Ring Homomorphism - showing Multiplicativity

1. Apr 5, 2012

RVP91

Hi,

I have the following map Q: A --> Z/2Z (where Z denotes the symbol for integers) defined by
Q(a + bi) = (a + b) + 2Z

where A = Z = {a + bi | a,b in Z} and i = √-1.

I need to show it is a ring homomorphism.

I have shown it is addivitivity by showing Q(a + b) = Q(a) + Q(b) by doing the following,
Q((a+bi)+(c+di)) = Q((a+c)+(b+d)i) = a+c+b+d+2Z = (a+b+2Z)+(c+d+2Z) = Q(a+bi) + Q(c+di)

Now for multiplicativity i know I have to show Q(ab) = Q(a)Q(b) but my working out seems to break down when i try to show LHS = RHS or RHS = LHS.

This is how I approached it thus far, LHS = RHS
Q((a+bi)(c+di)) = Q((ac-bd)+(ad+bc)i) = ac-bd+ad+bc+2Z and then i'm not sure where to go as nowhere seems to take me to what I need to show.

Any help would be most appreciated, thanks in advance

2. Apr 5, 2012

DonAntonio

Hi,

I have the following map Q: A --> Z/2Z (where Z denotes the symbol for integers) defined by
Q(a + bi) = (a + b) + 2Z

where A = Z = {a + bi | a,b in Z} and i = √-1.

I need to show it is a ring homomorphism.

I have shown it is addivitivity by showing Q(a + b) = Q(a) + Q(b) by doing the following,
Q((a+bi)+(c+di)) = Q((a+c)+(b+d)i) = a+c+b+d+2Z = (a+b+2Z)+(c+d+2Z) = Q(a+bi) + Q(c+di)

Now for multiplicativity i know I have to show Q(ab) = Q(a)Q(b) but my working out seems to break down when i try to show LHS = RHS or RHS = LHS.

This is how I approached it thus far, LHS = RHS
Q((a+bi)(c+di)) = Q((ac-bd)+(ad+bc)i) = ac-bd+ad+bc+2Z and then i'm not sure where to go as nowhere seems to take me to what I need to show.

Any help would be most appreciated, thanks in advance[/QUOTE]

Why do you think it'd be a good idea to post this question in General Math instead of Linear & Abstract Algebra?

Anyway, as you wrote:

$Q\left((a+bi)(c+di)\right)=ac-bd+ad+bc+2Z$ , whereas

$Q(a+bi)Q(c+di)=\left((a+b)+2Z\right)\left((c+d)+2Z\right)=ac+ad+bc+bd+2Z$.

In order to show both lines above are the same, we must show that

$ac-bd+ad+bc-(ac+ad+bc+bd)=0\pmod 2\Longleftrightarrow -2bd=0\pmod 2$ , which is trivially true and we're done.

DonAntonio