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Inmathematics, ifA is anassociative algebra overK, then an elementa ofA is analgebraic element overK, or justalgebraic overK, if there exists some non-zeropolynomial${\displaystyle g(x)\in K[x]}$ withcoefficients inK such thatg(a) = 0.^[1] Elements ofA that are not algebraic overK aretranscendental overK. A special case of an associative algebra over${\displaystyle K}$ is anextension field${\displaystyle L}$ of${\displaystyle K}$.

These notions generalize thealgebraic numbers and thetranscendental numbers (where the field extension isC/Q, withC being the field ofcomplex numbers andQ being the field ofrational numbers).

• Thesquare root of 2 is algebraic overQ, since it is the root of the polynomialg(x) =x² − 2 whose coefficients are rational.
• Pi is transcendental overQ but algebraic over the field ofreal numbersR: it is the root ofg(x) =x − π, whose coefficients (1 and −π) are both real, but not of any polynomial with only rational coefficients. (The definition of the termtranscendental number usesC/Q, notC/R.)

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. thedimension 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 thepolynomial evaluation${\displaystyle {\epsilon }_a:K[X]\to K(a),P\mapsto P(a)}$. This is ahomomorphism and itskernel is${\displaystyle \{P\in K[X]\mid P(a)=0\}}$. If${\displaystyle a}$ is algebraic, thisideal contains non-zero polynomials, but as${\displaystyle K[X]}$ is aeuclidean domain, it contains a unique polynomial${\displaystyle p}$ with minimal degree and leading coefficient${\displaystyle 1}$, which then also generates the ideal and must beirreducible. The polynomial${\displaystyle p}$ is called theminimal polynomial of${\displaystyle a}$ and it encodes many important properties of${\displaystyle a}$. Hence the ring isomorphism${\displaystyle K[X]/(p)\to \mathrm{i}\mathrm{m}({\epsilon }_a)}$ obtained by thehomomorphism theorem is an isomorphism of fields, where we can then observe that${\displaystyle \mathrm{i}\mathrm{m}({\epsilon }_a)=K(a)}$. Otherwise,${\displaystyle {\epsilon }_a}$ is injective and hence we obtain a field isomorphism${\displaystyle K(X)\to K(a)}$, where${\displaystyle K(X)}$ is thefield of fractions of${\displaystyle K[X]}$, i.e. thefield 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 calledalgebraically 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 thefield of algebraic numbers.

• Algebraic independence

• Lang, Serge (2002),Algebra,Graduate Texts in Mathematics, vol. 211 (Revised third ed.), New York: Springer-Verlag,ISBN 978-0-387-95385-4,MR 1878556,Zbl 0984.00001

• Abstract algebra

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