# Constructing the splitting field for a polynomial over Z/Z3

1. Oct 14, 2016

### PsychonautQQ

1. The problem statement, all variables and given/known data
Construct a splitting s for the polynomial x^3+2x+1 over Z/Z3

2. Relevant equations
4=1 Mod 3
:P

3. The attempt at a solution
So i'm actually quite confused. There are no roots for x+3+2x+1 over Z/Z3. I am used to constructing splitting fields with polynomials that have coefficients in a field of character zero, for example Q(i) is a splitting field of x+1 over Q.

I guess i am confused because I don't know how to extensions of Z/Z3 will work, will the element I add not be reduced modulus 3? Will the new field have a new character? Any insight appreciated :-)

Thanks PF!

2. Oct 14, 2016

### Staff: Mentor

You mean $x^3 + 2x + 1$
I agree that this polynomial is prime over the given field.
I think you mean $x^2 + 1$
Right. Maybe you're supposed to find an extension for this field so that the given polynomial is factorable. However, it's not a simple matter to factor the given cubic.

3. Oct 14, 2016

### Staff: Mentor

It is the same procedure as in any field.
Assume you have a root $\xi$. This means $\xi^3 + 2\xi + 1=0$.
Now do the long division by $x-\xi$.

Finally you could either go ahead, or compute the automorphisms over $\mathbb{Z}_3$ and apply them to $\xi$.
I guess the chances are high, that $\xi^2$ and $\xi^3$ are the other roots. However, it has to be verified or otherwise justified.

4. Oct 15, 2016

### PsychonautQQ

I'm going to call epsilon 'a'. So divide x^3+2x+1 by x-a? or a^3+2a+1 by x-a? It's possible to do the long division with 2 variables?

Also, the automorphisms of Z/Z3 are M(1)=1 or M(1)=2, so the only useful automorphism will take M(a)=a^2.

I'm sorry but i'm still quite confused.

5. Oct 15, 2016

### PsychonautQQ

Ah, thank you for the corrections. So, since the only non-trivial automorphism of Z/Z3={0,1,2} equivalence classes is M(1)=2.

Will the element that i'm going to extend Z/Z3 with be an equivalence class as well? It must be right?

6. Oct 15, 2016

### Staff: Mentor

What do you do in case of $\mathbb{R} \subseteq \mathbb{R} [ i ] \cong \mathbb{C}$ ? Here it is $x^2+1=(x-i)(x+i)$.
And, yes, it is possible. Simply follow strictly the rules of division and the fact, that for your remainder $a^3+2a+1=0$ holds.
As a hint: it is better to do the subtractions very explicitly like
\begin{align*} 238 : (-7) = (-3) ... \\ - ((-7) \cdot (-3)) \;\;\;\;\;\; \\ \text{_________________} \\ +23 - (+21) (8) \\ \text{_________________} \\ 28 \\ etc.\end{align*}
You need to consider all automorphisms $\sigma \, : \, \mathbb{Z}_3[a] \rightarrow \mathbb{Z}_3[a]$ which leave all elements of $\mathbb{Z}_3$ invariant, i.e. possible values of $\sigma(a)$ are the key here. The degree of $\mathbb{Z}_3 [ a ]$ over $\mathbb{Z}_3$ tells you how many there are. $id_{\mathbb{Z}_3 [ a ] }$ is one of them. But you don't need to use them, if you haven't read or learned about this yet. Just go on with the factorization of $x^3+2x+1$ to see whether all of your roots are already in $\mathbb{Z}_3 [ a ]$.
There is no need to think about equivalence classes anymore. $\mathbb{Z} / 3\mathbb{Z} \cong \mathbb{Z}_3 = \{0,1,2\}$ and $\mathbb{Z}_3 [ a ]$ consist of all values $p(a)$ with polynomials $p(x) \in \mathbb{Z}_3 [ x ]$, i.e. coefficients in $\{0,1,2\}$ and $a^3+2a+1=0$.
To interpret $a$ as an equivalence class leads to something of the form $\mathbb{Z}_3 [ x ] / (p(x))$. Not sure, whether this would be helpful.

7. Oct 16, 2016

### PsychonautQQ

Could I do the following:
x^3 + 2x + 1 = (x-a)(cx^2+dx+e) and then solve for the coefficients? I'm having trouble doing the long division >.< haha I hope I don't sound too helpless. I'm trying to divide x^3 + 2x + 1 by x-a but I don't know how to make things cancel out in the subtraction :-(.

8. Oct 16, 2016

### Staff: Mentor

It would probably work.
However, since it is important to learn this division (IMO), I'll show you how to do.
(Sorry, for the sub-optimal alignment.)

$(x^3+2x+1) : (x-a) \rightarrow x^2$
$-[(x-a)\cdot x^2] = -x^3+ax^2$
$\text{+________________________}$
$ax^2 + 2x +1 / : (x-a) \rightarrow ax$
$-[(x-a)\cdot ax] = -ax^2+a^2x$
$\text{+________________________}$
$2x+a^2x+1 / :(x-a) \rightarrow (2+a^2)$
$-[(x-a)\cdot (2+a^2)] = -2x+2a-a^2x+a^3$
$\text{+________________________}$
$2a+a^3+1$

Summarizing, we have $(x^3+2x+1) : (x-a) = x^2+ax+(2+a^2)$ with remainder $a^3+2a+1=0$ by definition of $a$.

9. Oct 17, 2016

### PsychonautQQ

Oh my goodness, you are a hero, thank you so much!

So now I have to see if I can make x^2 + ax + (2 + a^2) = 0 (mod 3) where x is either 0, 1, or 2 and a^3 + 2a + 1 = 0?

Is it possible that I plug in a for this polynomial and for it to be a root? This would happen if a is a repeated root I believe.

Or since there were no roots in Z3 in the original polynomial, I believe that neither 0,1, or 2 are roots of x^2 + ax + (2 + a^2) (mod 3), but now that i'm adjoining a to Z3 I also have elements 1*a,2*a, 1+a, 2+a,

Is my thinking on the right track here?

10. Oct 17, 2016

### Staff: Mentor

I don't expect it to split over $\mathbb{Z}_3$. I'd rather expect an image of $a$ under some automorphism to be also a root.
If you plug in $a$ you will get $0$ by the definition of $a$.
Yes, see above. And you also have $a^2$ and $a^3$, since all polynomials in $a$ are now part of the field.
I think so, although I would simply apply the known formula to $x^2 +ax+(2+a^2)=0$ for quadratic polynomials and see where it gets me to. Since $2$ and $4 = 1 (3)$ are both units of $\mathbb{Z}_3$, the formula doesn't make problems.

11. Oct 18, 2016

### PsychonautQQ

Wow man, thanks for the help so much! You're a great teacher. I feel like I have more tools available to me then I realize, I need to start thinking outside of the box... using the quadratic formula to get more roots seemed brilliant when you mentioned it, but I guess it was really quite obvious haha.

12. Oct 19, 2016

### PsychonautQQ

So in regards to constructing a splitting field, all I have to say is that u is a root and then from just show that by adding one root I add them all? The only definition of u that I need to give is that it is a root? I solved for u in terms of x by changing the variables i'm looking for in the quadratic (since it's a quadratic either way) and got u = (-x (+/-) 1)/2, is this necessary information or no?

13. Oct 19, 2016

### Staff: Mentor

No. Not in general. An important theorem here is the following:

Let $\mathbb{F}(\alpha_1, \dots , \alpha_n) \supset \mathbb{F}$ be a finite, algebraic extension and $\alpha_2 , \dots \alpha_n$ separable elements, i.e. their minimal polynomials have only separated and simple zeroes, then $\mathbb{F}(\alpha_1, \dots , \alpha_n) = \mathbb{F}(\beta)$ is a simple field extension.

[See that $\alpha_1$ and therewith the entire extension doesn't have to be separable! This theorem, e.g. can be used for your other example to show that $\mathbb{Q}(\sqrt{3}+i\sqrt{2})= \mathbb{Q}(\sqrt{3}\, , \, i\sqrt{2})]$
It is. It shows, that with the extension by $u$, both other roots $u-1$ and $u+1$ are automatically included as well.