Determine the greatest common divisor of polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$.

Question

Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.

Is it correct that I keep using long division until the result is $0$, and then the previous result would be a gcd?

x^2+1 / x^3 + 1 \ x
        x^3 + x
        ________-
        1 - x

1 - x / x^2 + 1 \ -x-1
        x^2 - x 
        ________-
        x + 1
        x - 1
        ________-
        2

2 / 1 - x \ -1/2x + 1/2 
      - x 
    ______-
    1
    1
    ______-
    0

Conclusion: A gcd is $2$?

I'm not sure what's meant with "Write the gcd as a combination of the given polynomials.", so it would be great if someone could point me in the right direction.

Answer 1

The Euclidean algorithm steps are \begin{align*} x^3+1&=x (x^2+1) + (-x+1)\\ x^2+1&=-x (-x+1)+ (x+1)\\ -x+1&=-(x+1)+2 \end{align*}

So the GCD is $1$.

We can go back up \begin{align*} 2 &= (-x+1) +(x+1)\\ &= (-x+1) + (x^2+1)+x (-x+1)\\ &= (-x+1) (x+1)+(x^2+1)\\ &= (x^3+1 - x(x^2+1))(x+1)+(x^2+1)\\ &= (x+1) (x^3+1) + (-x^2-x+1) (x^2+1) \end{align*}

and so $1 = \frac{1}{2} [(x+1) (x^3+1) + (-x^2-x+1) (x^2+1)]$

Answer 2

Notice that $x^2+1$ is irreducible over $\mathbb{Q}$. Let $d = \gcd(x^2+1, x^3+1)$. Then $d \mid x^2+1$ so $d$ is a nonzero constant, or $d$ is an associate of $x^2+1$.

However, $x^2+1 \not\mid x^3+1$ so $d$ must be constant. We can choose $d = 1$ since $\gcd$ is usually supposed to be monic.

Finally, notice that

$$1=\left[\frac{1}{2}(x+1) \right](x^3+1) + \left[\frac{1}{2}(-x^2-x+1)\right] (x^2+1)$$

Answer 3

$$ \left( x^{3} + 1 \right) $$

$$ \left( x^{2} + 1 \right) $$

$$ \left( x^{3} + 1 \right) = \left( x^{2} + 1 \right) \cdot \color{magenta}{ \left( x \right) } + \left( - x + 1 \right) $$ $$ \left( x^{2} + 1 \right) = \left( - x + 1 \right) \cdot \color{magenta}{ \left( - x - 1 \right) } + \left( 2 \right) $$ $$ \left( - x + 1 \right) = \left( 2 \right) \cdot \color{magenta}{ \left( \frac{ - x + 1 }{ 2 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( x \right) } \Longrightarrow \Longrightarrow \frac{ \left( x \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( - x - 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( - x^{2} - x + 1 \right) }{ \left( - x - 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - x + 1 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{3} + 1 }{ 2 } \right) }{ \left( \frac{ x^{2} + 1 }{ 2 } \right) } $$ $$ \left( x^{3} + 1 \right) \left( \frac{ - x - 1 }{ 2 } \right) - \left( x^{2} + 1 \right) \left( \frac{ - x^{2} - x + 1 }{ 2 } \right) = \left( -1 \right) $$

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