Algebraic element

Examples

• The square root of 2 is algebraic over Q, since it is the root of the polynomial whose coefficients are rational.
• Pi is transcendental over Q but algebraic over the field of real numbers R: it is the root of , whose coefficients (1 and −) are both real, but not of any polynomial with only rational coefficients. (The definition of the term transcendental number uses C/Q, not C/R.)

Properties

The following conditions are equivalent for an element a of an extension field L of K:

• a is algebraic over K,
• the field extension K(a)/K is algebraic, i.e. every element of K(a) is algebraic over K (here K(a) denotes the smallest subfield of L containing K and a),
• the field extension K(a)/K has finite degree, i.e. the dimension of K(a) as a K-vector space is finite,
• K[a] = K(a), where K[a] is the set of all elements of L that can be written in the form g(a) with a polynomial g whose coefficients lie in K.

To make this more explicit, consider the polynomial evaluation \varepsilon_a: K[X] \rightarrow K(a),, P \mapsto P(a). This is a homomorphism and its kernel is {P \in K[X] \mid P(a) = 0 }. If a is algebraic, this ideal contains non-zero polynomials, but as K[X] is a euclidean domain, it contains a unique polynomial p with minimal degree and leading coefficient 1, which then also generates the ideal and must be irreducible. The polynomial p is called the minimal polynomial of a and it encodes many important properties of a. Hence the ring isomorphism K[X]/(p) \rightarrow \mathrm{im}(\varepsilon_a) obtained by the homomorphism theorem is an isomorphism of fields, where we can then observe that \mathrm{im}(\varepsilon_a) = K(a). Otherwise, \varepsilon_a is injective and hence we obtain a field isomorphism K(X) \rightarrow K(a), where K(X) is the field of fractions of K[X], i.e. the field of rational functions on K, by the universal property of the field of fractions. We can conclude that in any case, we find an isomorphism K(a) \cong K[X]/(p) or K(a) \cong K(X). Investigating this construction yields the desired results.

This characterization can be used to show that the sum, difference, product and quotient of algebraic elements over K are again algebraic over K. For if a and b are both algebraic, then (K(a))(b) is finite. As it contains the aforementioned combinations of a and b, adjoining one of them to K also yields a finite extension, and therefore these elements are algebraic as well. Thus set of all elements of L that are algebraic over K is a field that sits in between L and K.

Fields that do not allow any algebraic elements over them (except their own elements) are called algebraically closed. The field of complex numbers is an example. If L is algebraically closed, then the field of algebraic elements of L over K is algebraically closed, which can again be directly shown using the characterisation of simple algebraic extensions above. An example for this is the field of algebraic numbers.

References

References

1. Roman, Steven. (2008). "Advanced Linear Algebra". Springer New York Springer e-books.