# Invertible ideal

• Jan 19th 2010, 09:08 AM
Carl
Invertible ideal
Hi there

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

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

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

Did I get that right?
• Jan 19th 2010, 10:30 PM
NonCommAlg
Quote:

Originally Posted by Carl
Hi there

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

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

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

Did I get that right?

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

if $I=(2,1+\sqrt{-5}),$ then $I^2=(2)$ and therefore $I$ is invertible. (why?)
• Jan 20th 2010, 12:53 PM
Carl
Quote:

Originally Posted by NonCommAlg
if $I=(2,1+\sqrt{-5}),$ then $I^2=(2)$ and therefore $I$ 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 $1+\sqrt{-5}$ are both irreducible elements in the ring?
• Jan 22nd 2010, 07:07 AM
NonCommAlg
Quote:

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 $1+\sqrt{-5}$ are both irreducible elements in the ring?

sorry for the late response. from $I^2=(2),$ you get $\frac{1}{2}II=R,$ which means that $I^{-1}=\frac{1}{2}I.$
• Jan 22nd 2010, 09:23 AM
Carl
Quote:

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!