algebraic element

• The square root of 2 is algebraic over {{math|Q}}, since it is the root of the polynomial {{math|g(x) {{=}} x² − 2}} whose coefficients are rational.
• Pi is transcendental over {{math|Q}} but algebraic over the field of real numbers {{math|R}}: it is the root of {{math|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 {{math|C/Q}}, not {{math|C/R}}.)

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 $\backslash varepsilon\_a:\; K[X]\; \backslash rightarrow\; K(a),\backslash ,\; P\; \backslash mapsto\; P(a)$. This is a homomorphism and its kernel is $\backslash \{P\; \backslash in\; K[X]\; \backslash mid\; P(a)\; =\; 0\; \backslash \}$. 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 $K[X]/(p)\; \backslash rightarrow\; \backslash mathrm\{im\}(\backslash varepsilon\_a)$ obtained by the homomorphism theorem is an isomorphism of fields, where we can then observe that $\backslash mathrm\{im\}(\backslash varepsilon\_a)\; =\; K(a)$. Otherwise, $\backslash varepsilon\_a$ is injective and hence we obtain a field isomorphism $K(X)\; \backslash rightarrow\; K(a)$, 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 $K(a)\; \backslash cong\; K[X]/(p)$ or $K(a)\; \backslash cong\; K(X)$. 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.

• Algebraic independence

{{reflist}}

• {{Lang Algebra | edition=3r}}

Category:Algebraic properties of elements