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Kernel of this homomorphism

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data

    1) Show that the kernel of the homomorphism [itex] \theta: \mathbb{Z} \rightarrow \mathbb{Z}_{10} [/itex] defined by [itex] \theta(a+bi) = [a+3b]_{10}, a,b \in \mathbb{Z} [/itex] is [itex] <1+3i> [/itex] (i.e. the ideal generated by 1+3i).


    3. The attempt at a solution

    My answer confuses me. It shows that any element of [itex] <1+3i> [/itex] is indeed in the kernel (one way proof), then to show the other way it says;

    Conversely let [itex] a+bi \in ker(\theta) \Rightarrow a+3b \equiv 0mod10 \Rightarrow 3a \equiv bmod10 [/itex]. Therefore [itex] a+bi \in <1+3i> [/itex] .

    I'm not sure how it concludes the "therefore" part so easily. For example, [itex] 4+2i [/itex] would be in the kernel, because [itex] 12 \equiv 2mod10 [/itex] but from this proof it's not intuitively obvious to me that that [itex] 4+2i [/itex] is contained in [itex] <1+3i> [/itex]. It is though, because [itex] (1+3i)(1-i) = 4+2i [/itex]. But exactly how does the latter half of the proof show that, (just as an example), [itex] 4+2i [/itex] is contained within [itex] <1+3i> [/itex]?

    Basically, how does [itex]3a \equiv bmod10 \Rightarrow a+bi \in <1+3i> [/itex]?


    I've found a way to prove this, but it requires a bit of work;

    An element is in the kernel if

    [itex] 3a - b = 10k [/itex]

    [itex] b = 3a - 10k [/itex]

    So any complex number of the form;

    [itex] a + (3a-10k)i [/itex]

    Is in the kernel.

    Is [itex] a + (3a-10k)i \subseteq <1+3i> [/itex]?

    This means, is there some [itex] c+di [/itex] such that;

    [itex] (c+di)(1+3i) = (a+(3a-10k))i [/itex]?

    Yes, re-arranging shows that.

    [itex] c = a - 3k, d = -k [/itex]

    So [itex] (a+(3a-10k)) \in <1+3i> [/itex]

    So [itex] ker(\theta) \subseteq <1+3i> [/itex]

    This proves it for me, but the way my given answers simply say [itex]3a \equiv bmod10 \Rightarrow a+bi \in <1+3i> [/itex] suggests there is a clear way of noticing this without doing any working out. What is it?
     
  2. jcsd
  3. Mar 26, 2012 #2

    Dick

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    Science Advisor
    Homework Helper

    What you need for a+bi to be in <1+3i> is that it be evenly divisible by 1+3i. Multiplying numerator and denominator of (a+bi)/(1+3i) by (1-3i) gives you (a+3b)/10+i*(-3a+b)/10, hence the divisibility by 10 conditions.
     
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