# Challenge: Check The Proof, Please

1. Feb 15, 2014

### NoodleDurh

1. The problem statement, all variables and given/known data

Show that if $|\mathbb{Z}_p/(f)| = p^{deg(f(x))}$ where the $deg(f(x)) = n$ and $f(x) \in \mathbb{Z}_{p}$ is irreducible over $\mathbb{Z}_{p}[x]$, then $\mathbb{Z}_{p}[x]/(f) \rightarrow_{\phi}^{\cong} \bigoplus_{i \in I} \mathbb{Z}_{p_{i}}$ where $|I| = n$

2. Relevant equations

3. The attempt at a solution

(An attempt at rigor)
Surjection is trival. Obviously, the size of $\bigoplus_{i \in I} \mathbb{Z}_{p_{i}}$ and $\mathbb{Z}_{p}[x]/(f)$ are the same, so $\bigoplus_{i \in I} \mathbb{Z}_{p_{i}}$ is covered by $\mathbb{Z}_{p}[x]/(f)$ completely, and thus $im( \phi ) = \bigoplus_{i \in I} \mathbb{Z}_{p_{i}}$. Injectivity is less so. Since $\phi$ is a homomorphism we can see that $\phi (a) = \phi (b)$ iff $\phi (a) - \phi (b) = 0$ when $\phi (a-b) = 0$, precisely when $a-b \in ker (\phi )$ hence, $a = b$, So now we must check the order of the elements. (this is where I am stuk)

Last edited: Feb 15, 2014
2. Feb 17, 2014

### NoodleDurh

[Bump] Sorry it has been 3 days.

3. Feb 17, 2014

### gopher_p

There seem to be a few issues with your problem statement. In particular, are all instances of $\mathbb{Z}_{p}$ except the very last meant to be $\mathbb{Z}_{p}[x]$? Also is $\rightarrow_{\phi}^{\cong}$ meant to denote an isomorphism? Is $\phi$ some sort of canonical homoomorphism?

and
Also, it seems a bit odd to me that you were able to get this far without using the fact that $f(x)$ is irreducible, although it's possible that's an extraneous piece of information given $|\mathbb{Z}_p[x]/(f)| = p^{deg(f(x))}$.

It's been a while since I've done any "real" abstract algebra, so there's a good chance I'm asking obvious/irrelevant questions.

4. Feb 18, 2014

### NoodleDurh

I think I might have figured it out. But, still I do not know when a proof is done. Also, my notation might be strange... $\mathbb{Z}_{p}$ is $\mathbb{Z} / p$

Since we know that $|\mathbb{Z}_{p}| = p$, in addtion to knowing that $|\mathbb{Z}_{p}[x]/(f)| = p^{deg(f(x))}$ and $f(x) \in \mathbb{Z}_{p}$ is irreducible over $\mathbb{Z}_{p}$. Clearly, we can see that $| \bigoplus_{i \in I} \mathbb{Z}_{p_{i}}| = \prod p_{i}$ and since we choose a $p$ which is "universal", or the same for both $\bigoplus_{i \in I} \mathbb{Z}_{p_{i}}$ and $\mathbb{Z}_{p}[x]/(f)$, their sizes must be the same too. In addition to choosing a $f(x)$ we have chosen one which is irreducible, and by definition is prime. So, since we the the polynomial is the prime, the one we mod out by in $\mathbb{Z}_{p}$ and is of finite size, we can see that $\mathbb{Z}_{p}[x]/ (f)$ must be finite also. If we take a homorphism $\phi$ we can see that this is one-to-one and onto, therefore there exist an isomorphism between $\bigoplus_{i \in I} \mathbb{Z}_{p_{i}}$ and $\mathbb{Z}_{p}[x]/(f)$. By the rule, $(a_{0},a_{2},\cdots , a_{p-1}) \mapsto f(x)$ where $f(x)$ is some polynomial... QED(?)

I can give an example where it is true.

$\mathbb{Z}_{2}[x]/(x^{2}+x+1) \rightarrow_{\phi} \mathbb{Z}_{2} \oplus \mathbb{Z}_{2}$

the multiplication tables for both

$\begin{pmatrix} + & (0,0) & (0,1) & (1,0) & (1,1) \\ (0,0) & (0,0) & (0,1) & (1,0) & (1,1) \\ (0,1) & (0,1) & (0,0) & (1,1) & (1,0)\\ (1,0) & (1,0) & (1,1) & (0,0) & (0,1)\\ (1,1) & (1,1) & (1,0) & (0,1) & (0,0) \end{pmatrix}$

$\begin{pmatrix} + & 0 & 1 & x & x+1 \\ 0 & 0 & 1 & x & x+1 \\ 1 & 1 & 0 & x+1 & x \\ x & x & x+1 & 0 & 1 \\ x+1 & x+1 & x & 1 & 0 \end{pmatrix}$

we see that each duple gets mapped to the coeff of the monomial

Last edited: Feb 18, 2014
5. Feb 18, 2014

### gopher_p

Let $G_1,G_2$ be groups such that $|G_1|=|G_2|>1$, and define $\phi:G_1\rightarrow G_2$ by $\phi(g)=e_{G_2}$ for all $g\in G_1$, where $e_{G_2}$ is the identity element of $G_2$. Then $\phi$ is a homomorphism (prove it, it's easy) that is not injective or surjective (prove it, it's easy). So there's a homomorphism between two same-sized groups which is not an isomorphism.

So I ask again, what is $\phi$ in your problem, why is it a homomorphism, and why is it surjective? If you answer these questions, then you are done. Everything you need to prove your claim is in the answers to these three questions.

Now it looks like you have a function $\phi:\mathbb{Z}_p[x]\rightarrow\bigoplus\limits_{1\leq i\leq n} \mathbb{Z}_{p}$, $\sum\limits_{i\in\mathbb{N}} a_ix^i\mapsto (a_{n-1},a_{n-2},\ldots,a_1,a_0)$. And it seems like this should be a homomorphism (prove it, it should be easy). But it is unlikely that a member of $f\in\mathbb{Z}_p[x]$ will be in the kernel of this homomorphism, and so there is no natural (to me) way to consider $\phi$ as a map from $\mathbb{Z}_p[x]/f\rightarrow\bigoplus\limits_{1\leq i\leq n} \mathbb{Z}_{p}$.

6. Feb 18, 2014

### NoodleDurh

So, I do not understand how it is not one, or the other. Unless you mean to say it is either injective or surjective, because if $G_1$ is $S_2$ and $G_2$ is $\mathbb{Z}_{2}$, there is a natural injection isomophism between the two. Yet, if we do take that map $\phi$ and look at $S_2 \rightarrow_{\phi} \mathbb{Z}_{2}$ and $\phi(x) = 1$ $\forall x \in S_2$, then this must be a surjection, right?

$\phi$ is a homorphism, which happens to be a isomorhpism between fields. $\phi$ needs to be surjective so we know that they are in a correspondence. And since we now know that every element has a particular element which it maps to, and there is an inverse for the element.

^ I am going to prove this in a bit, it seems like something interesting.

But this $f(x)$ isn't any regular $f(x)$ it is irreducible and by definition akin to a prime, and when we mod out by a prime we "make" it live in the kernel, correct?. Like, if we have $\mathbb{Z} / p$ this creates an ID(integal domain). We can do the same thing for the $\mathbb{Z}_{p}[x] / (f(x))$ to create an integral domain, and thus since we choose a p-prime which is the same, the $f(x)$ has a $deg \ge 0$. We see that they have the same size and can "induce" an isomorphism between these guys. Maybe I am not wording myself right, but It seems clear... doesn't it?

Last edited: Feb 18, 2014
7. Feb 18, 2014

### gopher_p

Sorry. I should have said "neither ... nor". Surjective is just another word for onto. If every element in the first group/ring/field is sent to the identity of the second, then every member of the second group that is not the identity is not in the image. And I'm not sure if you're clear on this (an please excuse me if you are), but the fact that two groups/rings/fields are isomorphic does not mean that every homomorphism from one to the other is necessarily an isomorphism.

This is the part that I am asking you to prove. I think the original claim is that $\mathbb{Z}_{p}[x] / (f(x))$ is isomorphic to $\bigoplus_{1\leq i\leq n} \mathbb{Z}_{p}$, and please correct me if that is not the case. If that is, in fact, the problem, then you can't just pull a $\phi$ out of your pocket and say that it's an isomorphism. You either have to define a function (i.e. tell me where it maps each element of the domain) and demonstrate that it is an isomorphism. Or you need to demonstrate, using some pertinent theorems and maybe a sprinkling of abstract nonsense, that an isomorphism exists.

Now it has just occurred to me that you might be using a (possibly nonexistent) fact regarding rings or integral domains of prime-power size to justify why these objects must be isomorphic. Like I said, my algebra is a bit rusty, especially with regards to ring theory. So you'll need to be fairly explicit when using any "big" theorems to justify your claims.

In general, an irreducible element of a ring need not be prime. It's possible that they are the same here.

This isn't entirely true. There should be a definition of "modding out" in every case that is rooted in some kind of equivalence relation. In the case of $\mathbb{Z} / n$, it is the set consisting of all equivalence classes of $\mathbb{Z}$ under the $\mod n$ equivalence relation. So here we've "modded out" by $n$ without appealing to any sort of function. Since there is no function, there's no kernel to speak of.

Now in each instance of "modding out", where we have a set $X$ and an equivalence relation $\equiv$, there is a canonical map $\pi:X\to X/\equiv$, $x\mapsto [x]$ where $[x]=\{y\in X\ |\ y\equiv x\}$. In the case of groups, we can talk about $\text{ker}(\pi)=\{x\ |\ \pi(x)=0\}=\{x\ |\ x\equiv 0\}$. And if you look back, you'll see that all of the fussing over modding out by normal subgroups was due to the desire for this canonical map to "pass on" a group operation from $X$ to $X/\equiv$ and make $\pi$ a group homomorphism.

My whole point in this little diatribe is to try and explain that, while "mod[ing] out by a prime ... 'make' it live in the kernel" is true in a certain sense, it's not the kernel of anything that is happening in your problem ... yet. In fact, $f$ is not even an element of either of the objects that you are trying to show are isomorphic, so it can't be in the kernel of any map between the two.

Now it might be (probably is if the main claim is to be believed) the case that, for irreducible $f\in \mathbb{Z}_p[x]$ of degree $n$ there is some canonical homomorphism $\phi:\mathbb{Z}_p[x]\rightarrow\bigoplus_{1\leq i\leq n} \mathbb{Z}_{p}$ such that the kernel of $\phi$ is precisely some "nice" set (prime ideal maybe?) containing/generated by $f$ so that (by some isomorphism theorem, maybe) $\mathbb{Z}_p[x]/f\cong \text{im}(\phi)=\bigoplus_{1\leq i\leq n} \mathbb{Z}_{p}$. But I think it's your job to either figure out what that canonical homomorphism is (i.e. write down what it "does" to elements of $\mathbb{Z}_p[x]$) or invoke some big general (i.e. abstract nonsense) theorem about rings and ideals to the specific case of the ring $\mathbb{Z}_p[x]$ and the (prime?) ideal generated by $f$.

8. Feb 19, 2014

### gopher_p

I think I've changed my mind on this. It's a group homomorphism; that I'm fairly sure of. I don't think that it's a ring homomorphism in general. Foiled once again by FOIL.

9. Feb 19, 2014

### NoodleDurh

Thank for you help... It seems I need to head back to the drawing board. Originally, The problem was:
Suppose $f(x) \in \mathbb{Z}_{p}[x]$ and $f(x)$ is irreducible over $\mathbb{Z}_{p}$. If $deg f(x) = n$, then $\mathbb{Z}_{p}[x]/ (f(x))$ is a field with $p^{deg (f(x))}$ elements. I didn't know how to prove this, so I modified it. To the problem above. I guess I can't prove that one either. I guess it was a good attempt for the first proof. Although I didn't prove it...

10. Feb 19, 2014

### gopher_p

If $\mathbb{Z}_{p}[x]/ (f(x))$ is a field, then it can't possibly be isomorphic to $\bigoplus_{1\leq i\leq n} \mathbb{Z}_{p}$, because $\bigoplus_{1\leq i\leq n} \mathbb{Z}_{p}$ is not a field (if $n>1$, under the standard component-wise multiplication); it has zero divisors.

As far as your actual problem goes; with regards to showing that $\mathbb{Z}_{p}[x]/ (f(x))$ is a field, so you know any "big" theorems that tell you when taking certain kinds of rings and modding out by certain kinds of ideals gives you a field?

11. Feb 19, 2014

### NoodleDurh

Yeah, I know this. But if think about them as Rings then, there should exist a isomophism between them. Also, I am not sure I know what a nonzero element would look like. For example, when we are in the Ring $\mathbb{Z}_{2} \oplus \mathbb{Z}_{2}$ we get elements like $(0,1)$ and $(1,0)$ but how do we know that they are nonzero elements?

12. Feb 19, 2014

### gopher_p

If two rings are isomorphic (as rings), and one of them is a field, then the second one must also be a field. This statement is still true if you replace the word "field" with "integral domain", "UFD", "PID", etc. There isn't any additional structure involved; they're all just rings that happen to have nice properties. And two things being (fill in the blank)-isomorphic basically means that they are the same (fill in the blank). So if two rings are ring-isomorphic, they're essentially the same ring. They might look different to you and I, but in the the land of rings, no one can tell them apart.

Something that you either just learned or are about to learn (spoiler alert) is that all finite fields have order $p^n$ for some prime $p$ and that any two same-sized finite fields are isomorphic. So once you've found a finite field of size $p^n$, you can examine it and understand everything there is to know about all of the other fields of that size. This exercise that you've been given basically tells you how to construct a field of any possible finite size; just find an irreducible polynomial $f$ of degree $n$ in $\mathbb{Z}_p[x]$ and look at $\mathbb{Z}_p[x]/(f)$. Of course it's your job to prove once and for all that this constructions works; i.e. that it does, in fact, give you a field of order $p^n$.

Well, this might sound a bit silly, but ... they are nonzero elements because they are elements that aren't the zero element $(0,0)$. Remember that every ring $R$ is first and foremost an abelian group under its addition operation $+_R$, and the zero of the ring is the additive identity; i.e. it is the unique element $0$ such that $0+_Rr=r$ for all $r\in R$. Once you've identified the $0$ in a given ring, every other element is automatically, by definition, a nonzero element.

13. Feb 20, 2014

### NoodleDurh

This is what I thought, and ended up forgetting about the zero divisors. So as you said $\oplus \lim{i \ in I} \mathbb{Z}_{p_{i}}$ isn't a field. But still a Ring.

I think this is what I am suppose to prove, that theorem you are talking about, because it essentially leads me there. I came to a conclusion from it, but not the proper one, haha*uneasy laugh*.