Writing elements of field extension in terms of the basis determined by a root of a polynomial

Question

Let $\alpha \in \mathbb{C}$ be a root of the irreducible polynomial $$f(X) = X^3 + X + 3$$

Write the elements of $\mathbb {Q}(\alpha)$ in terms of the basis $\{1, \alpha, \alpha^2\}$.

The first part is to work out $\alpha^3$ in terms of the basis, but I can't work out if I need to explicitly find the roots to calculate what $\alpha$ is or can it be answered from a relation between the basis elements?

Answer 1

Since $\alpha$ is a root of $f(X)$, by definition we have $$0 = f(\alpha) = \alpha^3 + \alpha + 3.$$ Now, solve for $\alpha^3$...

Answer 2

$\{1,\alpha,\alpha^2\}$ form a basis of $\Bbb{Q}(\alpha)$ viewed as a $\Bbb{Q}$ vector space. Now think to the analogy of $V$ viewed as a $K-$vector space, where $K$ is a field and $V$ a vector space.

In the latter example if $(v_1,\dots,v_n)$ was a basis for $V$, then every element $w$ of $V$ could have been written like $$w=\lambda_1v_1+\cdots + \lambda_nv_n,$$

where the $\lambda_i$'s where elements in $K$.

So in your example if $\{1,\alpha,\alpha^2\}$ form a basis of $\Bbb{Q}(\alpha)$ viewed as a $\Bbb{Q}$ vector space you can write every element $\beta$ of $\Bbb{Q}(\alpha)$ as $$\beta=q_1\cdot 1 + q_2 \cdot \alpha + q_3 \cdot \alpha^2,$$

where the $q_i$'s are in $\Bbb{Q}$.

Notice also that since $\alpha$ is a root of the polynomial $f(x) = x^3+x+3$, we have $f(\alpha)=\alpha^3+\alpha +3=0$, so $\alpha^3 = -\alpha -3$

(And so the dimension of $\Bbb{Q}(\alpha)$ viewed as a $\Bbb{Q}$ vector space is $3$).

Answer 3

To write the elements of $\mathbb{Q}(\alpha)$, take any polynomial $p(x) \in \mathbb{Q}[x]$ and use the Division Algorithm, i.e., there exist $q(x),r(x) \in \mathbb{Q}[x]$ such that

$$p(x) = q(x)f(x) + r(x)$$

where $r(x) = 0$ or $\partial r(x) < \partial f(x) = 3$, then notice that

$$p(\alpha) = q(\alpha)\underbrace{f(\alpha)}_{=0} + r(\alpha) = a_0 + a_1\alpha + a_2 \alpha^2$$

where $a_0,a_1,a_2 \in \mathbb{Q}$ and there you go.