaystyle (2, 1+\sqrt{-5})$.

Did I get that right?
if $\displaystyle I=(a), \ a \neq 0,$ then $\displaystyle I^{-1}=Ra^{-1}.$ so, in general, every non-zero principal idea of an integral domain is invertible.

if $\displaystyle I=(2,1+\sqrt{-5}),$ then $\displaystyle I^2=(2)$ and therefore $\displaystyle I$ is invertible. (why?)

3. Originally Posted by NonCommAlg
if $\displaystyle I=(2,1+\sqrt{-5}),$ then $\displaystyle I^2=(2)$ and therefore $\displaystyle I$ is invertible. (why?)
Well, I find it hard to see why this property will make the ideal invertible. I would say that it has something to do with the fact that 2 and $\displaystyle 1+\sqrt{-5}$ are both irreducible elements in the ring?

4. Originally Posted by Carl
Well, I find it hard to see why this property will make the ideal invertible. I would say that it has something to do with the fact that 2 and $\displaystyle 1+\sqrt{-5}$ are both irreducible elements in the ring?
sorry for the late response. from $\displaystyle I^2=(2),$ you get $\displaystyle \frac{1}{2}II=R,$ which means that $\displaystyle I^{-1}=\frac{1}{2}I.$

5. Originally Posted by NonCommAlg
sorry for the late response.[/tex]
No problem, the exercise is first for next week.
Ok, I got it now, thanks!