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Ring Homomorphism - showing Multiplicativity

  1. Apr 5, 2012 #1
    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. jcsd
  3. Apr 5, 2012 #2
    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:

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

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

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

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

    DonAntonio
     
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