# Kernel of this homomorphism

1. Mar 26, 2012

### Silversonic

1. The problem statement, all variables and given/known data

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

3. The attempt at a solution

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

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

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

Basically, how does $3a \equiv bmod10 \Rightarrow a+bi \in <1+3i>$?

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

An element is in the kernel if

$3a - b = 10k$

$b = 3a - 10k$

So any complex number of the form;

$a + (3a-10k)i$

Is in the kernel.

Is $a + (3a-10k)i \subseteq <1+3i>$?

This means, is there some $c+di$ such that;

$(c+di)(1+3i) = (a+(3a-10k))i$?

Yes, re-arranging shows that.

$c = a - 3k, d = -k$

So $(a+(3a-10k)) \in <1+3i>$

So $ker(\theta) \subseteq <1+3i>$

This proves it for me, but the way my given answers simply say $3a \equiv bmod10 \Rightarrow a+bi \in <1+3i>$ suggests there is a clear way of noticing this without doing any working out. What is it?

2. Mar 26, 2012

### Dick

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.