# Norms and common divisors

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• May 27th 2010, 06:25 AM
lucyannepalmer
Norms and common divisors
Define the norm N($\displaystyle \alpha$) for an element $\displaystyle \alpha= a + b\sqrt{-5} in Z\sqrt{-5}$. Show that any common divisor of $\displaystyle 1 + \sqrt{-5}$ and 2 in the ring $\displaystyle Z\sqrt{-5}$ is a unit.

I know that
N($\displaystyle 1 + \sqrt{-5})= 6$ and that
N(2)= 4

But how do you show that any common divisor is in the ring?
• May 27th 2010, 06:35 AM
Swlabr
Quote:

Originally Posted by lucyannepalmer
Define the norm N($\displaystyle \alpha$) for an element $\displaystyle \alpha= a + b\sqrt{-5} in Z\sqrt{-5}$. Show that any common divisor of $\displaystyle 1 + \sqrt{-5}$ and 2 in the ring $\displaystyle Z\sqrt{-5}$ is a unit.

I know that
N($\displaystyle 1 + \sqrt{-5})= 6$ and that
N(2)= 4

But how do you show that any common divisor is in the ring?

You need to show that there exists no element with norm 2.

Simples!

(In general, this is made much easier when you use modulo arithmetic to eliminate big numbers.)
• May 27th 2010, 06:50 AM
lucyannepalmer
Thanks for your quick reply.

Why is it that you have to show there exists no element with norm 2? and how would I go about doing that?
• May 27th 2010, 06:58 AM
Swlabr
Quote:

Originally Posted by lucyannepalmer
Thanks for your quick reply.

Why is it that you have to show there exists no element with norm 2? and how would I go about doing that?

Well, how is a norm defined? What is the property that it has?

To prove that no element has norm 2, assume one exists. Assume there exists $\displaystyle \alpha, \beta$ such that $\displaystyle N(\alpha + \beta \sqrt{-5}) = 2$, and just follow it through...
• May 27th 2010, 07:25 AM
TheArtofSymmetry
Quote:

Originally Posted by lucyannepalmer
Define the norm N($\displaystyle \alpha$) for an element $\displaystyle \alpha= a + b\sqrt{-5} in Z\sqrt{-5}$. Show that any common divisor of $\displaystyle 1 + \sqrt{-5}$ and 2 in the ring $\displaystyle Z\sqrt{-5}$ is a unit.

I know that
N($\displaystyle 1 + \sqrt{-5})= 6$ and that
N(2)= 4

But how do you show that any common divisor is in the ring?

N(xy)=N(x)N(y) where $\displaystyle x, y \in \mathbb{Z}[\sqrt{-5}]$.

There are three things to consider:

1. N(x) >= 0
2. if N(x) = 1, then x is a unit.
3. As Swlabr said, there is no element x such that N(x)=2.

If the common divisor of your example, let's say k, has the norm N(k)=1, then k is the unit in your ring. Can you conclude now?
• May 27th 2010, 07:42 AM
lucyannepalmer
So the only common divisor of $\displaystyle 1 + \sqrt{-5}$ and 2, in the ring $\displaystyle Z\sqrt{-5}$ is 1?
• May 27th 2010, 07:45 AM
Swlabr
Quote:

Originally Posted by lucyannepalmer
So the only common divisor of $\displaystyle 1 + \sqrt{-5}$ and 2, in the ring $\displaystyle Z\sqrt{-5}$ is 1?

No. Any common divisor will have norm 1, and $\displaystyle N(a) = 1 \Leftrightarrow a$ is a unit. This may be in your notes (the $\displaystyle N(a) = 1 \Leftrightarrow a$ is a unit bit). Otherwise, it is good exercise to prove it.

You need to prove that if a|2 and a| $\displaystyle (1 + \sqrt{-5})$ then N(a)=1. To do this, you must notice that as $\displaystyle N(2) = N(1 + \sqrt{-5})=4$ and so if a|2 and a|$\displaystyle (1 + \sqrt{-5})$ then N(a)|4. There are only three numbers which divide 4. They are 1, 2 and 4.

If N(a)=1 then $\displaystyle a$ is a unit.

N(a)=2 cannot happen, you must prove this.

If N(a)=4 then you need to prove that $\displaystyle a$ cannot divide both ring elements. To do this, notice that there must exist $\displaystyle b_1, b_2$ such that $\displaystyle ab_1 = 1 + \sqrt{-5}$ and $\displaystyle ab_2 = 2$ $\displaystyle \Rightarrow N(ab_1) = N(ab_2) = 4 \Rightarrow N(b_1) = N(b_2) = 1 \Rightarrow 2 {b_{2}}^{-1}b_1 = 1 + \sqrt{-5}$. Is this possible?
• May 27th 2010, 07:47 AM
TheArtofSymmetry
Quote:

Originally Posted by lucyannepalmer
So the only common divisor of $\displaystyle 1 + \sqrt{-5}$ and 2, in the ring $\displaystyle Z\sqrt{-5}$ is 1?

+1 and -1
• May 27th 2010, 07:51 AM
lucyannepalmer
Thank you both, that was very helpful.