Algebraic independence

In abstract algebra, a subset S {displaystyle S} of a field L {displaystyle L} is algebraically independent over a subfield K {displaystyle K} if the elements of S {displaystyle S} do not satisfy any non-trivial polynomial equation with coefficients in K {displaystyle K} . In particular, a one element set { α } {displaystyle {alpha }} is algebraically independent over K {displaystyle K} if and only if α {displaystyle alpha } is transcendental over K {displaystyle K} . In general, all the elements of an algebraically independent set S {displaystyle S} over K {displaystyle K} are by necessity transcendental over K {displaystyle K} , and over all of the field extensions over K {displaystyle K} generated by the remaining elements of S {displaystyle S} . The two real numbers π {displaystyle {sqrt {pi }}} and 2 π + 1 {displaystyle 2pi +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 { π } {displaystyle {{sqrt {pi }}}} and { 2 π + 1 } {displaystyle {2pi +1}} are algebraically independent over the field Q {displaystyle mathbb {Q} } of rational numbers. However, the set { π , 2 π + 1 } {displaystyle {{sqrt {pi }},2pi +1}} is not algebraically independent over the rational numbers, because the nontrivial polynomial is zero when x = π {displaystyle x={sqrt {pi }}} and y = 2 π + 1 {displaystyle y=2pi +1} . Although both π {displaystyle pi } and e are known to be transcendental,it is not known whether the set of both of them is algebraically independent over Q {displaystyle mathbb {Q} } . In fact, it is not even known if π + e {displaystyle pi +e} is irrational.Nesterenko proved in 1996 that: The Lindemann–Weierstrass theorem can often be used to prove that some sets are algebraically independent over Q {displaystyle mathbb {Q} } . It states that whenever α 1 , … , α n {displaystyle alpha _{1},ldots ,alpha _{n}} are algebraic numbers that are linearly independent over Q {displaystyle mathbb {Q} } , then e α 1 , … , e α n {displaystyle e^{alpha _{1}},ldots ,e^{alpha _{n}}} are also algebraically independent over Q {displaystyle mathbb {Q} } . Given a field extension L / K {displaystyle L/K} which is not algebraic, Zorn's lemma can be used to show that there always exists a maximal algebraically independent subset of L {displaystyle L} over K {displaystyle K} . Further, all the maximal algebraically independent subsets have the same cardinality, known as the transcendence degree of the extension.