Difference between revisions of "Algebraically closed field"
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A field $k$ is algebraically closed if any polynomial of non-zero degree over $k$ has at | A field $k$ is algebraically closed if any polynomial of non-zero degree over $k$ has at |
Latest revision as of 21:31, 5 March 2012
2020 Mathematics Subject Classification: Primary: 12Exx Secondary: 12Fxx [MSN][ZBL]
A field $k$ is algebraically closed if any polynomial of non-zero degree over $k$ has at
least one root in $k$. In fact, it follows that for an algebraically closed
field $k$ each polynomial of degree $n$ over $k$ has exactly $n$ roots
in $k$, i.e. each irreducible polynomial from the ring of polynomials
$k[x]$ is of degree one. A field $k$ is algebraically closed if and only
if it has no proper algebraic extension (cf.
Extension of a field). For any field $k$,
there exists a unique (up to isomorphism) algebraic extension of $k$
that is algebraically closed; it is called the algebraic closure of
$k$ and is usually denoted by $\bar k$. Any algebraically closed field
containing $k$ contains a subfield isomorphic to $k$.
The field of complex numbers is the algebraic closure of the field of real numbers. This is the fundamental theorem of algebra (cf. Algebra, fundamental theorem of).
References
[La] | S. Lang, "Algebra", Addison-Wesley (1974) MR0783636 Zbl 0712.00001 |
[ZaSa] | O. Zariski, P. Samuel, "Commutative algebra", 1, Springer (1975) MR0384768 Zbl 0313.13001 |
Algebraically closed field. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Algebraically_closed_field&oldid=21550