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# algebraic extension

###### Definition 1.

Let $L/K$ be an extension of fields. $L/K$ is said to be an
*algebraic extension* of fields if every element of $L$ is
algebraic over $K$. If $L/K$ is not algebraic then we say that it is a transcendental extension of fields.

Examples:

1. Let $L=\mathbb{Q}(\sqrt{2})$. The extension $L/\mathbb{Q}$ is an algebraic extension. Indeed, any element $\alpha\in L$ is of the form

$\alpha=q+t\sqrt{2}\in L$ for some $q,t\in\mathbb{Q}$. Then $\alpha\in L$ is a root of

$X^{2}-2qX+q^{2}-2t^{2}=0$ 2. The field extension $\mathbb{R}/\mathbb{Q}$ is not an algebraic extension. For example, $\pi\in\mathbb{R}$ is a transcendental number over $\mathbb{Q}$ (see pi). So $\mathbb{R}/\mathbb{Q}$ is a transcendental extension of fields.

3. Let $K$ be a field and denote by $\overline{K}$ the algebraic closure of $K$. Then the extension $\overline{K}/K$ is algebraic.

4. In general, a finite extension of fields is an algebraic extension. However, the converse is not true. The extension $\overline{\mathbb{Q}}/\mathbb{Q}$ is

*far*from finite.5. The extension $\mathbb{Q}(\pi)/\mathbb{Q}$ is transcendental because $\pi$ is a transcendental number, i.e. $\pi$ is not the root of any polynomial $p(x)\in\mathbb{Q}[x]$.

## Mathematics Subject Classification

12F05*no label found*

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## Recent Activity

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo