Since $\alpha^3=2$, $\alpha^4=\alpha^3\cdot\alpha=2\alpha$.
Inverses require a lot more cleverness. Here's one possible approach. Note that $X$ is a finite-dimensional vector space over the field $\mathbb{Q}$, and that for any $x\in X$, the map $\mu_x(y)=xy$ is a $\mathbb{Q}$-linear map $X\to X$. If $x\not=0$, furthermore, $\mu_x$ is injective. But any injective linear map from a finite-dimensional vector space to itself is also surjective. It follows that $1$ is in the image of $\mu_x$, which says exactly that $x$ has an inverse.
(In principle, using Cramer's rule to compute the inverse of the linear map $\mu_x$, you can use this argument to explicitly write down a formula for $x^{-1}$, but it will be quite complicated!)
Your comment has one grave mistake in the second equation, which would have made life easier.
You should have the pair of equations :
$$
\begin{cases}
0 = a^2d^2 - 7b^2c^2 + abd^2s + tb^2d^2 \\
0 = 2abcd + b^2cds = 2a + bs \tag{*}
\end{cases}
$$
(you made a mistake in the comment, and let us assume $c \neq 0$, because the first part of your comment covers the case of it being zero i.e which real numbers are in the algebraic closure) which sounds like plenty to work with. However, after the mistake in the comment you get $s = \frac{-2a}{b}$, which when you put in the first equation gives you after simplification that $t = \frac{a^2}{b^2} - \frac{7c^2}{d^2}$.
But, $s$ and $t$ are integers, because $f$ was chosen to have integer coefficients! Thus, the above two quantities must be integers.
Now we will go in hidden steps, because hide and seek is my (second) favourite game.
So $s = \frac{-2a}{b}$ is an integer. Then, $-2a = bs$. Recall $a$ and $b$ are chosen coprime.
- So $b$ divides $-2a$. What can you conclude from coprimality?
$b$ must divide $2$.
- Write down a set consisting of every possible value that $b$ can take.
$b$ can be any of $2,1,-1,-2$ i.e. a divisor of $2$.
Look at the answers before going to the next section.
Now note that $t = \frac{a^2}{b^2} - \frac{7c^2}{d^2}$ is an integer. Recall $c,d$ are co prime.
- Suppose $|b| = 1$. Then $d^2$ divides $7c^2$(why?) so what values can $d$ take?
Well, $d^2$ must divide $7$, but that forces $d = \pm 1$.
- Suppose $|b| = 2$. Conclude that $d^2$ divides $28$. What values can $d$ take now?
Well, we have $a^2-4t = \frac{28c^2}{d^2}$ after rearrangement, and the RHS must be an integer, now the claim follows from coprimality and $d$ may be any of $2,1,-1,-2$.
- Show in fact that $|d| = 2$ if $|b| = 2$, by noticing something extra in the argument in the above yellow box.
Well, we have $28$ divides $d^2(a^2 - 4t)$. However, $b$ is coprime to $a$, so $a$ is odd, so $a^2 - 4t$ is odd, and therefore $d^2$ is even, hence $d$ is even so $|d| = 2$.
Thus , we conclude that an algebraic integer is either of the form $a + c\alpha$ or $\frac{a+c \alpha}{2}$ , where $a,c$ are integers in the former case, and odd integers in the latter case. Let us show that these are sufficient conditions.
For this, we must show that every element of any of the above two forms , call it $z$, satisfies some
polynomial.
First, I will do a preliminary step.
- Let $a,c$ be odd integers. Show that $a^2 + 7c^2$ is a multiple of $4$.
Write $a^2 + 7c^2 = (a+c)(a-c) + 8c^2$, and recognize the RHS as the sum of two multiples of $4$.
- Show that $z + \bar z$ and $z \bar{z}$ are both integers, where $\bar z$ is the conjugate of $z$. (Note that the conjugate of $p+q\alpha$ is $p-q\alpha$ for $p,q $ real).
We have $a -c\alpha + a + c \alpha = 2a$, so clearly $z+\bar z$ is integral in both cases. Their product is $a^2 + 7c^2$ in the first case, of course an integer, and $\frac{a^2+7c^2}{4}$ in the second case, which is an integer by the first step.
Conclude that if $z + \bar z = -s$ and $z\bar z = t$ then $z^2 + sz+t = 0$. Consequently, $z$ satisfies a polynomial with integer coefficients.
Easy verification, let $f(x) = x^2 + sx+t$, then $f(z) = 0$.
This concludes the description of the algebraic closure. Is there a better description.
Exercise : Show that the algebraic closure can be expressed as $\{a + bl : a,b \in \mathbb Z\}$ where $l = \frac{1+\alpha}{2}$. Thus, the algebraic closure is $\mathbb Z[\frac{1+\alpha}{2}]$
Best Answer
Let $L$ be the $\Bbb Q$-span of $1$, $\alpha$ and $\alpha^2$. Then $L$ is a ring, and also a three-dimensional $\Bbb Q$-vector space. If $\beta=p+q\alpha+r\alpha^2$ is a nonzero element of $L$, then $f:u\mapsto\beta u$ is a $\Bbb Q$-linear map from $L$ to $L$. As the real numbers form a field, $f$ is injective. By rank-nullity, $f$ is surjective, so there is $u\in L$ with $f(u)=1$. Then $u=1/\beta$.