norm function,irreducible help.

**benjyboy** · May 14th 2009, 09:57 AM

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.

**Gamma** · May 14th 2009, 06:37 PM

2*3
$(1-\sqrt{-5})(1+\sqrt{-5})$

**Swlabr** · May 14th 2009, 11:18 PM

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 $\mathbb{N} \cup \{0\}$ .

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

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

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 " $N(a+b \sqrt{-n})=a^2+n b^2$ 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...

**benjyboy** · May 15th 2009, 06:23 AM

$<br /> <br /> (1-\sqrt{-5})(1+\sqrt{-5})<br />$
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.

**Swlabr** · May 15th 2009, 06:41 AM

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.

Originally Posted by benjyboy

why does this =6=2.3?

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

Originally Posted by benjyboy

...and why is it not prime?

The result follows immediatly from the definition...

**Gamma** · May 15th 2009, 08:54 AM

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 $\exists \alpha , \beta \in R$ so that $\alpha \beta = 2$ then either $\alpha$ or $\beta$ is a unit.

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

Suppose $2 = \alpha \beta$

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

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

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

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

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

Do you see why this is impossible?

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

But this means $a^2 + 5*0^2 = a^2 = 2$ .
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 $+/- \sqrt2$ 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 $\alpha$ or $\beta$ 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.

**Gamma** · May 15th 2009, 01:13 PM

I missed this question that you asked, my apologies.

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

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

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

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