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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.

Bilbottom
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Sana Muzammal
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    It will be helpful if you let us know what $t$ is. – Bilbottom May 14 '18 at 22:38
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    Welcome to stackexchange. To get help here you should edit your question so that it has the full context of what you are asking about. In this case there must be some ring and some $t$. Then tell us what you do understand, and where you are stuck. Please use mathjax: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Ethan Bolker May 14 '18 at 22:40
  • t is just theta in my notes. (I apologise in advance for my ignorance if this is not what you are asking!) – Sana Muzammal May 14 '18 at 22:40
  • Whether it is $t$ or $\theta$, we still need to know what it is. Is it a homomorphism of rings? If so, between which two rings? Perhaps you could share the definitions that you have? – Xander Henderson May 14 '18 at 22:41
  • Ah, thanks so much. I am a newbie to this. – Sana Muzammal May 14 '18 at 22:42
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    Good edits to improve the question! Can you figure out which cosets in the domain map to the $0$ element of the codomain? Then you'll know the kernel. Find out which elements of the codomain are actually values of $\theta$ and you'll know the image. – Ethan Bolker May 14 '18 at 23:00
  • @EthanBolker I found my cosets to be [0]_8 and [4]_8? – Sana Muzammal May 14 '18 at 23:19
  • @BillWallis Thank you for the edits. – Sana Muzammal May 14 '18 at 23:20
  • Right. So those two cosets make up the kernel (that's the definition of the kernel). Now find the cosets of the kernel in the domain ring to finish the answer. Then post all of this as an answer to your own question (that's allowed here). – Ethan Bolker May 14 '18 at 23:27
  • @EthanBolker I believe that to be just [0]_8? Thanks so much for your help. – Sana Muzammal May 14 '18 at 23:31
  • Good work on this exercise. You might find this somewhat philosophical answer helpful: https://math.stackexchange.com/questions/2039702/what-is-an-homomorphism-isomorphism-saying/2039715#2039715 – Ethan Bolker May 14 '18 at 23:49
  • Lovely analogy, thanks!! – Sana Muzammal May 14 '18 at 23:59

2 Answers2

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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 :)

G Tony Jacobs
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    Wow, thank you so much. I can sometimes answer a question but I find it 'uncomfortable' answering without understanding the 'why' behind it. That makes so much sense, appreciate your explanation. – Sana Muzammal May 14 '18 at 23:42
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$\ker(\theta) = \{[i]_8 \in \mathbb{Z}_8 : \theta([i]_8) = [0]_4\}$

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 $3$.

Xablau123
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  • Amazing, thank you! – Sana Muzammal May 14 '18 at 23:34
  • You can also see that the image of $\theta$ is $\mathbb{Z}_4$ observing that $\mathbb{Z}_4 \subset \mathbb{Z}_8$, so the map $\theta$ is just a projection. – Xablau123 May 14 '18 at 23:42
  • It's not entirely correct to say that $\Bbb Z_4\subset\Bbb Z_8$. Elements of the former are cosets of $4\Bbb Z$, and elements of the latter are cosets of $8\Bbb Z$. Indeed, there are embeddings of $\Bbb Z_4$ into $\Bbb Z_8$, but this map doesn't reverse either of them. – G Tony Jacobs May 15 '18 at 00:03
  • Ok, you're right, it's just that I usually imagine $\mathbb{Z}_n = {0,1,\ldots,n-1}$, but of course, their elements are different in the sense you said. – Xablau123 May 15 '18 at 00:13