Remark

It is clear that in an extension E of K an algebraically free family over K consists of linearly independent elements over $K$ (hence pairwise distinct); in other words, it is also a free family for the vector space structure of $E$ (with respect to K). But the converse is false, for if E is an algebraic extension of K, any non-empty family of elements of E (and a fortiori a non-empty family of linearly independent elements over K) is never algebraically free over K. When there is a risk of confusion, we shall say that a subset of an extension E of K which is free for the vector space structure of E with respect to K is linearly free over K.

Let E be an extension of a field K. A subset S of E is said to be algebraically free (over $K$) if the family defined by the identity mapping of S onto itself is algebraically free. The elements of an algebraically free subset of E are also called algebraically independent. If a subset of E is not algebraically free, it is said to be algebraically related and that its elements are algebraically dependent. For a family $(x_i)_{i \in I}$ of elements of E to be algebraically free it is necessary and sufficient that $i \mapsto x_i$ should be a bijection of I onto an algebraically free subset of E.

Every subset of an algebraically free set is algebraically free. Furthermore: