Finding the Kernel and Image of $\mathbb Z \to \mathbb Z[i]/(1+3i)$, $x\mapsto x+(1+3i)$

Question

I'm trying to apply the homomorphism theorem to the following function:

$$h:\mathbb Z \to \mathbb Z[i]/(1+3i)$$ $$x\mapsto x+(1+3i)$$

Where $(1+3i)$ is the ideal generated by $1+3i$.

I know that Because $\ker(h)$ is an ideal in $\mathbb Z$ which is an Principal Ideal Domain, then $\ker(h)=m\mathbb Z$ for some $m \in \mathbb Z$, but I'm having some trouble finding that $m$.

And I have no clue how I can find the elements of $\text{im}(h)$.

How can I do this?

Answer 1

Let us find the kernel first. Note that an integer $x\in\mathbb{Z}$ has $x\in\text{ker}(h)$ if and only if $$x\in(1+3i).$$ Note that for any $\mathbb{Z}$-multiple of $10$, say $10m$, we have $$10m=m\cdot 10=m\cdot (1-3i)(1+3i)\in (1+3i),$$ so we deduce that $10\mathbb{Z}\subseteq\text{ker}(h)$. Because $\text{ker}(h)$ is of the form $n\mathbb{Z}$ for some $n\in\mathbb{N}$, we must have $n|10$, which means that $\text{ker}(h)$ must be one of: $\mathbb{Z},2\mathbb{Z},5\mathbb{Z},10\mathbb{Z}$. It is easy to check that $2,5\notin (1+3i)$, so it must be that $\ker(h)=10\mathbb{Z}$.

Now the image of $h$ is by definition $\{a+(1+3i):a\in\mathbb{Z}\}\subseteq \mathbb{Z}[i]/(1+3i)$. Because $a+(1+3i)=b+(1+3i)$ whenever $a-b$ is a multiple of $10$, we then have a more explicit description of the image: $$\text{im}(h)=\{0+(1+3i),1+(1+3i),...,9+(1+3i)\}.$$ Edit: To be more careful, at this point we have only shown that $\text{im}(h)$ consists of at most ten distinct elements, but the first isomorphism theorem implies that $\text{im}(h)$ consists of exactly $10$ elements. So we know that our description of $\text{im}(h)$ is correct. Moreover, $h$ is surjective by Brian Shin's comment, so we can deduce from the first isomorphism theorem $$\mathbb{Z}/10\mathbb{Z}\cong \mathbb{Z}[i]/(1+3i)$$ as rings.

Answer 2

Note that $\ker{h}=\{n \in \mathbb{Z} : n \in \langle 1+3i \rangle\}=\langle 1+3i \rangle \cap \mathbb{Z}$. $n \in \ker{h} \implies n=(1+3i)(a+bi)=a+bi+3ai-3b=(a-3b)+i(b+3a)$. Since $n \in \mathbb{Z}$, $b=-3a$. Thus, every $n \in \ker{h}$ is of the form $a(1+3i)(1-3i)=10a$ for some $a \in \mathbb{Z}$. Clearly, the least such element is $10$ because $|10a|=|10||a|=10|a|\geq 10$ and for $a=1$, equality is satisfied. So, $\ker{h}=\langle 10 \rangle$.