# Show that N is multiplicative

Printable View

• January 26th 2008, 12:22 PM
tttcomrader
Show that N is multiplicative
Define $N:Z[w] \rightarrow Z by N(a+bw) = a^2 - ab + b^2$ with $w = \frac {-1}{2} + \frac {3^{1/2}}{2}i$.

Show that N is multiplicative.

I'm a bit rusty on this, showing it is multiplicative means closure, associative, inverse, identity, right? Or N(ab) = N(a)N(b)?
• January 26th 2008, 01:04 PM
Opalg
Quote:

Originally Posted by tttcomrader
Define $N:Z[w] \rightarrow Z by N(a+bw) = a^2 - ab + b^2$ with $w = \frac {-1}{2} + \frac {3^{1/2}}{2}i$.

Show that N is multiplicative.

I'm a bit rusty on this, showing it is multiplicative means closure, associative, inverse, identity, right? Or N(ab) = N(a)N(b)?

You need to show that N((a+bw)(c+dw)) = N(a+bw)N(c+dw).
• January 26th 2008, 02:07 PM
tttcomrader
Alright, that would be easy, thank you!
• January 26th 2008, 07:30 PM
tttcomrader
I was trying to work this out, but here is what I got.

Let x=a+bw and y=c+dw.

N(xy) = N(ac+adw+cbw+bdw), I can't get it into a the form of u+vw, how would I do it? Thanks.
• January 27th 2008, 12:32 AM
Opalg
Quote:

Originally Posted by tttcomrader
I was trying to work this out, but here is what I got.

Let x=a+bw and y=c+dw.

N(xy) = N(ac+adw+cbw+bdw), I can't get it into a the form of u+vw, how would I do it? Thanks.

If x=a+bw and y=c+dw then $xy = ac +adw +bcw +bdw^2$. But $w^2 = -1-w$, so $xy = (ac-bd) + (ad+bc-bd)w$.

Thus $N(xy) = (ac-bd)^2 -(ac-bd)(ad+bc-bd) + (ad+bc-bd)^2$. You need to verify that this is equal to $N(x)N(y) = (a^2-ab+b^2)(c^2-cd+d^2)$. Good luck!