I'm learning ring theory but just wanted some more information as my notes are limited

Question

Let $\theta : \mathbb{Z}_8 \to \mathbb{Z}_4$ 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. Find $\mathrm{Ker}(\theta)$ and exhibit the partition of $\mathbb{Z}_8$ into cosets of $\mathrm{Ker}(\theta)$.

Can anyone please explain to me what $\mathrm{Image}(\theta)$ and $\mathrm{Kernel}(\theta)$ are. I have the definition and I understand the mapping but how is it used? Why is it relevant? I don't even know if these are the correct questions to ask.

Answer 1

$\ker(\theta) = \{[i]_8 \in \mathbb{Z}_8 : \theta([i]_8) = [0]_4\}$

So, you have to calculate $\theta([i]_8)$ and analyze the element modulo $4$.

Doing that, we have: $\theta([i]_8) = [i]_4 = [0]_4$, but that happens iff $i = 4k$ for $k \in \mathbb{Z}$. But you know that the only elements in $\mathbb{Z}_8$ that are multiple of $4$ are $[4]_8$ and $[0]_8$, so, $\ker(\theta) = \{[0]_8,[4]_8\}$.

The image of $\theta$ is $\mathbb{Z}_4$ itself, because $\theta$ is surjective.

To see that, let $[i]_4 \in \mathbb{Z}_4$. Then $[i]_4 = \theta([i]_8)$. Note that $[i]_4 = 0, 1, 2$ or $3$.

Answer 2

How is it used? Why is it relevant? Why do we care about kernels and images, anyway?

When we look at algebraic maps, i.e., homomorphisms between groups, rings, modules, etc., we want to understand their structure. Why? Well, it's pretty much what there is to study. The image and kernel of a homomorphism tell you, respectively, how much of the target object is "covered" or "reached" by the map, and the kernel tells you how much of the source object is folded up in order to make it fit in that image.

In this case, the source object is the ring $\Bbb Z_8$, and the target object is the ring $\Bbb Z_4$. Since the target object is smaller, (only four elements, as opposed to eight), something from the original object will have to be folded together. Thus, if you're trying to find out which $x\in\Bbb Z_8$ satisfy $\theta(x)=b$ in $\Bbb Z_4$, then you'll expect to have multiple solutions (if you have any), because more than one element of the domain is mapped to the same place in the image. The number of multiple solutions you'll have is determined by the size of the kernel.

Whether you have any solutions at all depends on whether $b\in\operatorname{Im}(\theta)$. The image of a map is precisely the same as the "range" of a function, as we talk about it in elementary precalculus or calculus.

There's more than that, but hopefully that's enough motivation to keep you calculating these things. Finding the specific image and kernel in your case appears to be something you're well on your way to doing, so I'll not get in the way of that :)