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Constructing the splitting field for a polynomial over Z/Z3

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!
 
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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.
You mean ##x^3 + 2x + 1##
I agree that this polynomial is prime over the given field.
PsychonautQQ said:
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 think you mean ##x^2 + 1##
PsychonautQQ said:
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?
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.
PsychonautQQ said:
Will the new field have a new character? Any insight appreciated :-)

Thanks PF!
 

fresh_42

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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 :-)
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.
 
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.
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.
 
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.
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?
 

fresh_42

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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?
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
$$\begin{align*} 238 : (-7) = (-3) ... \\ - ((-7) \cdot (-3)) \;\;\;\;\;\; \\ \text{_________________} \\ +23 - (+21) (8) \\ \text{_________________} \\ 28 \\ etc.\end{align*}$$
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.
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 ] ##.
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?
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.
 
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 :-(.
 

fresh_42

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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 :-(.
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##.
 
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##.
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?
 

fresh_42

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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?
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.
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.
If you plug in ##a## you will get ##0## by the definition of ##a##.
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,
Yes, see above. And you also have ##a^2## and ##a^3##, since all polynomials in ##a## are now part of the field.
Is my thinking on the right track here?
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.
 
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.
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.
 
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.
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?
 

fresh_42

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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?
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})]##
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?
It is. It shows, that with the extension by ##u##, both other roots ##u-1## and ##u+1## are automatically included as well.
 

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