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.

  • 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

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 $$1/(1-\alpha) = r(\alpha)$$

  • $\begingroup$ 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

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