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
EDIT: As pointed out in the comments, I swapped $q_i$ and $p_i$. After this edit, this change is at least consistent throughout.
First, let $K = \mathbb Q(\sqrt[q_1]{p_1}, \dots, \sqrt[q_n]{p_n})$, $\alpha = \sum \sqrt[q_i]{p_i}$. $\mathbb Q$ has characteristic 0, so this extension $K/\mathbb Q$ is separable. We can then consider a field extension $L/K$ called the Galois closure of $K/\mathbb Q$. We require that $L/\mathbb Q$ is the smallest Galois extension containing $\mathbb Q$. Essentially, we just adjoin all the conjugates of the $\sqrt[q_i]{p_i}$. We can describe it even more explicitly, as the conjugates of $\sqrt[q_i]{p_i}$ are of the form $\zeta^k \sqrt[q_i]{p_i}$ where $\zeta$ is a primitive $q_i^{th}$ root of unity. Then $L = \mathbb Q(\sqrt[q_1]{p_1}, \dots, \sqrt[q_n]{p_n}, \zeta_1, \dots, \zeta_n)$ where $\zeta_i$ is a primitive $q_i^{th}$ root of unity.
Anyways, the point of doing this is that our initial extension $K/\mathbb Q$ was not Galois (unless all $q_i = 2$), so we are trying to replace it with the Galois closure and work within that. The reason we want a Galois extension is the following:
Proof: Let $H = G(K/F(\alpha))$. For $\sigma \in G(K/F)$, $\sigma(\alpha)=\alpha$ iff $\sigma \in H$. Then $[K:F(\alpha)] = |H|$ and $F(\alpha) = K$ iff $|H| = 1$.
Now, our extension $K/\mathbb Q$ is not Galois so we cannot immediately try to apply this result. However, it gives us the idea of applying every element of $G(L/\mathbb Q)$ to $\alpha$ and observing where it is fixed. To do this, we need to understand this Galois group $G = G(L/\mathbb Q)$ a bit better so let me first define $K_i = \mathbb Q(\sqrt[q_i]{p_i})$ and $L_i = K_i(\zeta_i)$ be the Galois closure. Then $K = K_1 \cdots K_n$ and $L = L_1 \cdots L_n$. We'll work first with these slices $L_i/K_i/\mathbb Q$.
Let's first consider $G_i = G(L_i/\mathbb Q)$. I won't show this, but $G_i$ is generated by the following two elements:
$$ \sigma_i: \begin{cases} \sqrt[q_i]{p_i} \mapsto \zeta_i \sqrt[q_i]{p_i}\\ \zeta_i \mapsto \zeta_i \end{cases} \\ \tau_i: \begin{cases} \sqrt[q_i]{p_i} \mapsto \sqrt[q_i]{p_i}\\ \zeta_i \mapsto \zeta_i^{k_i} \end{cases} $$
where $k_i$ and $q_i$ are relatively prime. In fact, by doing some conjugation computations you can show that every element of $G_i$ is of the form $\sigma_i^{a_i} \tau_i^{b_i}$ and that this expression is unique for $a_i$ modulo $q_i$ and $b_i$ modulo $\phi(q_i) = q_i - 1$. Essentially, we are realizing $G_i$ as a semidirect product of $\langle \sigma_i \rangle$ and $\langle \tau_i \rangle$.
What then is the fixed field of $K$? Well if we let $\phi \in G(L/\mathbb Q)$ then $\phi$ is determined by its restrictions $\phi|_{L_i} \in G_i$ as $L = L_1 \cdots L_n$. Each $\phi|_{L_i} \in G_i$, so it can be written uniquely as $\sigma_i^{a_i} \tau_i^{b_i}$. With this representation, we can see that $\phi$ fixes $K_i$ iff $\phi|_{L_i}$ is purely a power of $\tau_i$. Thus, $\phi \in G(L/K)$ iff $\phi|_{L_i}$ is a power of $\tau_i$ for all $i$.
We are trying to show that $\mathbb Q(\alpha) = K$. By Galois theory, this is the same as showing that $G(L/K) = G(L/\mathbb Q(\alpha))$. All the work we did above allowed us to figure out what $G(L/K)$ is. Furthermore, as $\mathbb Q(\alpha) \subseteq K$, we have $G(L/K) \subseteq G(L/\mathbb Q(\alpha))$. Take now some $\phi \in G(L/\mathbb Q(\alpha))$. Then by definition, $\phi(\alpha) = \alpha$. We can rewrite this as $\sum \zeta_i^{m_i}\sqrt[q_i]{p_i} = \sum \sqrt[q_i]{p_i}$ for some $m_i \in \mathbb Z$. Hence, $\sum \sqrt[q_i]{p_i} = \sum Re(\zeta_i^{m_i} \sqrt[q_i]{p_i}) = \sum \sqrt[q_i]{p_i} Re(\zeta_i^{m_i})$. For this to happen, each $\zeta_i^{m_i} = 1$. In other words, $\phi|_{L_i}$ contains no powers of $\sigma_i$ and therefore fixes $K_i$. Hence, $\phi$ fixes $K$ and $G(L/K) = G(L/\mathbb Q(\alpha))$.
Bonus: Using the same ideas as above, you can show that $L = \mathbb Q(\sum \sqrt[q_i]{p_i} + \sum \zeta_i)$.