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Difference between revisions of "Algebraically closed field"

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A field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a0117501.png" /> in which any polynomial of non-zero degree over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a0117502.png" /> has at least one root. In fact, it follows that for an algebraically closed field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a0117503.png" /> each polynomial of degree <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a0117504.png" /> over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a0117505.png" /> has exactly <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a0117506.png" /> roots in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a0117507.png" />, i.e. each irreducible polynomial from the ring of polynomials <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a0117508.png" /> is of degree one. A field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a0117509.png" /> is algebraically closed if and only if it has no proper algebraic extension (cf. [[Extension of a field|Extension of a field]]). For any field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a01175010.png" />, there exists a unique (up to isomorphism) algebraic extension of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a01175011.png" /> that is algebraically closed; it is called the algebraic closure of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a01175012.png" /> and is usually denoted by <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a01175013.png" />. Any algebraically closed field containing <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a01175014.png" /> contains a subfield isomorphic to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/a/a011/a011750/a01175015.png" />.
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{{TEX|done}}
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{{MSC|12Exx|12Fxx}}
  
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|Algebra, fundamental theorem of]]).
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A field $k$ is algebraically closed if any polynomial of non-zero degree over $k$ has at
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least one root in $k$. In fact, it follows that for an algebraically closed
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field $k$ each polynomial of degree $n$ over $k$ has exactly $n$ roots
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in $k$, i.e. each irreducible polynomial from the ring of polynomials
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$k[x]$ is of degree one. A field $k$ is algebraically closed if and only
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if it has no proper algebraic extension (cf.
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[[Extension of a field|Extension of a field]]). For any field $k$,
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there exists a unique (up to isomorphism) algebraic extension of $k$
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that is algebraically closed; it is called the algebraic closure of
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$k$ and is usually denoted by $\bar k$. Any algebraically closed field
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containing $k$ contains a subfield isomorphic to $k$.
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The field of complex numbers is the algebraic closure of the field of
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real numbers. This is the fundamental theorem of algebra (cf.
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[[Algebra, fundamental theorem of|Algebra, fundamental theorem of]]).
  
 
====References====
 
====References====
<table><TR><TD valign="top">[1]</TD> <TD valign="top"O. Zariski,   P. Samuel,   "Commutative algebra" , '''1''' , Springer (1975)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  S. Lang,  "Algebra" , Addison-Wesley  (1974)</TD></TR></table>
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{|
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|-
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|valign="top"|{{Ref|La}}||valign="top"| S. Lang, "Algebra", Addison-Wesley (1974)
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|-
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|valign="top"|{{Ref|ZaSa}}||valign="top"| O. Zariski, P. Samuel, "Commutative algebra", '''1''', Springer (1975)
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|-
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|}

Revision as of 18:13, 17 February 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)
[ZaSa] O. Zariski, P. Samuel, "Commutative algebra", 1, Springer (1975)
How to Cite This Entry:
Algebraically closed field. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Algebraically_closed_field&oldid=11228
This article was adapted from an original article by O.A. Ivanova (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article