Algebraically independent

In abstract algebra, a subset S of a field L is algebraically independent over a subfield K if the elements of S do not satisfy any non-trivial polynomial equation with coefficients in K.

In particular, a one element set {α} is algebraically independent over K if and only if α is transcendental over K. In general, all the elements of an algebraically independent set S over K are by necessity transcendental over K, and over all of the field extensions over K generated by the remaining elements of S.

The two real numbers ${\displaystyle {\sqrt {\pi }}}$ and ${\displaystyle 2\pi +1}$ are each transcendental numbers: they are not the roots of any nontrivial polynomial whose coefficients are rational numbers. Thus, each of the two singleton sets ${\displaystyle \{{\sqrt {\pi }}\}}$ and ${\displaystyle \{2\pi +1\}}$ are algebraically independent over the field ${\displaystyle \mathbb {Q} }$ of rational numbers.

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