# Expressing a polynomial in terms of basis

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

PROBLEM:

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?

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