What is the kernel of the homomorphism defined by \theta(a+bi) = [a+3b]_{10}?

[Silversonic]

Homework Statement

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

The Attempt at a Solution

My answer confuses me. It shows that any element of &lt;1+3i&gt; 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 &lt;1+3i&gt; .

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 &lt;1+3i&gt;. 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 &lt;1+3i&gt;?

Basically, how does 3a \equiv bmod10 \Rightarrow a+bi \in &lt;1+3i&gt;?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 &lt;1+3i&gt;?

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 &lt;1+3i&gt;

So ker(\theta) \subseteq &lt;1+3i&gt;

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

 

[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.