Algebraically closed field

In mathematics, a field $F$ is said to be algebraically closed if every polynomial in one variable of degree at least 1, with coefficients in $F$, has a zero (root) in $F$.

As an example, the field of real numbers is not algebraically closed, because the polynomial equation

$3x^{2}+1=0$

has no solution in real numbers, even though both of its coefficients (3 and 1) are real. The same argument proves that the field of rational numbers is not algebraically closed. Also, no finite field $F$ is algebraically closed, because if $a_{1}$, $a_{2}$, …, $a_{n}$ are the elements of $F$, then the polynomial

$(x-a_{1})(x-a_{2})$ ··· $(x-a_{n})+1$

has no zero in $F$. By contrast, the field of complex numbers is algebraically closed: this is stated by the fundamental theorem of algebra. Another example of an algebraically closed field is the field of algebraic numbers.

Given a field $F$, the assertion “$F$ is algebraically closed” is equivalent to each one of the following:

• Every polynomial $p(x)$ of degree $n$ ≥ 1, with coefficients in $F$, splits into linear factors. In other words, there are elements $k$, $x_{1}$, $x_{2}$, …, $x_{n}$ in $F$ such that

$p(x)=k(x-x_{1})(x-x_{2})$ ··· $(x-x_{n})$.

• The field $F$ has no proper algebraic extension.

• For each natural number $n$, every linear map from $F^{n}$ into itself has some eigenvector.

• Every rational function in one variable $x$, with coefficients in $F$, can be written as the sum of a polynomial function with rational functions of the form $a/(x-b)^{n}$, where $n$ is a natural number, and $a$ and $b$ are elements of $F$.

If $F$ is an algebraically closed field, $a$ is an element of $F$, and $n$ is a natural number, then $a$ has an $n$^th root in $F$, since this is the same thing as saying that the equation $x^{n}-a=0$ has some root in $F$. However, there are fields in which every element has an $n$^th root (for each natural number $n$) but which are not algebraically closed. In fact, even assuming that every polynomial of the form $x^{n}-a$ splits into linear factors is not enough to assure that the field is algebraically closed.

Every field $F$ has an "algebraic closure", which is the smallest algebraically closed field of which $F$ is a subfield.

References

• S. Lang, Algebra, Springer-Verlag, 2004, ISBN 0-387-95385-X

• B.L. van der Waerden, Algebra I, Springer-Verlag, 1991, ISBN 0-387-97424-5