Partition of $\mathbb{Z}_8$ into cosets of Ker(θ)

Question

Question: Let $\theta : \mathbb{Z}_8 → \mathbb{Z}_4$ be defined by $\theta([i]_8) = [i]_4$ for all $i \in \mathbb{Z}$. You may assume that this is a well-defined ring homomorphism.

1. Find Ker($\theta$)

2. Exhibit the partition of $\mathbb{Z}_8$ into cosets of Ker($\theta$).

My Answer: So I found the Ker($\theta$) as Ker($\theta$) $=\{[0]_8, [4]_8\}$ as $\theta[0]_8=[0]_4$ and $\theta[4]_8=[4]_4=[0]_4$. But I'm stuck on part 2. I was thinking that I would need to show that the partition of $\mathbb{Z}_8$ are equal to $[0]_4$ or $[4]_4$.

Sorry about how I have typed it, haven't really asked questions on the site before.

Answer 1

A partition of a set $X$ is a collection of subsets $X_i\subset X$ such that $X_i\cap X_j=\varnothing$ whenever $i\neq j$, and such that $\bigcup_{i\in I}X_i=X$. That is, a collection of non-overlapping subsets that together cover $X$.

For a (commutative) ring $R$ a coset of an ideal $I\subset R$ is defined as a subset of $R$ of the form $$r+I:=\{r+i:\ i\in I\},$$ for some $r\in R$. It is a nice exercise to prove that the cosets of an ideal form a partition of the ring.

With this information; what are the cosets of $\ker\theta\subset\Bbb{Z}/8\Bbb{Z}$? Can you show that they form a partition of $\Bbb{Z}/8\Bbb{Z}$?

Answer 2

I think you have found the kernel correctly. The cosets of an ideal $I$ in a ring are defined to be the translates $r+I$, where $r$ is some element. In particular, if $r \in I$, then $r+I=I$. Applying this, we see that we have 4 cosets, given by

1. $0+ \text{Ker}(\theta) = \text{Ker}(\theta) = 4+ \text{Ker}(\theta)$
2. $1+ \text{Ker}(\theta) = \{[1]_8,[5]_8\} = 5+ \text{Ker}(\theta)$
3. $2+ \text{Ker}(\theta) = \{[2]_8,[6]_8\} = 6+ \text{Ker}(\theta)$
4. $3+ \text{Ker}(\theta) = \{[3]_8,[7]_8\} = 7+ \text{Ker}(\theta)$

I'm not sure if you talked about quotient rings yet, but if you look at the left hand side, the translates $\{0,1,2,3\}$ end up having a structure isomorphic to $\mathbb{Z}_4$.

(Edited)

Answer 3

You've misunderstood what a partition is, I think. A partition is any bunch of disjoint subsets that cover the whole set. For example, the following is a partition of $\mathbb{Z}_8$:

$$\{1\}, \{2, 4, 6\}, \{3, 0\}, \{5\}, \{7\}.$$

Here's another:

$$\{1, 2, 3, 4, 6, 7\}, \{0,5\}.$$

Here's the crucial fact. Whenever you have a ring $R$, and an ideal $I$ of that ring, the cosets $a+I$ (for various $a\in R$) partition $R$. That is (assuming for simplicity that things are appropriately finite), there exist some $a_1, \dots, a_n\in R$ such that

$$a_1 + I, \dots, a_n + I$$

is a partition of $R$.

Now let $R = \mathbb{Z}_8$ and $I = \ker \theta$. Can you find such $a_1, \dots, a_n$? (Note that taking $0 + \ker\theta$ and $4 + \ker\theta$ gives you the same thing, for the reasons you said. On the other hand, $1+\ker\theta$ is something new.)