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  1. MHB Master
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    #1
    Hello!!!

    I have applied a lot of times the euclidean division of $x^6-1$ with $x^2- \alpha^{a+1} (\alpha+1)x+ \alpha^{2a+3}, a \geq 0$, $\alpha$ a primitive $6$-th root of unity. But I don't get the right result...

    We are over $\mathbb{F}_7$.

    I got that $x^6-1=(x^2- \alpha^{a+1}(\alpha+1) x+ \alpha^{2a+3}) (x^4+ \alpha^{a+1} (\alpha+1) x^3+ \alpha^{2a+2}[(\alpha+1)^2- \alpha] x^2+ \alpha^{3a+3}(\alpha+1)[\alpha^2+1]x+ \alpha^{4a}[\alpha^4(\alpha+1)^2(\alpha^2+1)-[(\alpha+1)^2- \alpha]])+ \alpha^{5a}(\alpha+1)[\alpha^2+ \alpha+1+\alpha^5]x-2 \alpha^4-2\alpha^3-2 \alpha^8-\alpha^7+\alpha^4+\alpha^3-1$.

    But the rest should be $0$. Have I done something wrong?

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    #2
    Quote Originally Posted by evinda View Post
    Hello!!!

    I have applied a lot of times the euclidean division of $x^6-1$ with $x^2- \alpha^{a+1} (\alpha+1)x+ \alpha^{2a+3}, a \geq 0$, $\alpha$ a primitive $6$-th root of unity. But I don't get the right result...

    We are over $\mathbb{F}_7$.

    I got that $x^6-1=(x^2- \alpha^{a+1}(\alpha+1) x+ \alpha^{2a+3}) (x^4+ \alpha^{a+1} (\alpha+1) x^3+ \alpha^{2a+2}[(\alpha+1)^2- \alpha] x^2+ \alpha^{3a+3}(\alpha+1)[\alpha^2+1]x+ \alpha^{4a}[\alpha^4(\alpha+1)^2(\alpha^2+1)-[(\alpha+1)^2- \alpha]])+ \alpha^{5a}(\alpha+1)[\alpha^2+ \alpha+1+\alpha^5]x-2 \alpha^4-2\alpha^3-2 \alpha^8-\alpha^7+\alpha^4+\alpha^3-1$.

    But the rest should be $0$. Have I done something wrong?
    Hey evinda!!


    Did you take into account that for instance $\alpha^3 = -1$?
    And that $\alpha^5 = \overline \alpha$?
    And that $\alpha + \overline \alpha = 1$?



    Oh, and how about writing it as:
    $$\frac{x^6-1}{x^2- \alpha^{a+1} (\alpha+1)x+ \alpha^{2a+3}}
    = \frac{(x-1)(x+1)(x^2-x+1)(x^2+x+1)}{(x-\alpha^{a+1})(x-\alpha^{a+2})}
    = \frac{(x-1)(x+1)(x-\alpha)(x-\overline\alpha)(x-\alpha^2)(x-\overline\alpha^2)}{(x-\alpha^{a+1})(x-\alpha^{a+2})}
    $$


    Now we can see that for instance with $a=1$, it simplifies to:
    $$(x-1)(x-\alpha)(x-\overline\alpha)(x-\overline\alpha^2) = (x-1)(x^2-x+1)(x-\overline\alpha^2)
    $$
    Btw, that is not in $\mathbb F_7$ is it?

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    #3 Thread Author
    Quote Originally Posted by I like Serena View Post

    Did you take into account that for instance $\alpha^3 = -1$?

    And that $\alpha + \overline \alpha = 1$?
    How do we get these relations?


    Quote Originally Posted by I like Serena View Post
    Oh, and how about writing it as:
    $$\frac{x^6-1}{x^2- \alpha^{a+1} (\alpha+1)x+ \alpha^{2a+3}}
    = \frac{(x-1)(x+1)(x^2-x+1)(x^2+x+1)}{(x-\alpha^{a+1})(x-\alpha^{a+2})}
    = \frac{(x-1)(x+1)(x-\alpha)(x-\overline\alpha)(x-\alpha^2)(x-\overline\alpha^2)}{(x-\alpha^{a+1})(x-\alpha^{a+2})}
    $$
    Why does it hold that $(x-1)(x+1)(x^2-x+1)(x^2+x+1)=(x-1)(x+1)(x-\alpha)(x-\overline\alpha)(x-\alpha^2)(x-\overline\alpha^2) $ ?

    Quote Originally Posted by I like Serena View Post
    Now we can see that for instance with $a=1$, it simplifies to:
    $$(x-1)(x-\alpha)(x-\overline\alpha)(x-\overline\alpha^2) = (x-1)(x^2-x+1)(x-\overline\alpha^2)
    $$
    Btw, that is not in $\mathbb F_7$ is it?
    Why isn't this in $\mathbb{F}_7$ ?

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    #4
    Quote Originally Posted by evinda View Post
    How do we get these relations?
    Since $\alpha$ is a roots of unity.
    The solutions are $1, \alpha, \alpha^2, ..., \alpha^5$.
    It means that we can write:
    $$x^6 - 1 = (x-1)(x-\alpha)(x-\alpha^2)...(x-\alpha^5)$$
    We can verify that the stated equality holds.


    Quote Quote:
    Why isn't this in $\mathbb{F}_7$ ?
    Isn't $\mathbb F_7 = \{0,1,2,3,4,5,6\}$?
    The primitive root of unity $\alpha$ is not in there...

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    #5 Thread Author
    Quote Originally Posted by I like Serena View Post
    Since $\alpha$ is a roots of unity.
    I thought that we get from the fact that $\alpha$ is a $6$-th root of unity that $\alpha^6=1$ and $\alpha^i \neq \alpha^j, i \neq j, 1 \leq i, j \leq 6$.


    Quote Originally Posted by I like Serena View Post

    Isn't $\mathbb F_7 = \{0,1,2,3,4,5,6\}$?
    The primitive root of unity $\alpha$ is not in there...
    If we take a field of the form $\mathbb{F}_{7^m}$ for some $m \in \mathbb{N}$ ?

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    #6
    Quote Originally Posted by evinda View Post
    I thought that we get from the fact that $\alpha$ is a $6$-th root of unity that $\alpha^6=1$ and $\alpha^i \neq \alpha^j, i \neq j, 1 \leq i, j \leq 6$.

    If we take a field of the form $\mathbb{F}_{7^m}$ for some $m \in \mathbb{N}$ ?
    Hold on!

    Actually, $\mathbb{F}_{7}$ does have a primitive 6-th root of unity.
    We have either $\alpha=3$ or $\alpha=4$.
    Note that the numbers $3^k$ are all distinct for $1 \le k \le 6$, and $3^6=1$.

    Btw, we still get $\alpha^3 = -1$, since it's a primitive 2-nd root of unity.

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    #7 Thread Author
    Quote Originally Posted by I like Serena View Post

    Actually, $\mathbb{F}_{7}$ does have a primitive 6-th root of unity.
    We have either $\alpha=3$ or $\alpha=4$.
    How did you deduce this?

    Quote Originally Posted by I like Serena View Post

    Note that the numbers $3^k$ are all distinct for $1 \le k \le 6$, and $3^6=1$.
    I see...

    Quote Originally Posted by I like Serena View Post

    Btw, we still get $\alpha^3 = -1$, since it's a primitive 2-nd root of unity.
    Why is it a 2-nd root of unity?

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    #8
    Quote Originally Posted by evinda View Post
    How did you deduce this?

    I see...
    By checking all numbers in $\mathbb F_7$, to see if they satisfied the criterion for a primitive root of unity.


    Quote Quote:
    Why is it a 2-nd root of unity?
    Because $(\alpha^3)^2 = 1$, so $x=\alpha^3$ satisfies $x^2=1$.
    We also know that $\alpha^3 \ne 1$, which follows from the fact that $\alpha$ is a primitive 6-th root of unity.
    And finally, because the only numbers that satisfy $x^2=1$ are $x=\pm 1$.

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    #9 Thread Author
    Quote Originally Posted by I like Serena View Post
    By checking all numbers in $\mathbb F_7$, to see if they satisfied the criterion for a primitive root of unity.




    Because $(\alpha^3)^2 = 1$, so $x=\alpha^3$ satisfies $x^2=1$.
    We also know that $\alpha^3 \ne 1$, which follows from the fact that $\alpha$ is a primitive 6-th root of unity.
    And finally, because the only numbers that satisfy $x^2=1$ are $x=\pm 1$.
    Ah I see... Do we use this at my initial calculations or at the calulations that you did?

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    #10
    Quote Originally Posted by evinda View Post
    Ah I see... Do we use this at my initial calculations or at the calculations that you did?
    Well... if we divide $x^6-1$ by $x-\alpha^{a+1}$ first, we get:
    $$
    \frac{x^6-1}{x-\alpha^{a+1}}
    = x^5 + \alpha^{a+1}x^4 + \alpha^{2(a+1)}x^3 + \alpha^{3(a+1)}x^2 + \alpha^{4(a+1)}x + \alpha^{5(a+1)}
    +\frac{\alpha^{6(a+1)} - 1}{x-\alpha^{a+1}}
    $$
    Note that the remainder term $\alpha^{6(a+1)} - 1=0$, because $\alpha$ is a 6-th root of unity.

    That leaves dividing what we found by $x-\alpha^{a+2}$...



    Alternatively, if I redo your calculation, I get the same terms for $x^4$, $x^3$, and $x^2$, but a different term for $x$.
    Mine is $\alpha^{a+1}(\alpha^{2a+2} - 2\alpha^{a+2}) x = \alpha^{2a+3}(\alpha^a - 2)x$.
    Oh, and my final remainder term has a constant of $1+\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5$, which is indeed $0$. It suggests that my calculation is correct.

    Perhaps there is a mistake in your calculation.

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