5
$\begingroup$

Let's consider the polynomial $$f(X)=X^3+aX^2-(3+a)X+1\in\mathbb{Q}[X]$$ where $a\in\mathbb Z$. Simple observations show that it is irreducible and has 3 real roots. If $\alpha$ is one root we can even see that the splitting field is $\mathbb Q(\alpha)$ since a second root is $1/(1-\alpha)$.

My question: Is there a way to write $1/(1-\alpha)$ as a linear combination of $\alpha$? And if so, does there exist a general method or trick to find it?

So far I've tried to expand the fraction until I have an integer demoninator but without any success.

$\endgroup$
  • 2
    $\begingroup$ Perhaps you could write (at least formally) $1/(1 - \alpha) = \sum_n \alpha^n$ and use the polynomial to reduce terms involving $\alpha^3$. $\endgroup$ – user296602 Apr 12 '18 at 15:13
  • $\begingroup$ See math.stackexchange.com/questions/1767252/… $\endgroup$ – lhf Apr 12 '18 at 15:49
2
$\begingroup$

Note that $\alpha$ satisfies $\alpha^3 = -a\alpha^2 + (3+a)\alpha - 1$. Under the basis $\{1,\alpha,\alpha^2\}$, multiplication by $1-\alpha$ corresponds to the matrix $$A=\begin{pmatrix} 1 & 0 & 1 \\ -1 & 1 & -3-a \\ 0 & -1 & 1+a \end{pmatrix}$$

Now find $A^{-1}$, it exists because original equation is irreducible. Let $c_0,c_1,c_2$ be entries of the 1st column of $A^{-1}$, then $$1/(1-\alpha) = c_0 + c_1 \alpha + c_2 \alpha^2 $$


Alternatively, use Euclidean alogorithm to find polynomials $r(x),s(x)$ such that $$(1-x) r(x) + f(x) s(x) = 1$$ Then $$1/(1-\alpha) = r(\alpha)$$

$\endgroup$
  • $\begingroup$ Thanks! That's exactly the answer I've been looking for. The only thing I don't understand is how the first column yields the desired combination. Could you explain that in a little more detail? $\endgroup$ – Buh Apr 12 '18 at 16:40
  • 2
    $\begingroup$ The first basis vector is $1$, so the first column corresponds to $(1-\alpha)^{-1}\cdot1$. $\endgroup$ – David Hill Apr 12 '18 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.