Euclidean ordered field

• Every Euclidean field is an ordered Pythagorean field, but the converse is not true.^[1]
• If E/F is a finite extension, and E is Euclidean, then so is F. This "going-down theorem" is a consequence of the Diller–Dress theorem.^[2]

• The real constructible numbers, those (signed) lengths which can be constructed from a rational segment by ruler and compass constructions, form a Euclidean field.^[3]

Every real closed field is a Euclidean field. The following examples are also real closed fields.

• The real numbers ${\displaystyle \mathbb {R} }$ with the usual operations and ordering form a Euclidean field.
• The field of real algebraic numbers ${\displaystyle \mathbb {R} \cap \mathbb {\overline {Q}} }$ is a Euclidean field.
• The field of hyperreal numbers is a Euclidean field.

• The rational numbers ${\displaystyle \mathbb {Q} }$ with the usual operations and ordering do not form a Euclidean field. For example, 2 is not a square in ${\displaystyle \mathbb {Q} }$ since the square root of 2 is irrational.^[4] By the going-down result above, no algebraic number field can be Euclidean.^[2]
• The complex numbers ${\displaystyle \mathbb {C} }$ do not form a Euclidean field since they cannot be given the structure of an ordered field.

The Euclidean closure of an ordered field K is an extension of K in the quadratic closure of K which is maximal with respect to being an ordered field with an order extending that of K.^[5] It is also the smallest subfield of the algebraic closure of K that is a Euclidean field and is an ordered extension of K.