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The polynomial $p(x):=x^{3}-2x-2\in\mathbb{Q}[x]$ is irreducible hence $\mathbb{K=Q}[x]/\langle p(x)\rangle$ is a field, let $\theta\in\mathbb{K}$ be a root of $p(x)$.

Can someone please help me find the inverse of the element $\theta^{2}+\theta+1$ in $\mathbb{Q}(\theta)$ ?

I know I should somehow use the euclidean algorithm, but I don't know how (also, why can I do this in general to compute the inverse of elements in $\mathbb{Q}(\theta)$ ?)

Help is appreciated!

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up vote 5 down vote accepted

This is the answer using Euclid's algorithm:

In $\mathbb{Q}[x]$ , apply the algorithm on $x^3-2x-2$ and $x^2+x+1$. Since the first is irreducible, and does not divide the second, the gcd is 1. Thus working back through the algorithm, we get some polynomials $f$ and $g$ such that $f(x)(x^3-2x-2)+g(x)(x^2+x+1) = 1$. Substituting, $g(\theta)(\theta^2+\theta+1) = 1$, so that is your inverse. Notice that this works in general, as long as the irreducible you quotient by does not divide the other polynomial, that is the given polynomial has non-zero image under the quotienting.

We have $$(x^3-2x-2) = (x-1)(x^2+x+1) +(-2x-1) $$

$$(x^2+x+1) = (-\frac{x}2-\frac14)(-2x-1)+\frac34$$

So working backwards, $$ 1 = \frac43\left[\left(x^2+x+1\right) - \left(-\frac{x}2-\frac14\right)\left(-2x-1\right)\right]\\ = \frac43\left[\left(x^2+x+1\right) - \left(-\frac{x}2-\frac14\right)\left((x^3-2x-2) - (x-1)(x^2+x+1)\right)\right]\\ = \frac13\left[(2x+1)(x^3-2x-2) + (-2x^2+x+5)(x^2+x+1)\right] $$

Substituting, the answer is $\frac13(-2\theta^2+\theta+5)$.

Alternate method:

Notice that all elements of $\mathbb{Q}(\theta)$ have an unique representative of degree $\leq 2$. So assume that the inverse is $a\theta^2+b\theta+1$ (the constant terms must be inverses). Multiply out by $(\theta^2+\theta+1)$, set equal to 1, and solve for $a$ and $b$.

Alternate method 2:

The extension is degree 3, therefore for any element $\alpha$, it's powers $1,\alpha,\alpha^2,\alpha^3$ must be linearly dependent over $\mathbb{Q}$. Substitute for $\alpha$ the given polynomial, and find an explicit linear dependence. This will be some cubic polynomial in $\alpha$ that vanishes. Now factor out an appropriate power of $\alpha$ so that the constant term is non-zero. Then just take that to one side and write the rest of the polynomial as $\alpha \times (\text{some quadratic in} \ \alpha)$. This gives you the inverse of $\alpha$, as the right hand side is an unit.

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Can you please show me hwo to use Euclid's algorithms to compute $f$ and $g$ ? – Belgi Sep 9 '12 at 15:58

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