Find a generator of the ideal $I = ⟨x^4 + x^3 − 3x^2 − 5x − 2, x^3 + 3x^2 − 6x − 8⟩ ⊆ \mathbb{Q}[x]$

Question

Recall that every ideal of $\mathbb{Q}[x]$ is a principal ideal. Find a generator of the ideal $I = ⟨x^4 + x^3 − 3x^2 − 5x − 2, x^3 + 3x^2 − 6x − 8⟩ ⊆ \mathbb{Q}[x]$.

My general idea was that in order for something to be a generator of $I$ it must be able to generate both $x^4 + x^3 − 3x^2 − 5x − 2$ and $x^3 + 3x^2 − 6x − 8$ If you were able to find a common factor between the two of them, the multiplication of those factors would produce the generator of the ideal. I'm not sure if I am on the right track.

Answer 1

To summarise the comments, by using the Euclidean algorithm we compute that $$ d=gcd(f,g)=x^2-x-2=(x-2)(x+1). $$ So the ideal $I$ is generated by $d$. Note that your $x-1$ is not a linear factor.

Answer 2

This problem is simple because it is easy to decompose into factors the polynomials: $$x^4 + x^3 − 3x^2 − 5x − 2= (x-2)(x+1)^3\\ x^3 + 3x^2 − 6x − 8=(x-2)(x+1)(x+4)$$

We have $(x-2)(x+1)$ a common factor. So any polynomial in the ideal is divisible by $(x-2)(x+1)$, in other words, the ideal is contained in the ideal generated by $(x-2)(x+1)$. To show that it is equal, we need to show that $(x-2)(x+1)$ is in the ideal. Equivalently, we need to show that $1$ is in the ideal generated by $(x+1)^2$ and $(x+4)$, which is true-- for instance we have $$(x+1)^2=(x+4-3)^2= (x+4)^2-6(x+4)+9$$ so
$$1 = \frac{1}{9}(x+1)^2-(\frac{1}{9}(x+4)-\frac{2}{3})(x+4)$$

Answer 3

You have good answers; this is just about an algorithmic approach for the extended Euclidean algorithm that mimics continued fractions

$$ \left( x^{4} + x^{3} - 3 x^{2} - 5 x - 2 \right) $$

$$ \left( x^{3} + 3 x^{2} - 6 x - 8 \right) $$

$$ \left( x^{4} + x^{3} - 3 x^{2} - 5 x - 2 \right) = \left( x^{3} + 3 x^{2} - 6 x - 8 \right) \cdot \color{magenta}{ \left( x - 2 \right) } + \left( 9 x^{2} - 9 x - 18 \right) $$ $$ \left( x^{3} + 3 x^{2} - 6 x - 8 \right) = \left( 9 x^{2} - 9 x - 18 \right) \cdot \color{magenta}{ \left( \frac{ x + 4 }{ 9 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( x - 2 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x - 2 \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ x + 4 }{ 9 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{2} + 2 x + 1 }{ 9 } \right) }{ \left( \frac{ x + 4 }{ 9 } \right) } $$ $$ \left( x^{2} + 2 x + 1 \right) \left( \frac{ 1}{9 } \right) - \left( x + 4 \right) \left( \frac{ x - 2 }{ 9 } \right) = \left( 1 \right) $$ $$ \left( x^{4} + x^{3} - 3 x^{2} - 5 x - 2 \right) = \left( x^{2} + 2 x + 1 \right) \cdot \color{magenta}{ \left( x^{2} - x - 2 \right) } + \left( 0 \right) $$ $$ \left( x^{3} + 3 x^{2} - 6 x - 8 \right) = \left( x + 4 \right) \cdot \color{magenta}{ \left( x^{2} - x - 2 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x^{2} - x - 2 \right) } $$ $$ \left( x^{4} + x^{3} - 3 x^{2} - 5 x - 2 \right) \left( \frac{ 1}{9 } \right) - \left( x^{3} + 3 x^{2} - 6 x - 8 \right) \left( \frac{ x - 2 }{ 9 } \right) = \left( x^{2} - x - 2 \right) $$