R=Zsqrt(-5) where Z=intergers.
Define a norm function N on R with values in non negative integers by
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.
Suppose then taking norms (remember, ) we find that . W.l.o.g. we have either or and .
Now, simply prove that no element has norm 2 and that implies that is a unit ( ).
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 " 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...
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 so that then either or is a unit.
The point of the norm argument is to take you into a space that we know stuff about.
But the only way to get 4 from a product of positive integers ( .
4=1*4 (then is a unit, it must be +/- 1)
4=4*1 (then is a unit, it must be +/- 1)
Lets show this is impossible. WLOG, suppose .
for INTEGERS .
Do you see why this is impossible?
It is the sum of two positive things, if then just the right term alone is too big so equality is not possible. Thus b=0.
But this means .
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 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 or 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.