# Eisenstein Integers, need help with a proof

• Mar 5th 2011, 08:43 AM
Aprshaw
Eisenstein Integers, need help with a proof
Hey all, I am having a problem with an exercise from the book "A Classical Introduction to Modern Number Theory". It involves the ring $\mathbb{Z}[\omega]$, where
$
\omega = \frac{-1 + \sqrt{-3}}{2}
$

The question is
For any $\alpha \in \mathbb{Z}[\omega]$, show that $\alpha$ is congruent to 1, -1 or 0 mod 1 - $\omega$

I figured the best thing to use was Euclidean division, so I set
$\alpha = a + b\omega$
and worked out that
$\frac{\alpha}{1 - \omega} = \frac{2a - b + (a + b)\omega}{3}$

I can see how it would work, and have tested it a few times on random Eisenstein integers, but cannot work out how to actually prove it!

Any help would be much appreciated
• Mar 5th 2011, 10:04 AM
tonio
Quote:

Originally Posted by Aprshaw
Hey all, I am having a problem with an exercise from the book "A Classical Introduction to Modern Number Theory". It involves the ring $\mathbb{Z}[\omega]$, where
$
\omega = \frac{-1 + \sqrt{-3}}{2}
$

The question is
For any $\alpha \in \mathbb{Z}[\omega]$, show that $\alpha$ is congruent to 1, -1 or 0 mod 1 - $\omega$

I figured the best thing to use was Euclidean division, so I set
$\alpha = a + b\omega$
and worked out that
$\frac{\alpha}{1 - \omega} = \frac{2a - b + (a + b)\omega}{3}$

I can see how it would work, and have tested it a few times on random Eisenstein integers, but cannot work out how to actually prove it!

Any help would be much appreciated

What does it mean "to be congruent to" in a general ring??

Tonio
• Mar 5th 2011, 10:50 AM
Aprshaw
I believe
$\alpha$ is congruent to $\beta \pmod{1 - \omega}$
is the same as saying
$\alpha = (1 - \omega) \cdot \kappa + \beta$
Which is Euclidean division in the ring $\mathbb{Z}[\omega]$

However, I am not completely certain of this, but this is the assumption I have made.
• Mar 5th 2011, 01:19 PM
tonio
Quote:

Originally Posted by Aprshaw
I believe
$\alpha$ is congruent to $\beta \pmod{1 - \omega}$
is the same as saying
$\alpha = (1 - \omega) \cdot \kappa + \beta$
Which is Euclidean division in the ring $\mathbb{Z}[\omega]$

However, I am not completely certain of this, but this is the assumption I have made.

Of course. I just missed the "mod" thingy in your original post...

1) Check first that $a+b\omega=0\!\!\pmod {(1-\omega )}\Longleftrightarrow a+b=0\!\!\mod 3$

2) Now deduce from the above your problem's solution...or

Directly, show that $a+b=i\!\!\pmod {(1-\omega )}\Longleftrightarrow a+b=i\!\!\pmod 3\,,\,\,for\,\,i=-1,0,1$

Tonio