Ring Homomorphism - showing Multiplicativity

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The discussion centers on demonstrating the multiplicativity of the ring homomorphism Q: A --> Z/2Z, where A = Z[i]. The user successfully established additivity but struggled with multiplicativity, specifically showing that Q(ab) = Q(a)Q(b). The left-hand side was calculated as Q((a+bi)(c+di)) = ac-bd+ad+bc+2Z, but the user could not progress further. The solution involves considering cases based on the parity of a and b to complete the proof.

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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
 
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Q(a+bi) is really pretty simple. If both a and b are even then Q(a+bi)=a+b+2Z=0+2Z, since a+b is even. If a is even and b is odd then Q(a+bi)=1+2Z. Can you work out the other cases and show it's a multiplicative homomorphism by considering cases?
 
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