*(This is part of my series on algebraic number theory.)*

In the introductory post of my series on field theory, I remarked that $\mathbf{Q}$ is a lot more complicated than $\mathbf{R}$ from a field-theoretic standpoint. The reader who has worked through the questions at the end of each post will probably already have some idea of what was meant by that. This first post on algebraic number theory will elaborate on this, and introduc *number fields*, which are the main objects of study in this discipline.

**8.1 $\mathbf{R}$ is simple**

$\mathbf{R}$ is *extremely* simple: it has only two algebraic extensions, which are $\mathbf{R}$ itself and $\mathbf{C}$. Indeed, for any non-real complex number $z$, the field $\mathbf{R}(z)$ is exactly $\mathbf{C}$. Since a root of a polynomial in real coefficients is always a complex, we see that any proper algebraic extension of $\mathbf{R}$ must be of the form $\mathbf{R}(z)$ for a non-real complex $z$, and thus must be $\mathbf{C}$. Then, since $\mathbf{C}$ is algebraically closed, all the elements which are algebraic over $\mathbf{C}$ are in $\mathbf{C}$ already, and so we can’t construct a bigger algebraic extension. *What this means is that if we start from $\mathbf{R}$, we can only add one element until we get an algebraically closed field.* Then when we have an algebraically closed field, the only way to obtain a bigger field is by adding a transcendental element.

**8.2 $\mathbf{Q}$ is complicated (that’s why it’s so fun!)**

So, if we start from $\mathbf{Q}$, how many elements do we need to add in order to obtain an algebraically closed field? Let’s try. We can add for example $i$, whose minimal polynomial over $\mathbf{Q}$ is $x^2+1$ and which is thus of degree $2$, to obtain $\mathbf{Q}(i) = \{a+bi | a,b \in \mathbf{Q}\}$. Is this field algebraically closed? No, because it does not contain $\sqrt{2}$ or $-\sqrt{2}$, and so the polynomial $x^2-2$ has no root in it. So we construct $(\mathbf{Q}(i))(\sqrt{2}) = \mathbf{Q}(i,\sqrt{2})$. Algebraically closed yet? Nope, it does not contain $\sqrt{3}$.

Hopefully this will have convinced you that in order to obtain an algebraically closed field starting from $\mathbf{Q}$, we need to add not one, nor two, nor even a billion, but *infinitely many* elements. The field we then obtain is called the *field of algebraic numbers* and noted $\overline{\mathbf{Q}}$, it consists of all elements of $\mathbf{C}$ which are roots of *some* polynomial with rational coefficients. Also, all the fields we obtain each time we add one extra element, on the way from $\mathbf{Q}$ to $\overline{\mathbf{Q}}$, are distinct. *This means that there exist infinitely many different fields which lie between $\mathbf{Q}$ and $\overline{\mathbf{Q}}$.* (Compare with $\mathbf{R}$, where we have $\mathbf{R}$, $\mathbf{C}$, and that’s it.)

**8.3 Number fields**

Those fields are so important that they have been collectively given a special name: they are called (*algebraic*)* number fields*, and certainly not all is known about them, and as stated above they are the main objects of study in the field of algebraic number theory. A number field is defined very simply as a field of the form $\mathbf{Q}(\alpha)$, for some element $\alpha$ algebraic over $\mathbf{Q}$. You might wonder why this is so, because in the discussion above we sometimes added several elements before reaching a given “intermediate” field between $\mathbf{Q}$ and $\overline{\mathbf{Q}}$. A classic result in field theory called the primitive element theorem, interpreted to our present situation, asserts that if we can construct a field by adding to $\mathbf{Q}$ a finite number of algebraic elements, then we can also construct *the same field* by adding to $\mathbf{Q}$ *only one* algebraic element, and so our definition of a number field does indeed cover all number fields.

**Next:** Rings