Uniqueness of algebraic closure of $\Bbb{Q}$ in $\Bbb{C}$. I ask 2 questions.
1) How can I prove the following sentence? If $\Bbb{Q}'$ is the subset of $\Bbb{C}$ consisting of all algebraic elements over $\Bbb{Q}$, then $\Bbb{Q}'$ is an algebraic closure of $\Bbb{Q}$.
2) Is $\Bbb{Q}'$ the unique algebraic closure of $\Bbb{Q}$ in $\Bbb{C}$?
Because I am an undergraduate student, I may not know the answers of them...
 A: Let's take as assumed the following fact:

$\mathbb C$ is algebraically closed.

This implies that every polynomial with coefficient in $\mathbb Q$ have root in $\mathbb C$ (and so in $\mathbb Q'$ by definition).
The key theorem to do all the work is the following

Let $K$ be a field and $\alpha_1,\dots,\alpha_n \in L$ be elements of an extension $L \supseteq K$. Then $K(\alpha_1,\dots,\alpha_n)$ is finite dimensional $K$-vector space iff all the $\alpha_i$'s are algebraic over $K$, and so iff $K(\alpha_1,\dots,\alpha_n)$ is algebraic over $K$. 

Clearly $\mathbb Q'$ is an algebraic extension of $\mathbb Q$, by definition. 
To see that is also algebraically close consider a polynomial $p \in \mathbb Q'[x]$, such polynomial has in $\mathbb C$: let's call $\alpha_1,\dots,\alpha_m \in \mathbb Q'$ the coefficient of $p$ and $\beta_1,\dots,\beta_n$ the root of $p$.
By the above mentioned theorem $\mathbb Q(\alpha_1,\dots,\alpha_m)$ is finite dimensional $\mathbb Q$-vector space (and an algebraic extension over $\mathbb Q$ of course).
The $\beta_i$'s are algebraic over $\mathbb Q(\alpha_1,\dots,\alpha_m)$ and so the extension $\mathbb Q(\alpha_1,\dots,\alpha_m,\beta_1,\dots,\beta_n)$ is a finite dimensional $\mathbb Q(\alpha_1,\dots,\alpha_m)$-vector space, which is a finite dimensional $\mathbb Q$-vector space. From this it follows that $\mathbb Q(\alpha_1,\dots,\alpha_m,\beta_1,\beta_n)$ is a finite dimensional
$\mathbb Q$-vector space (by the multiplicativity formula for degrees) and so all the $\beta_i$'s are algebraic over $\mathbb Q$, so they belong to $\mathbb Q'$.
This proves that $\mathbb Q'$ is algebraic closed and it's clearly the smallest subfield of $\mathbb C$ which is algebraically closed and contains the algebraic elements over $\mathbb Q$: being formed by all the elements of $\mathbb C$ which are algebraic over $\mathbb Q$ it must be contained in every algebraically closed subfield of $\mathbb C$, so it is the algebraic closure of $\mathbb Q$ in $\mathbb C$.
(The use of the the above denote that it is the unique subfield of $\mathbb C$ with such property).
