On $\mathbb{Q}[x]$ find a generator of $(x^7+2x^4+x^3+x+3, x^4+1)$

Question

Prove that $\forall f,g \in \mathbb{Q}[x]$ we have $I:=(f,g)=(f+g,f-g)=:J$, is it true on $\mathbb{Z}[x]$? Then on $\mathbb{Q}[x]$ find a generator of $(x^7+2x^4+x^3+x+3, x^4+1)$

About the first point, my solution is not that elegant but I think it works:

since $f+g \in I$ and $f-g\in I$ then $J \subseteq I$. Moreover $f=\frac{1}{2} (f+g)+\frac{1}{2} (f-g)\in J$ and $g=\frac{1}{2} (f+g)-\frac{1}{2} (f-g)\in J$ then $I \subseteq J$. So $I=J$.

It doesn't work on $\mathbb{Z}[x]$: if $f=1$ and $g=2$ then $f-g=-1$ which is invertible.

Using the fact that in a commutative ring $R$ we have $(a,b)=\{ax+by\,\, |\,\, x,y\in R\}$. So about the second point should I write the generator in that way? Which means the gcd (?).

Answer 1

Hint: $(f,g)=(\gcd(f,g))$, which can be found with the Euclidean algorithm.

Answer 2

$$ \left( x^{7} + 2 x^{4} + x^{3} + x + 3 \right) $$

$$ \left( x^{4} + 1 \right) $$

$$ \left( x^{7} + 2 x^{4} + x^{3} + x + 3 \right) = \left( x^{4} + 1 \right) \cdot \color{magenta}{ \left( x^{3} + 2 \right) } + \left( x + 1 \right) $$ $$ \left( x^{4} + 1 \right) = \left( x + 1 \right) \cdot \color{magenta}{ \left( x^{3} - x^{2} + x - 1 \right) } + \left( 2 \right) $$ $$ \left( x + 1 \right) = \left( 2 \right) \cdot \color{magenta}{ \left( \frac{ x + 1 }{ 2 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( x^{3} + 2 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{3} + 2 \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( x^{3} - x^{2} + x - 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{6} - x^{5} + x^{4} + x^{3} - 2 x^{2} + 2 x - 1 \right) }{ \left( x^{3} - x^{2} + x - 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ x + 1 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{7} + 2 x^{4} + x^{3} + x + 3 }{ 2 } \right) }{ \left( \frac{ x^{4} + 1 }{ 2 } \right) } $$ $$ \left( x^{7} + 2 x^{4} + x^{3} + x + 3 \right) \left( \frac{ x^{3} - x^{2} + x - 1 }{ 2 } \right) - \left( x^{4} + 1 \right) \left( \frac{ x^{6} - x^{5} + x^{4} + x^{3} - 2 x^{2} + 2 x - 1 }{ 2 } \right) = \left( -1 \right) $$