Number field

In algebra, a number field is a subclass of fields that consists solely of numbers or numerical elements. In number theory, however, a number field refers to a finite extension of the field of rational numbers $ \mathbb{Q} $.

In algebra, a number field is considered a subclass of the general field, but with certain specific properties.

General field: an abstract algebraic structure defined over sets that do not necessarily contain numbers.
Number field: a subclass of general fields that includes numbers or numerical elements, such as rational, real, complex numbers, or their algebraic extensions.

Therefore, from an algebraic point of view, a number field is a specific type of general field, though not all fields are number fields.

For example, the field of real numbers $ \mathbb{R} $, complex numbers $ \mathbb{C} $, and rational numbers $ \mathbb{Q} $ are all number fields.

In number theory, a number field is defined as a finite extension of the field of rational numbers $ \mathbb{Q} $.

In other words, it’s a field that contains $ \mathbb{Q} $ and is generated by adding a finite number of elements to $ \mathbb{Q} $, where these new elements are solutions to polynomial equations with rational coefficients.

More formally, if $ K $ is a field that contains $ \mathbb{Q} $, then $ K $ is a number field if its dimension as a vector space over $ \mathbb{Q} $ is finite. This dimension is known as the degree of the extension of $ K $ over $ \mathbb{Q} $.

For example, the field $ \mathbb{Q}(\sqrt{2}) $, which consists of all numbers of the form $ a + b\sqrt{2} $ with $ a, b \in \mathbb{Q} $, is a number field with degree 2 over $ \mathbb{Q} $, because $ 1 $ and $ \sqrt{2} $ form a basis of $ \mathbb{Q}(\sqrt{2}) $ as a vector space over $ \mathbb{Q} $.

A practical example

The field $ \mathbb{Q}(\sqrt{2}) $ is a finite extension of the rational numbers $ \mathbb{Q} $, containing all numbers of the form $ a + b\sqrt{2} $, where $ a $ and $ b $ are rational.

This field is generated by adding $ \sqrt{2} $ to the rational numbers.

For instance, if we take two elements from $ \mathbb{Q}(\sqrt{2}) $, such as $ (1 + \sqrt{2}) $ and $ (3 - 2\sqrt{2}) $, we can add them:

$$ (1 + \sqrt{2}) + (3 - 2\sqrt{2}) = 4 - \sqrt{2} $$

We can also multiply them:

$$ (1 + \sqrt{2}) \times (3 - 2\sqrt{2}) = 1 \times 3 + 1 \times (-2\sqrt{2}) + \sqrt{2} \times 3 + \sqrt{2} \times (-2\sqrt{2}) $$

$$ = 3 - 2\sqrt{2} + 3\sqrt{2} - 4 = -1 + \sqrt{2} $$

The field $ \mathbb{Q}(\sqrt{2}) $ is a finite extension of $ \mathbb{Q} $ because it is generated by adding a finite number of elements to the base field. In this case, we add only $ \sqrt{2} $.

This field contains all rational numbers $ \mathbb{Q} $ along with a finite number of new elements, like $ \sqrt{2} $, which do not belong to $ \mathbb{Q} $.

And so on.