Algebraic element

In mathematics, if is an associative algebra over , then an element of is an algebraic element over , or just algebraic over , if there exists some non-zero polynomial g(x) \in K /math> with coefficients in such that . Elements of that are not algebraic over are transcendental over . A special case of an associative algebra over $K$ is an extension field $L$ of $K$.

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

Examples

* The square root of 2 is algebraic over , since it is the root of the polynomial whose coefficients are rational.
* Pi is transcendental over but algebraic over the field of real numbers : 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 , not .)

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 = K(a), where K /math> 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 \rightarrow K(a),\, P \mapsto P(a). This is a homomorphism and its kernel is $\$. If $a$ is algebraic, this ideal contains non-zero polynomials, but as K /math> 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 (p) \rightarrow \mathrm(\varepsilon_a) obtained by the homomorphism theorem is an isomorphism of fields, where we can then observe that $\mathrm(\varepsilon_a) = K(a)$. 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 /math>, 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 (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.

See also

* Algebraic independence

References

Further reading

*{{Lang Algebra , edition=3r

Algebraic properties of elements