MHB Greatest common divisor of polynomials

[evinda]

Hello!

I want to find the greatest common divisor of $x^4+1$ and $x^2+x+1$.
I applied the Euclidean division and found that $x^4+1=(x^2+x+1) \cdot (x^2-x)+(x+1)$.
So,isn't it like that: $gcd(x^4+1,x^2+x+1)=x+1$ ?
But.. in my textbook,the result is $1$! Which of the both results is right?? (Blush) (Thinking)

 

[Evgeny.Makarov]

You did not finish the Euclidean algorithm, which consists of several Euclidean divisions.

 

[evinda]

You did not finish the Euclidean algorithm, which consists of several Euclidean divisions.

Oh,yes!You are right! (Giggle) I found now that the greatest common divisor is equal to $1$! Thank you! :)

 

[chisigma]

Hello!

I want to find the greatest common divisor of $x^4+1$ and $x^2+x+1$.
I applied the Euclidean division and found that $x^4+1=(x^2+x+1) \cdot (x^2-x)+(x+1)$.
So,isn't it like that: $gcd(x^4+1,x^2+x+1)=x+1$ ?
But.. in my textbook,the result is $1$! Which of the both results is right?? (Blush) (Thinking)

If they are 'ordinary polynomials' then their factorisations are...

$\displaystyle 1 + x^{4} = (x - e^{i\ \frac{\pi}{4}})\ (x - e^{- i\ \frac{\pi}{4}})\ (x - e^{i\ \frac{3\ \pi}{4}})\ (x - e^{- i\ \frac{3\ \pi}{4}})$

$\displaystyle 1 + x + x^{2} = (x - e^{i\ \frac{2\ \pi}{3}})\ (x - e^{- i\ \frac{2\ \pi}{3}})$

... and the two polynomial have no common factors so that the g.c.d. is 1.

If they are 'integer modulo 2 polynomials' however then their factorisations are...$\displaystyle 1 + x^{4} = (1 + x)\ (1 + x)\ (1 + x + x^{2})$$\displaystyle 1 + x + x^{2} = 1 + x + x^{2}$ ... so that the g.c.d. is $\displaystyle 1 + x + x^{2}$...

Kind regards $\chi$ $\sigma$