Result of Euclidean division

[evinda]

Hello! (Wave)

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... (Sweating)

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?

 

[I like Serena]

Hello! (Wave)

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... (Sweating)

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! (Smile)Did you take into account that for instance $\alpha^3 = -1$?
And that $\alpha^5 = \overline \alpha$?
And that $\alpha + \overline \alpha = 1$? (Wondering)
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? (Thinking)

 

[evinda]

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

And that $\alpha + \overline \alpha = 1$? (Wondering)

How do we get these relations?

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) $ ?

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? (Thinking)

Why isn't this in $\mathbb{F}_7$ ? (Thinking)

 

[I like Serena]

How do we get these relations?

Since $\alpha$ is a primitive root of unity, isn't it:
$$\alpha = e^{\pm i \pi / 3}$$
(Wondering)
Then it would follow that $\alpha^3 = e^{i\pi} = -1$.

Those roots of unity correspond to points on the unit circle in $\mathbb C$:
View attachment 5588

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) $?

The equation $x^6 - 1 = 0$ has 6 solutions in $\mathbb C$, which are so called 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. (Thinking)

Why isn't this in $\mathbb{F}_7$ ? (Thinking)

Isn't $\mathbb F_7 = \{0,1,2,3,4,5,6\}$?
The primitive root of unity $\alpha$ is not in there... (Worried)

 

[evinda]

Since $\alpha$ is a primitive root of unity, isn't it:
$$\alpha = e^{\pm i \pi / 3}$$
(Wondering)
Then it would follow that $\alpha^3 = e^{i\pi} = -1$.

Those roots of unity correspond to points on the unit circle in $\mathbb C$:The equation $x^6 - 1 = 0$ has 6 solutions in $\mathbb C$, which are so called 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 $\alpha^i \neq \alpha^j, i \neq j, 1 \leq i, j \leq 6$. (Sweating)

Isn't $\mathbb F_7 = \{0,1,2,3,4,5,6\}$?
The primitive root of unity $\alpha$ is not in there... (Worried)

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

 

[I like Serena]

I thought that we get from the fact that $\alpha$ is a $6$-th root of unity that $\alpha^6=1$ and $\alpha^i \neq \alpha^j, i \neq j, 1 \leq i, j \leq 6$. (Sweating)

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

Hold on! (Wait)

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 $3^6=1$. (Emo)

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

 

[evinda]

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?

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

I see...

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

Why is it a 2-nd root of unity?

 

[I like Serena]

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. (Sweating)

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 $x=\pm 1$. (Whew)

 

[evinda]

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

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 $x=\pm 1$. (Whew)

Ah I see... Do we use this at my initial calculations or at the calulations that you did?

 

[I like Serena]

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 $x-\alpha^{a+2}$... (Thinking)
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. (Worried)

 

[evinda]

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 $x-\alpha^{a+2}$... (Thinking)

So we divide $x^5 + \alpha^{a+1}x^4 + \alpha^{2(a+1)}x^3 + \alpha^{3(a+1)}x^2 + \alpha^{4(a+1)}x + \alpha^{5(a+1)}$ by $x- \alpha^{a+2}$ , right?I got that $$ x^5 + \alpha^{a+1}x^4 + \alpha^{2(a+1)}x^3 + \alpha^{3(a+1)}x^2 + \alpha^{4(a+1)}x + \alpha^{5(a+1)}=(x- \alpha^{a+2}) (x^4+ \alpha^{a+1}(\alpha+1) x^3+ \alpha^{2a+2}[\alpha^2+ \alpha-1] x^2+ \alpha^{3a+3}[1+ \alpha [\alpha^2+ \alpha-1]]x+ \alpha^{4a+4}[1+ \alpha(1+ \alpha(\alpha^2+\alpha-1) )])+ \alpha^{5a+5}[\alpha [1+ \alpha (1+ \alpha(\alpha^2+ \alpha+1))]]$$

Is it again wrong? (Sweating)

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. (Worried)

Ok, I will retry them... (Nerd)

 

[evinda]

Again I get the same result as before... (Sweating)

Also why does it hold that $1+\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5$ is equal to $0$ ? (Thinking)

 

[evinda]

So what did you find $\frac{x^6-1}{x^2- \alpha^{a+1} (\alpha+1) x+ \alpha^{2a+3}}$ to be equal to?

 

[I like Serena]

Also why does it hold that $1+\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5$ is equal to $0$ ? (Thinking)

Isn't it a geometric series? What would its sum be? (Wondering)

 

[I like Serena]

So what did you find $\frac{x^6-1}{x^2- \alpha^{a+1} (\alpha+1) x+ \alpha^{2a+3}}$ to be equal to?

Let $B=\alpha^{a+1}(\alpha+1)$ and $C=\alpha^{2a+3}$.

Then:
\begin{aligned}x^6-1 &= \Big(x^2-Bx+C\Big)\
\Big(x^4 + Bx^3 + (B^2-C)x^2 + B(B^2-2C)x + (B^4 - 3B^2C + C^2)\Big) \\
&+ \Big[B(B^4-3B^2C+C^2) - BC(B^2-2C)\Big]x - C(B^4-3B^2C+C^2) - 1
\end{aligned}

I veried that $- C(B^4-3B^2C+C^2) - 1 = 0$ by filling in $\alpha$ and making use of its properties.
I also filled in $\alpha$ in $Bx^3$ and in $(B^2-C)x^2$ and found the same thing you did.