Suppose $\alpha$ is a root of the irreducible polynomial $x^3 -2x-2 \in \mathbb{Q}[x]$.


Express $(\alpha +1)(\alpha^2+\alpha+1)^{-1}$ in terms of $\{1, \alpha, \alpha^2\}$ for $\mathbb{Q} (\alpha)$ over $\mathbb Q$.

My attempt: I tried to find $(\alpha^2+\alpha+1)^{-1}$, $(a+b\alpha+c\alpha^2)(\alpha^2+\alpha+1)=1$. Implies that $a=1,b=0$ and $c=0$ . From here I dont know what to do or did I do something wrong here?

Any help and hints would be much appreciated!


You did not do something correct there, but your intention seems to be along the right path.

Since $\alpha$ is a root of the polynomial, you know that $\alpha^3 - 2\alpha - 2 = 0$, or in other words, $\alpha^3 = 2\alpha + 2$.
This means you can express any polynomial in $\alpha$ in terms of $\{1, \alpha, \alpha^2\}$, by substituting (repeatedly, if necessary) $\alpha^3$ in powers of $\alpha$ greater than $2$.

Now, if you can express $(\alpha^2+\alpha+1)^{-1}$ as a polynomial (not a rational function) in $\alpha$, then by the observation above you are done. Of course, by the observation itself, if such a polynomial exists, it is expressible in terms of $\{1, \alpha, \alpha^2\}$. Hence, we're looking for $a$, $b$ and $c$ such that

$$(\alpha^2+\alpha+1)^{-1} = a + b\alpha + c\alpha^2 \iff 1 = (\alpha^2+\alpha+1)(a + b\alpha + c\alpha^2)$$

We can expand that to find that

$$\begin{align} 1 &= a + (a+b)\alpha + (a+b+c)\alpha^2+(b+c)\alpha^3 + c\alpha^4 \\&= a + (a+b)\alpha + (a+b+c)\alpha^2+(b+c)(2\alpha+2) + c\alpha(2\alpha+2) \\&= (a+2b+2c) + (a+3b+4c)\alpha + (a+b+3c)\alpha^2 \end{align}$$

This yields the following system of linear equations:

$$\begin{align} a+2b+2c &= 1 \\a+3b+4c &= 0 \\a+b+3c &= 0 \end{align}$$

Do you think you can take it from here?

  • $\begingroup$ Thank you very much! I already tried that before my bad it was my substitution that failed I didn't check it twice. $\endgroup$ – Tokita Ohma Jan 21 at 3:55
  • $\begingroup$ You're welcome! Glad to have helped. $\endgroup$ – Fimpellizieri Jan 21 at 14:05

Note that $\{1, \alpha, \alpha^2\}$ is a basis of $\Bbb Q(\alpha)$ over $\Bbb Q$. Thus you should try to determine $a, b c$ from $(a+b \alpha+c\alpha^2)(1+\alpha+\alpha^2)=1$. When expanding this expression you should use $\alpha^3-2\alpha-2 =0$. This results for example in $\alpha^4=\alpha \alpha^3=\alpha(2\alpha+2)=2\alpha^2+2\alpha$.


There is another technique based on the proof that $\mathbb{Q} [x] /\langle x^3-2x-2\rangle$ is a field.

Let us compute the GCD of polynomials $x^2+x+1$ and $x^3-2x-2$. We have $$x^3-2x-2=(x-1)(x^2+x+1)- 2x-1$$ and $$4(x^2+x+1)=(2x+1)^2+3$$ And then going backwards $$3=4(x^2+x+1)-(2x+1)^2=(5+x-2x^2) (x^2+x+1)+(2x+1)(x^3-2x-2)$$ Replacing $x$ by $\alpha$ we get $$\frac{1}{1+\alpha+\alpha^2}=\frac{5+\alpha-2\alpha^2}{3}$$ Next we can multiply by $(\alpha+1)$ on both sides and replace $\alpha^3$ on right side by $2\alpha+2$.

The above is a direct method for inverting any polynomial in $\alpha$ and helps in proving that $\mathbb{Q} [\alpha] $ is a field. The technique however is not any more efficient compared to other methods.


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