$R = \mathbb{Z}[i]\ /\ (1+3i)$ - Ring homomorphisms

Question

Let $R = \mathbb{Z}[i]\ /\ (1+3i)$

(a) Conclude that R is a ring and $|R|=10$.

(b) Define $\phi:\mathbb{Z} \rightarrow R$ by $\phi(n) = 1_R + \ldots + 1_R$ ($n$ times) if $n>0$ and $\phi(-n) = -\phi(n)$. Show this is a ring homomorphism.

(c) Show that $\mathbb{Z}/10\mathbb{Z} \cong \mathbb{Z}[i]\ /\ (1+3i)$

(a) I think I need to use $(1+3i)(1-3i)=10$ but not sure how.

(b) I'm not sure how to show this.

(c) Maybe I need to use the kernel but I'm not sure how to start.

Have you seen the concept of Euclidean ring ? – Jean Marie Feb 26 '16 at 18:22 — Jean Marie, Feb 26 '16 at 18:22

score 0 · Answer 1 · answered Feb 26 '16 at 18:41

0

For (c), consider $\phi$ as in (b). We have to prove that $\ker \phi = 10 \mathbb Z$:

$10 \in \ker \phi$ because $10 = (1-3i)(1+3i)$. Thus $10 \mathbb Z \subseteq \ker \phi$.
If $n \ker \phi$, then $n=(a+bi)(1+3i)=(a-3b)+(3a+b)i$ and so $b=-3a$ and $n=10a$. Thus $\ker \phi \subseteq 10 \mathbb Z$.

answered Feb 26 '16 at 18:41

lhf

216,483

how do u do part (b). And do you have to show the map $\phi(n)$ is surjective for part (c) also. – joeman619 Feb 26 '16 at 20:26
@joeman619, (b) is standard. Surjectiveness comes from (a) and what I've written above. – lhf Feb 26 '16 at 20:53
how do I conclude it is a ring? – joeman619 Mar 01 '16 at 22:17
and why is $|R|=10$? – joeman619 Mar 01 '16 at 22:17

$R = \mathbb{Z}[i]\ /\ (1+3i)$ - Ring homomorphisms

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