MHB Did I Make an Error in My Euclidean Division Calculation?

AI Thread Summary
The discussion revolves around the Euclidean division of the polynomial \(x^6-1\) by \(x^2 - \alpha^{a+1}(\alpha+1)x + \alpha^{2a+3}\) over the field \(\mathbb{F}_7\), where \(\alpha\) is a primitive 6th root of unity. Participants analyze the calculations and identify potential errors, particularly in the treatment of roots and the properties of \(\alpha\). It is clarified that \(\mathbb{F}_7\) does indeed have a primitive 6th root of unity, specifically \(\alpha = 3\) or \(\alpha = 4\). The conversation also touches on the simplification of the polynomial division and the significance of the remainder being zero. Ultimately, the participants aim to verify their calculations and understand the underlying algebraic structures involved.
evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
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?
 
Mathematics news on Phys.org
evinda said:
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)
 
I like Serena said:
Did you take into account that for instance $\alpha^3 = -1$?

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

How do we get these relations?
I like Serena said:
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) $ ?

I like Serena said:
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)
 
evinda said:
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)
 

Attachments

  • sixth-roots-of-unity.JPG
    sixth-roots-of-unity.JPG
    12.4 KB · Views: 104
I like Serena said:
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)
I like Serena said:
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 $m \in \mathbb{N}$ ?
 
evinda said:
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)
 
I like Serena said:
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?

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

I see...

I like Serena said:
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?
 
evinda said:
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)
 
I like Serena said:
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?
 
  • #10
evinda said:
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)
 
  • #11
I like Serena said:
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)
I like Serena said:
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)
 
  • #12
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)
 
  • #13
So what did you find $\frac{x^6-1}{x^2- \alpha^{a+1} (\alpha+1) x+ \alpha^{2a+3}}$ to be equal to?
 
  • #14
evinda said:
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)
 
  • #15
evinda said:
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.
 
Back
Top