Algebraic element

In mathematics, if $L$ is an extension field of $K$, then an element $a$ of $L$ is called an algebraic element over $K$, or just algebraic over $K$, if there exists some non-zero polynomial $g(x)$ with coefficients in $K$ such that $g(a) = 0$. Elements of $L$ that are not algebraic over $K$ are called transcendental over $K$.

These notions generalize the algebraic numbers and the transcendental numbers (where the field extension is $C/Q$, with $C$ being the field of complex numbers and $Q$ being the field of rational numbers).

Examples

 * The square root of 2 is algebraic over $Q$, since it is the root of the polynomial $g(x) = x^{2} − 2$ whose coefficients are rational.
 * Pi is transcendental over $Q$ but algebraic over the field of real numbers $R$: it is the root of $g(x) = x − π$, whose coefficients (1 and −$\pi$) 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 $$L$$:
 * $$a$$ 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.