TITLE: An amateur doubt in Galois theory
QUESTION [7 upvotes]: Good morning,
today I have read that "number theory is nothing but the study of $\mathrm{Gal}(\mathbb{\bar{Q}}/\mathbb{Q})$", here http://www.math.uconn.edu/~alozano/elliptic/finding%20points.pdf
can anyone give a very naive layman definition of what it actually means?
Furthermore, I got this doubt that $\bar {\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$, and the thing that confuses me is 
the field of rational numbers $\mathbb{Q}$ is not a algebraically closed as there exists a polynomial with $a_{1},a_{2},\dotsc,a_{n}\in \mathbb{Q}$ and  $(x-a_{1})(x-a_{2})\cdots(x-a_{n})+1$  has no zero in $\mathbb{Q}$.
Then why are we considering the field extension of $\bar {\mathbb{Q}}/\mathbb{Q}$ when $\mathbb{Q}$ is not algebraically closed, won't it contradict the definition of algebraic closure?
But I am not getting an answer i was looking for ,i want what are the things going on behind the $\mathrm{Gal}(\mathbb{\bar{Q}}/\mathbb{Q})$ like what is the thing we get if we take the $\mathbb{\bar{Q}/\mathbb{Q}}$ and what does taking the $\mathrm{Gal}(\mathbb{\bar{Q}}/\mathbb{Q})$  give someone,
Thank you

REPLY [4 votes]: The object $\mathrm{Gal}(\mathbb{\bar{Q}}/\mathbb{Q})$ is not just an abstract group; it has a natural and intricate topology on it as well. Then one would consider various naturally occurring continuous actions of this topological group on various modules. Two examples are: The absolute Galois group of a number field acts naturally on the roots of unity contained in that field, or, given an elliptic curve over the field, the Galois group acts on the torsion points on the elliptic curve with co-ordinates contained in the field.
Also one considers some natural representations of this Galois group on some vector spaces. Examples can be constructed from the above description of Galois actions. A lot of number theory is encapsulated in the study of such Galois actions and Galois representations. This is perhaps what is meant by the statement that you quote.