Define thenormN()for an element . Show that any common divisor of and 2 in the ring is a unit.

I know that

N( and that

N(2)= 4

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

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- May 27th 2010, 07:25 AMlucyannepalmerNorms and common divisors
Define the

*norm*N*(**)*for an element . Show that any common divisor of and 2 in the ring is a unit.

I know that

N( and that

N(2)= 4

But how do you show that any common divisor is in the ring? - May 27th 2010, 07:35 AMSwlabr
- May 27th 2010, 07:50 AMlucyannepalmer
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, 07:58 AMSwlabr
- May 27th 2010, 08:25 AMTheArtofSymmetry
N(xy)=N(x)N(y) where .

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, 08:42 AMlucyannepalmer
So the only common divisor of and 2, in the ring is 1?

- May 27th 2010, 08:45 AMSwlabr
No. Any common divisor will have norm 1, and is a unit. This may be in your notes (the is a unit bit). Otherwise, it is good exercise to prove it.

You need to prove that if a|2 and a| then N(a)=1. To do this, you must notice that as and so if a|2 and a| then N(a)|4. There are only three numbers which divide 4. They are 1, 2 and 4.

If N(a)=1 then is a unit.

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

If N(a)=4 then you need to prove that cannot divide both ring elements. To do this, notice that there must exist such that and . Is this possible? - May 27th 2010, 08:47 AMTheArtofSymmetry
- May 27th 2010, 08:51 AMlucyannepalmer
Thank you both, that was very helpful.