Calculating the splitting field of a polynomial over F_13

[PsychonautQQ]

Homework Statement

Find the splitting field of x^9-1 over F_13 (the field of 13 elements)

Homework Equations

The Attempt at a Solution

Every element in the cyclic group F_13* will have order 13 since 13 is prime, and thus 1 is the only root of x^9-1 in F_13. Thus I did the long vision of Dividing (x^9-1)/(x-1) = x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 and thus in F_13 x^9 - 1 = (x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)(x-1)

Thus I was thinking that adjoining a primitive 9th root of unity to F_13 would allow x^9-1 to split completely because if c is a primitive nth root of unity then c^n-1 + c^n-2 + ... + c + 1 = 0, but then I was thinking that this is only true for if n is a prime number, and in our case 9 would be 9 so it would not be true. Now that I am typing this I am quite sure that adjoining a primitive ninth root of unity would not do the job since it's minimal polynomial (over Q at least) would be the 9th cyclotomic polynomial which is not x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1. Any of you PF genius's have advice? :D

 

[fresh_42]

Guess it's time to write a program ...
What is $3^3 (13)$ and then $3^9 (13)$ and $3^2\cdot(9^2)^4 (13)$?
I haven't split the rest, though.

 

[PsychonautQQ]

Is writing a program really the best way to do this? There's not some key insight that can help us?

 

[fresh_42]

Is writing a program really the best way to do this? There's not some key insight that can help us?

I don't know. But you come up with so many examples, that the thought is near. And to look for solutions of $a^9=1$ for $a\in \mathbb{F}_{13}$ I used a program, rather than using calc.exe or to compute it by hand. O.k., as I knew the solution this seemed pretty ridiculous, however, it didn't take my really long.
And in the example above, I get $\sqrt[3]{4} \,, \, \sqrt[3]{10}$ as additional roots. How do you prove, that one isn't contained in the extension of the other?

 

[PsychonautQQ]

I don't know. But you come up with so many examples, that the thought is near. And to look for solutions of $a^9=1$ for $a\in \mathbb{F}_{13}$ I used a program, rather than using calc.exe or to compute it by hand. O.k., as I knew the solution this seemed pretty ridiculous, however, it didn't take my really long.
And in the example above, I get $\sqrt[3]{4} \,, \, \sqrt[3]{10}$ as additional roots. How do you prove, that one isn't contained in the extension of the other?

I am not sure how to prove one is not contained in the extension of the other with cube roots. Looking for solutions to a^9 = 1 in F_13 is equivalent to finding solutions of a^9 = 1 in the cyclic group of order 12, is this correct? The units in F_13 form a cyclic group of order 13-1 = 12.

 

[fresh_42]

I am not sure how to prove one is not contained in the extension of the other with cube roots. Looking for solutions to a^9 = 1 in F_13 is equivalent to finding solutions of a^9 = 1 in the cyclic group of order 12, is this correct? The units in F_13 form a cyclic group of order 13-1 = 12.

That's the point where I regularly get confused, too. No, solutions in $\mathbb{F}_{13}$ cannot be the same as solutions in $\mathbb{Z}_{12}.$ $\mathbb{Z}_{12}$ are the units in $\mathbb{F}_{13}$, because every element except $0$ can be divided.
E.g. $3^3=27=1$ in $\mathbb{F}_{13}$ but $3^3=3$ in $\mathbb{Z}_{12}$. So if now $3^3=1$, what is $3^9$? Similar with $9^9.$ If you're looking for the splitting field of a polynomial, it's always a good idea to look whether some factors can be split and if we can factor the remaining polynomial.

$\mathbb{Z}_{12}$ comes into play, if we look for the Galois group of the $13-$th roots of unity but this isn't of interest here. Here we have to factor $x^9-1$ which has beside the obvious $1$ two other solutions in $\mathbb{F}_{13}$ and can thus be split into some factors by long polynomial division.

 

[PsychonautQQ]

That's the point where I regularly get confused, too. No, solutions in $\mathbb{F}_{13}$ cannot be the same as solutions in $\mathbb{Z}_{12}.$ $\mathbb{Z}_{12}$ are the units in $\mathbb{F}_{13}$, because every element except $0$ can be divided.
E.g. $3^3=27=1$ in $\mathbb{F}_{13}$ but $3^3=3$ in $\mathbb{Z}_{12}$. So if now $3^3=1$, what is $3^9$? Similar with $9^9.$ If you're looking for the splitting field of a polynomial, it's always a good idea to look whether some factors can be split and if we can factor the remaining polynomial.

$\mathbb{Z}_{12}$ comes into play, if we look for the Galois group of the $13-$th roots of unity but this isn't of interest here. Here we have to factor $x^9-1$ which has beside the obvious $1$ two other solutions in $\mathbb{F}_{13}$ and can thus be split into some factors by long polynomial division.

Ah, right, interesting though. So $Z_12$ is a multiplicative cyclic group in this case, not an additive one.

 

[fresh_42]

Ah, right, interesting though. So Z_{12} is a multiplicative cyclic group in this case, not an additive one.

Yes.
However, since there is only one binary operation, you're free to write it as you like. You could as well write $a ◊ b$. But usually one uses the one that fits best to our intuition: so $(\mathbb{Z}_{12}\; , \;\cdot\,)$ for units and automorphisms and $(\mathbb{Z}_{12}\; , \;+\,)$ for the hours on the clock. And of course things change, if we deal with the ring structure of $(\mathbb{Z}_{12}\; , \;+\; , \;\cdot\,)$. To change the signs here would only mean, that you're especially mean to your readers.

 

[PsychonautQQ]

Yes.
However, since there is only one binary operation, you're free to write it as you like. You could as well write $a ◊ b$. But usually one uses the one that fits best to our intuition: so $(\mathbb{Z}_{12}\; , \;\cdot\,)$ for units and automorphisms and $(\mathbb{Z}_{12}\; , \;+\,)$ for the hours on the clock. And of course things change, if we deal with the ring structure of $(\mathbb{Z}_{12}\; , \;+\; , \;\cdot\,)$. To change the signs here would only mean, that you're especially mean to your readers.

Would you be so kind to explain why the solutions to $x^9-1$ are different in $F_13$ and $Z_12$? I understand that they definitely are from the example you gave, but for some reason it's still weird to me. Is it really so simple that in $F_13$ we are working modulus 13 and in $Z_12$ we are working modulus 12? And the reason for this difference is because 0 is not a unit and so is not included in $Z_12$? Am I missing any key insights here?

 

[fresh_42]

I'm not sure, what exactly your question is. So maybe I'm telling you things you already know.

$\mathbb{F}_{13}$ and $\mathbb{Z}_{12}$ are two very different objects.

$\mathbb{F}_{13}$ is a field of characteristic $13$ and $\mathbb{Z}_{12}$ a ring with characteristic $12$ and many zero divisors.
Now, if we only look on the multiplicative structure of the field $\mathbb{F}_{13}$, then it is (always) a multiplicative group, and since $0$ is not in the multiplicative group of a field and all other elements are, it is a group with $12$ elements. E.g. $2$ generates this group, because the powers of $2$ give all elements of $\mathbb{F}_{13}^* = \mathbb{F}_{13} - \{0\}$. Thus the units of $\mathbb{F}_{13}$ are $\mathbb{F}_{13}^* \cong \mathbb{Z}_{12}\,$, the cyclic group with $12$ elements.
Since the automorphisms of $\mathbb{F}_{13}$ are also isomorphic to $\mathbb{F}_{13}^*$, we have a second interpretation of $\mathbb{Z}_{12}$ in connection with $\mathbb{F}_{13}$.

However, we are talking about an equation in one variable $x$ and coefficients in $\mathbb{F}_{13}$, namely $x^9-1=0$. This means we are looking for zeros in $\mathbb{F}_{13}$. There is no need to consider $\mathbb{Z}_{12}$ at all, neither as automorphism group, nor as the group of units. Especially we are not interested in any solutions in $\mathbb{Z}_{12}$, because $3\cdot 4= 12$ in our field $\mathbb{F}_{13}$, and $12 \neq 0$.

Is it really so simple that in $F_{13}$ we are working modulus $13$ and in $Z_{12}$ we are working modulus $12$?

Yes. And the latter makes no sense in $\mathbb{F}_{13}$.

What makes sense is, that $1^9=1\, , \,3^9=(3^3)^3=27^3=1^3=1\, , \,9^9=(3^2)^9=3^{18}=(3^3)^6=27^6=1^6=1$ in $\mathbb{F}_{13}$.
Thus we have three zeros $\{1,3,9\}$ in $\mathbb{F}_{13}$, i.e. $(x-1)(x-3)(x-9)\,\vert \,(x^9-1)$ and we have only to inspect a polynomial of degree $6$, which has to be split.

 

[PsychonautQQ]

I'm not sure, what exactly your question is. So maybe I'm telling you things you already know.

$\mathbb{F}_{13}$ and $\mathbb{Z}_{12}$ are two very different objects.

$\mathbb{F}_{13}$ is a field of characteristic $13$ and $\mathbb{Z}_{12}$ a ring with characteristic $12$ and many zero divisors.
Now, if we only look on the multiplicative structure of the field $\mathbb{F}_{13}$, then it is (always) a multiplicative group, and since $0$ is not in the multiplicative group of a field and all other elements are, it is a group with $12$ elements. E.g. $2$ generates this group, because the powers of $2$ give all elements of $\mathbb{F}_{13}^* = \mathbb{F}_{13} - \{0\}$. Thus the units of $\mathbb{F}_{13}$ are $\mathbb{F}_{13}^* \cong \mathbb{Z}_{12}\,$, the cyclic group with $12$ elements.
Since the automorphisms of $\mathbb{F}_{13}$ are also isomorphic to $\mathbb{F}_{13}^*$, we have a second interpretation of $\mathbb{Z}_{12}$ in connection with $\mathbb{F}_{13}$.

However, we are talking about an equation in one variable $x$ and coefficients in $\mathbb{F}_{13}$, namely $x^9-1=0$. This means we are looking for zeros in $\mathbb{F}_{13}$. There is no need to consider $\mathbb{Z}_{12}$ at all, neither as automorphism group, nor as the group of units. Especially we are not interested in any solutions in $\mathbb{Z}_{12}$, because $3\cdot 4= 12$ in our field $\mathbb{F}_{13}$, and $12 \neq 0$.

Yes. And the latter makes no sense in $\mathbb{F}_{13}$.

What makes sense is, that $1^9=1\, , \,3^9=(3^3)^3=27^3=1^3=1\, , \,9^9=(3^2)^9=3^{18}=(3^3)^6=27^6=1^6=1$ in $\mathbb{F}_{13}$.
Thus we have three zeros $\{1,3,9\}$ in $\mathbb{F}_{13}$, i.e. $(x-1)(x-3)(x-9)\,\vert \,(x^9-1)$ and we have only to inspect a polynomial of degree $6$, which has to be split.

Awesome, thanks for this. The automorphisms of $F_13$ are isomorphic to the group of it's units, that was very enlightening for me. It's interesting that 2 is a generator of F*_13 and that F*_13 is isomorphic to Z_12, because 2 is not a generator of Z_12. So any isomorphic mapping must take 2 to a generator of Z_12; can it take 2 to any generator of Z_12? In which case there would be 4 isomorphisms from F*13 to Z_12, one that takes 2 to each of the numbers relatively prime to 12?

 

[fresh_42]

Awesome, thanks for this. The automorphisms of $F_13$ are isomorphic to the group of it's units, that was very enlightening for me. It's interesting that 2 is a generator of F*_13 and that F*_13 is isomorphic to Z_12, because 2 is not a generator of Z_12. So any isomorphic mapping must take 2 to a generator of Z_12; can it take 2 to any generator of Z_12? In which case there would be 4 isomorphisms from F*13 to Z_12, one that takes 2 to each of the numbers relatively prime to 12?

I should have been clearer, for I think you got confused with the generators. As a group, $\mathbb{Z}_{12}$ is usually meant additive, because we regard its elements as remainders $0,\ldots , 11\,$, whereas $\mathbb{F}_{13}^*$ is meant multiplicative: the elements $1,\ldots , 12\,$ in the field which can be divided.

$2$ is coprime to $13$, which means $\{2^0,2^1,2^2,2^3,\ldots ,2^{11} \text{ in } \mathbb{F}_{13}\} =$(in another order)$\{1,2,\ldots ,12\} = \mathbb{F}_{13}^*$ and the multiplicative group $(\mathbb{F}_{13}^*, \cdot)= \langle \; 2\; , \; \cdot \; \rangle \overset{2 \mapsto 1}{\cong} (\mathbb{Z}_{12}\; , \;+ \;)$.

On the other hand is $\sigma_2 : \langle \mathbb{F}_{13},+ \rangle \longrightarrow \langle \mathbb{F}_{13},+ \rangle$ defined by $\sigma_2(a)=2a$ a group isomorphism $\sigma_2(a+b)=\sigma_2(a)+\sigma_2(b)$ of the additive group of $\mathbb{F}_{13}$, i.e. on thirteen elements.

Since this is true for every number $n$ which is coprime to $13$ (not only for $n=2$), which are all twelve numbers except zero, the automorphism group of $\langle \mathbb{F}_{13},+ \rangle$ can be written as either
$$Aut((\mathbb{F}_{13},+)) = \langle \{\sigma_2^0,\sigma_2^1, \ldots , \sigma_2^{11}\} ,\cdot \rangle \cong (\mathbb{Z}_{12},+)$$
with one generating automorphism $\sigma_2$ or as
$$Aut((\mathbb{F}_{13},+))= \langle \{\sigma_1,\sigma_2, \sigma_3, \ldots , \sigma_{12}\} ,\cdot \rangle \cong (\mathbb{Z}_{12},+)$$
with twelve automorphisms $\sigma_n$, one for each of the twelve coprime numbers $n$.
(If you like, you can try to find out, that in fact each $\sigma_n = \sigma_2^k$ for some $k$.)

Is it confusing? IMO, definitely yes. But it is as it is. If you're looking for someone to blame, take the exponential function. You remember that the roots of unity are equidistantly distributed on the unit circle, i.e. of the form $\exp(\frac{m}{n}2\pi i)$. Now the addition of angles is the addition in the exponent. But this addition is a multiplication of complex numbers: $\exp(\frac{m_1}{n}2\pi i)\,\cdot\,\exp(\frac{m_2}{n}2\pi i) = \exp(\frac{(m_1+m_2)}{n}2\pi i)$. So if we talk about groups, it should be mentioned according to which operation, because as you know: $\mathbb{F}_{13}$ is a field and has therefore both operations, $13$ elements for addition and $12$ for multiplication; and $\mathbb{Z}_{12}$ is a ring, that also has both operations, here on only $12$ elements.

 

[PsychonautQQ]

Wow, what an enlightening post, thank you so much. I got a little bit lost when you tried to explain how the field and group operations relate to adding angles through adding the exponents which is actually multiplying two complex exponential functions. I mean I understand that all is that is true, but I don't see how it relates to what we were talking about before exactly.

 

[fresh_42]

Well, an automorphism in the (multiplicative) group of $n-$th roots of unity is jumping from one angle (root) to another. You already used it, as you know that automorphisms of Galois groups are those which jump from one zero (root) to another.

Primitive roots (or generating automorphisms) are those which when constantly applied are jumps to all roots before hitting any for the second time again. So multiplication is succeeding application of automorphisms (a multiplication) or the complex multiplication of two roots: the starting point times another root (the jump).

And all these multiplications can as well be seen as adding angles, which are the exponents. It is an automorphism between a multiplicative written group and an additive written group: the same number of elements, one starting with $1$ like $\mathbb{F}_{13}^*=\{1,2,3, \ldots , 12\}$ and one starting with $0$ like $\mathbb{Z}_{12}=\{0,1,2, \ldots , 11\}$.

Please don't read the rest, it's confusing.
One can consider $\mathbb{F}_{13}^*$ as the (multiplicative) group of $12-$th roots $(\{ r_1=\exp(\frac{0}{12}2 \pi i) , \ldots , r_{12}=\exp(\frac{11}{12}2 \pi i)\}, \cdot )$, which is isomorphic to the additive group $\mathbb{Z}_{12}=(\{\frac{0}{12}, \ldots , \frac{11}{12}\}\, , \,+).$ Now this is the internal group structure of multiplication in the field $\mathbb{F}_{13}$ and / or the automorphisms of $(\mathbb{F}_{13},+)$.

But if we only look on the elements $a \in \mathbb{F}_{13}$ and ask, whether $a^9=1$ or not, which is what we want to know, then we don't need to bother any automorphisms or roots of unity at this stage. And if, then probably those of $9-$th roots of unity. And who knows how they look like in $\mathbb{F}_{13}$? So in order to find zeros of $x^9-1=0$ in $\mathbb{F}_{13}$, we are only interested in the laws that hold in our coefficient field $\mathbb{F}_{13}$.

 

[PsychonautQQ]

you're a hero
edit: Seriously though, your posts are making me smarter