Constructing the splitting field for a polynomial over Z/Z3

In summary, the conversation discusses constructing a splitting field for the polynomial x^3+2x+1 over Z/Z3, but it is determined that the polynomial is prime over the given field. The conversation also mentions finding an extension for the field in order for the polynomial to be factorable, but it is not a simple matter to factor the given cubic. The possibility of using automorphisms to find the other roots is also discussed. The conversation concludes with the suggestion to divide x^3+2x+1 by x-a and solve for the coefficients, but the individual is having trouble with the long division process.
  • #1
PsychonautQQ
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Homework Statement


Construct a splitting s for the polynomial x^3+2x+1 over Z/Z3

Homework Equations


4=1 Mod 3
:P

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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  • #2
PsychonautQQ said:

Homework Statement


Construct a splitting s for the polynomial x^3+2x+1 over Z/Z3

Homework Equations


4=1 Mod 3
:P

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!
 
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  • #3
PsychonautQQ said:
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.
 
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  • #4
fresh_42 said:
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.
 
  • #5
Mark44 said:
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?
 
  • #6
PsychonautQQ said:
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 ] ##.
PsychonautQQ said:
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.
 
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  • #7
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
PsychonautQQ said:
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##.
 
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  • #9
fresh_42 said:
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?
 
  • #10
PsychonautQQ said:
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.
 
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  • #11
fresh_42 said:
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.
 
  • #12
fresh_42 said:
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?
 
  • #13
PsychonautQQ said:
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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1. What is a splitting field for a polynomial over Z/Z3?

A splitting field for a polynomial over Z/Z3 is a field extension that contains all the roots of the polynomial. This means that the polynomial can be factored completely into linear factors in this field.

2. Why is constructing a splitting field important?

Constructing a splitting field is important because it allows us to find all the roots of a polynomial, which is necessary for solving equations and understanding the behavior of the polynomial.

3. How do you construct a splitting field for a polynomial over Z/Z3?

To construct a splitting field for a polynomial over Z/Z3, we first find the irreducible factors of the polynomial in Z/Z3. Then, we adjoin a root of each irreducible factor to Z/Z3 to create a new field. This process is repeated until all the roots of the polynomial are included in the field.

4. Can the splitting field for a polynomial over Z/Z3 be unique?

No, the splitting field for a polynomial over Z/Z3 is not unique. This is because there can be multiple ways to factor a polynomial into irreducible factors, leading to different splitting fields.

5. How do you know when you have constructed the complete splitting field for a polynomial over Z/Z3?

We know that we have constructed the complete splitting field when the polynomial is completely factored into linear factors in the field and there are no more roots to be adjoined. This means that the field contains all the roots of the polynomial.

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