# norm function,irreducible help.

• May 14th 2009, 09:57 AM
benjyboy
norm function,irreducible help.
R=Zsqrt(-5) where Z=intergers.
Define a norm function N on R with values in non negative integers by

N(a+bsqrt(-5))=(a^2)+5b^2

Using N or otherwise show 2 in reals is irreducible.

By finding two distinct factorizations in R of the number 6, show 2 in the reals is not prime.

Im pretty desperate for a correct solution to this question i can revise from, its driving me mad having no notes on it, be very grateful thanks.
• May 14th 2009, 06:37 PM
Gamma
2*3
$\displaystyle (1-\sqrt{-5})(1+\sqrt{-5})$
• May 14th 2009, 11:18 PM
Swlabr
Quote:

Originally Posted by benjyboy
R=Zsqrt(-5) where Z=intergers.
N(a+bsqrt(-5))=(a^2)+5b^2

Using N or otherwise show 2 in reals is irreducible.

Firstly, note that this mapping is actually into $\displaystyle \mathbb{N} \cup \{0\}$.

Suppose $\displaystyle 2 = \alpha \beta$ then taking norms (remember, $\displaystyle N(xy)=N(x)N(y)$) we find that $\displaystyle 4 = N(2) = N(\alpha)N(\beta)$. W.l.o.g. we have either $\displaystyle N(\alpha) = N(\beta) = 2$ or $\displaystyle N(\alpha) = 4$ and $\displaystyle N(\beta) = 1$.

Now, simply prove that no element has norm 2 and that $\displaystyle N(\beta) = 1$ implies that $\displaystyle \beta$ is a unit ( $\displaystyle \beta = \pm 1$ ).

It is probably good revision making up a few of these yourselves, especially where they map onto the integers not just the natural numbers. Also, I once did an exam that asked me lots of these types of questions - I would advise you to write something like "$\displaystyle N(a+b \sqrt{-n})=a^2+n b^2$ is a norm function because..." at the very start. I did that little proof at the start of every such question. It nearly killed me, and it meant didn't quite have time to finish the paper properly...
• May 15th 2009, 06:23 AM
benjyboy
$\displaystyle (1-\sqrt{-5})(1+\sqrt{-5})$
to prove 2 is not a norm,could i just show 2 is irrational.
why does this =6=2.3?
and why is it not prime?
thanks.
• May 15th 2009, 06:41 AM
Swlabr
Quote:

Originally Posted by benjyboy
to prove 2 is not a norm,could i just show 2 is irrational.

Do you perhaps mean root 2? Essentially, yes.

Quote:

Originally Posted by benjyboy
why does this =6=2.3?

do you mean why does $\displaystyle (1+\sqrt{-5})(1-\sqrt{-5}) = 6$? 2.3=6 by the definition - multiplication is as in the real numbers.

Quote:

Originally Posted by benjyboy
...and why is it not prime?

The result follows immediatly from the definition...
• May 15th 2009, 08:54 AM
Gamma
There is no proving 2 is not a norm, that doesn't even make sense. 2 is definitely not irrational either, it is clearly an integer.

The point of the second question is to show that in fact two is irreducible. This means if $\displaystyle \exists \alpha , \beta \in R$ so that $\displaystyle \alpha \beta = 2$ then either $\displaystyle \alpha$ or $\displaystyle \beta$ is a unit.

The point of the norm argument is to take you into a space that we know stuff about.

Suppose $\displaystyle 2 = \alpha \beta$

$\displaystyle N:R \rightarrow \mathbb{Z}^{nonneg}$

Then $\displaystyle 4=N(2) = N(\alpha \beta)= N(\alpha)N(\beta)$

But the only way to get 4 from a product of positive integers ($\displaystyle N(\alpha),N(\beta)$.
Is:
Case 1)
4=1*4 (then $\displaystyle \alpha$ is a unit, it must be +/- 1)

Case 2)
4=4*1 (then $\displaystyle \beta$ is a unit, it must be +/- 1)

Case 3)
4=2*2
Lets show this is impossible. WLOG, suppose $\displaystyle N(\alpha)=2$.
$\displaystyle N(a+b\sqrt{-5})=a^2 + 5b^2 = 2$ for INTEGERS $\displaystyle a,b \in \mathbb{Z}$.

Do you see why this is impossible?

It is the sum of two positive things, if $\displaystyle b\geq 1$ then just the right term alone is too big $\displaystyle 5b^2 \geq 5 > 2$ so equality is not possible. Thus b=0.

But this means $\displaystyle a^2 + 5*0^2 = a^2 = 2$.
Is there an integer such that when you square it you get two? Not a chance. About a million ways to show that take your pick $\displaystyle +/- \sqrt2$ is not rational so its not an integer.

Thus the only cases that are possible were case 1 and 2.
In both of these cases we showed either $\displaystyle \alpha$ or $\displaystyle \beta$ is a unit.

What is the point of all this? Really you should show all 4 of the elements of the two factorizations I gave you are irreducible. Then you would have shown R is an integral domain that is in fact not a Unique Factorization Domain as you have just found two unique factorizations of 6 into irreducibles.
• May 15th 2009, 01:13 PM
Gamma
Prime
I missed this question that you asked, my apologies.

An element $\displaystyle p\in R$ is said to be prime if $\displaystyle ab\in (p) \Rightarrow a\in (p)$ or $\displaystyle b \in (p)$

(p) is the ideal generated by p in R.

2 is not prime in your ring R because $\displaystyle (1-\sqrt{-5})(1+\sqrt{-5})=6=2*3 \in (2)$; however neither $\displaystyle 1-\sqrt{-5}$ nor $\displaystyle 1+\sqrt{-5}$ is (2).