Algebraic element

In mathematics, if A is an associative algebra over K, then an element a of A is an algebraic element over K, or just algebraic over K, if there exists some non-zero polynomial ${\displaystyle g(x)\in K[x]}$ with coefficients in K such that g(a) = 0.¹ Elements of A that are not algebraic over K are transcendental over K. A special case of an associative algebra over ${\displaystyle K}$ is an extension field ${\displaystyle L}$ of ${\displaystyle K}$. If all elements of ${\displaystyle L}$ are algebraic over ${\displaystyle K}$, then ${\displaystyle L/K}$ is called an algebraic extension.

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).

• The square root of 2 is algebraic over Q, since it is the root of the polynomial g(x) = x² − 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 −π) 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.)
• An indeterminate x in a field K(x) of rational functions or K((x)) of formal Laurent series is defined to be transcendental over K.
• Algebraic functions are the algebraic elements over a field of rational functions K(x₁, ..., x_m).

The following conditions are equivalent for an element ${\displaystyle a}$ of an extension field ${\displaystyle L}$ of ${\displaystyle K}$:

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

To make this more explicit, consider the polynomial evaluation ${\displaystyle \varepsilon _{a}:K[X]\rightarrow K(a),\,P\mapsto P(a)}$. This is a homomorphism and its kernel is ${\displaystyle \{P\in K[X]\mid P(a)=0\}}$. If ${\displaystyle a}$ is algebraic, this ideal contains non-zero polynomials, but as ${\displaystyle K[X]}$ is a euclidean domain, it contains a unique polynomial ${\displaystyle p}$ with minimal degree and leading coefficient ${\displaystyle 1}$, which then also generates the ideal and must be irreducible. The polynomial ${\displaystyle p}$ is called the minimal polynomial of ${\displaystyle a}$ and it encodes many important properties of ${\displaystyle a}$. Hence the ring isomorphism ${\displaystyle K[X]/(p)\rightarrow \mathrm {im} (\varepsilon _{a})}$ obtained by the homomorphism theorem is an isomorphism of fields, where we can then observe that ${\displaystyle \mathrm {im} (\varepsilon _{a})=K(a)}$. Otherwise, ${\displaystyle \varepsilon _{a}}$ is injective and hence we obtain a field isomorphism ${\displaystyle K(X)\rightarrow K(a)}$, where ${\displaystyle K(X)}$ is the field of fractions of ${\displaystyle K[X]}$, i.e. the field of rational functions on ${\displaystyle K}$, by the universal property of the field of fractions. We can conclude that in any case, we find an isomorphism ${\displaystyle K(a)\cong K[X]/(p)}$ or ${\displaystyle K(a)\cong K(X)}$. 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 ${\displaystyle K}$ are again algebraic over ${\displaystyle K}$. For if ${\displaystyle a}$ and ${\displaystyle b}$ are both algebraic, then ${\displaystyle (K(a))(b)}$ is finite. As it contains the aforementioned combinations of ${\displaystyle a}$ and ${\displaystyle b}$, adjoining one of them to ${\displaystyle K}$ also yields a finite extension, and therefore these elements are algebraic as well. Thus set of all elements of ${\displaystyle L}$ that are algebraic over ${\displaystyle K}$ is a field that sits in between ${\displaystyle L}$ and ${\displaystyle K}$.

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 ${\displaystyle L}$ is algebraically closed, then the field of algebraic elements of ${\displaystyle L}$ over ${\displaystyle K}$ 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.