Agebraically closed field equivalent definition












0












$begingroup$


Is this a valid equivalent definition of an algebraically closed field?



Let $ F $ be a field. Define a function $ f(x): Flongrightarrow F $ as a finite combination of the operations in $ F $ (addition, multiplication and their inverses). If the range of the function $ f $ is $ F $, then $ F $ is an algebraically closed field.



My reasoning is that such function $ f $ should be a rational function
$ f(x)=dfrac{p(x)}{q(x)} $ where $ p(x) $ and $ q(x) $ are polynomials with coefficients in $ F $, so for any fixed element $ yin F $ I can write



$ f(x)=y $



$ dfrac{p(x)}{q(x)}=y $



$ p(x)=yq(x) $



$ p(x)-yq(x)=0 $



As $ p(x)-yq(x) $ is a polynomial with coefficients in $ F $, if $ F $ is truly algebraically closed then there must be some root $ rin F $ an so



$ f(r)=y $



As $ y $ is an arbitrary element of $ F $, then the range $ R $ of the function $ f $ satisfies $ R=F $



This should work for the field of complex numbers $ mathbb{C} $ but does this work for any algebraically closed field?










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  • $begingroup$
    I'm not sure I entirely understand your question, but as it reads to me right now, the following is a very simple counterexample : $F: mathbb{Q} rightarrow mathbb{Q}, x mapsto x + 1$.
    $endgroup$
    – Adam Higgins
    Jan 7 at 1:01










  • $begingroup$
    I think you mean to say that you need this to be true for all possible such functions.
    $endgroup$
    – jgon
    Jan 7 at 1:03










  • $begingroup$
    What you have shown here is that if $mathbb{F}$ is algebraically closed field, and $F : mathbb{F} rightarrow mathbb{F}$ is a function (well, partial function) that is given by a combination of the field operations, then it is surjective.
    $endgroup$
    – Adam Higgins
    Jan 7 at 1:04










  • $begingroup$
    @jgon your right, I meant every possible function. Is this a valid definition of an algebraically closed field?
    $endgroup$
    – SFSplastic
    Jan 7 at 1:33










  • $begingroup$
    Well, if you remove inversion from the list of allowed operations, then it should be, yes. The problem with inversion is that e.g. $1/x$ is not surjective, since $1/xne 0$ for any $x$.
    $endgroup$
    – jgon
    Jan 7 at 1:58


















0












$begingroup$


Is this a valid equivalent definition of an algebraically closed field?



Let $ F $ be a field. Define a function $ f(x): Flongrightarrow F $ as a finite combination of the operations in $ F $ (addition, multiplication and their inverses). If the range of the function $ f $ is $ F $, then $ F $ is an algebraically closed field.



My reasoning is that such function $ f $ should be a rational function
$ f(x)=dfrac{p(x)}{q(x)} $ where $ p(x) $ and $ q(x) $ are polynomials with coefficients in $ F $, so for any fixed element $ yin F $ I can write



$ f(x)=y $



$ dfrac{p(x)}{q(x)}=y $



$ p(x)=yq(x) $



$ p(x)-yq(x)=0 $



As $ p(x)-yq(x) $ is a polynomial with coefficients in $ F $, if $ F $ is truly algebraically closed then there must be some root $ rin F $ an so



$ f(r)=y $



As $ y $ is an arbitrary element of $ F $, then the range $ R $ of the function $ f $ satisfies $ R=F $



This should work for the field of complex numbers $ mathbb{C} $ but does this work for any algebraically closed field?










share|cite|improve this question







New contributor




SFSplastic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    I'm not sure I entirely understand your question, but as it reads to me right now, the following is a very simple counterexample : $F: mathbb{Q} rightarrow mathbb{Q}, x mapsto x + 1$.
    $endgroup$
    – Adam Higgins
    Jan 7 at 1:01










  • $begingroup$
    I think you mean to say that you need this to be true for all possible such functions.
    $endgroup$
    – jgon
    Jan 7 at 1:03










  • $begingroup$
    What you have shown here is that if $mathbb{F}$ is algebraically closed field, and $F : mathbb{F} rightarrow mathbb{F}$ is a function (well, partial function) that is given by a combination of the field operations, then it is surjective.
    $endgroup$
    – Adam Higgins
    Jan 7 at 1:04










  • $begingroup$
    @jgon your right, I meant every possible function. Is this a valid definition of an algebraically closed field?
    $endgroup$
    – SFSplastic
    Jan 7 at 1:33










  • $begingroup$
    Well, if you remove inversion from the list of allowed operations, then it should be, yes. The problem with inversion is that e.g. $1/x$ is not surjective, since $1/xne 0$ for any $x$.
    $endgroup$
    – jgon
    Jan 7 at 1:58
















0












0








0





$begingroup$


Is this a valid equivalent definition of an algebraically closed field?



Let $ F $ be a field. Define a function $ f(x): Flongrightarrow F $ as a finite combination of the operations in $ F $ (addition, multiplication and their inverses). If the range of the function $ f $ is $ F $, then $ F $ is an algebraically closed field.



My reasoning is that such function $ f $ should be a rational function
$ f(x)=dfrac{p(x)}{q(x)} $ where $ p(x) $ and $ q(x) $ are polynomials with coefficients in $ F $, so for any fixed element $ yin F $ I can write



$ f(x)=y $



$ dfrac{p(x)}{q(x)}=y $



$ p(x)=yq(x) $



$ p(x)-yq(x)=0 $



As $ p(x)-yq(x) $ is a polynomial with coefficients in $ F $, if $ F $ is truly algebraically closed then there must be some root $ rin F $ an so



$ f(r)=y $



As $ y $ is an arbitrary element of $ F $, then the range $ R $ of the function $ f $ satisfies $ R=F $



This should work for the field of complex numbers $ mathbb{C} $ but does this work for any algebraically closed field?










share|cite|improve this question







New contributor




SFSplastic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Is this a valid equivalent definition of an algebraically closed field?



Let $ F $ be a field. Define a function $ f(x): Flongrightarrow F $ as a finite combination of the operations in $ F $ (addition, multiplication and their inverses). If the range of the function $ f $ is $ F $, then $ F $ is an algebraically closed field.



My reasoning is that such function $ f $ should be a rational function
$ f(x)=dfrac{p(x)}{q(x)} $ where $ p(x) $ and $ q(x) $ are polynomials with coefficients in $ F $, so for any fixed element $ yin F $ I can write



$ f(x)=y $



$ dfrac{p(x)}{q(x)}=y $



$ p(x)=yq(x) $



$ p(x)-yq(x)=0 $



As $ p(x)-yq(x) $ is a polynomial with coefficients in $ F $, if $ F $ is truly algebraically closed then there must be some root $ rin F $ an so



$ f(r)=y $



As $ y $ is an arbitrary element of $ F $, then the range $ R $ of the function $ f $ satisfies $ R=F $



This should work for the field of complex numbers $ mathbb{C} $ but does this work for any algebraically closed field?







abstract-algebra field-theory definition






share|cite|improve this question







New contributor




SFSplastic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




SFSplastic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




SFSplastic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Jan 7 at 0:56









SFSplasticSFSplastic

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1




New contributor




SFSplastic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





SFSplastic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






SFSplastic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    I'm not sure I entirely understand your question, but as it reads to me right now, the following is a very simple counterexample : $F: mathbb{Q} rightarrow mathbb{Q}, x mapsto x + 1$.
    $endgroup$
    – Adam Higgins
    Jan 7 at 1:01










  • $begingroup$
    I think you mean to say that you need this to be true for all possible such functions.
    $endgroup$
    – jgon
    Jan 7 at 1:03










  • $begingroup$
    What you have shown here is that if $mathbb{F}$ is algebraically closed field, and $F : mathbb{F} rightarrow mathbb{F}$ is a function (well, partial function) that is given by a combination of the field operations, then it is surjective.
    $endgroup$
    – Adam Higgins
    Jan 7 at 1:04










  • $begingroup$
    @jgon your right, I meant every possible function. Is this a valid definition of an algebraically closed field?
    $endgroup$
    – SFSplastic
    Jan 7 at 1:33










  • $begingroup$
    Well, if you remove inversion from the list of allowed operations, then it should be, yes. The problem with inversion is that e.g. $1/x$ is not surjective, since $1/xne 0$ for any $x$.
    $endgroup$
    – jgon
    Jan 7 at 1:58




















  • $begingroup$
    I'm not sure I entirely understand your question, but as it reads to me right now, the following is a very simple counterexample : $F: mathbb{Q} rightarrow mathbb{Q}, x mapsto x + 1$.
    $endgroup$
    – Adam Higgins
    Jan 7 at 1:01










  • $begingroup$
    I think you mean to say that you need this to be true for all possible such functions.
    $endgroup$
    – jgon
    Jan 7 at 1:03










  • $begingroup$
    What you have shown here is that if $mathbb{F}$ is algebraically closed field, and $F : mathbb{F} rightarrow mathbb{F}$ is a function (well, partial function) that is given by a combination of the field operations, then it is surjective.
    $endgroup$
    – Adam Higgins
    Jan 7 at 1:04










  • $begingroup$
    @jgon your right, I meant every possible function. Is this a valid definition of an algebraically closed field?
    $endgroup$
    – SFSplastic
    Jan 7 at 1:33










  • $begingroup$
    Well, if you remove inversion from the list of allowed operations, then it should be, yes. The problem with inversion is that e.g. $1/x$ is not surjective, since $1/xne 0$ for any $x$.
    $endgroup$
    – jgon
    Jan 7 at 1:58


















$begingroup$
I'm not sure I entirely understand your question, but as it reads to me right now, the following is a very simple counterexample : $F: mathbb{Q} rightarrow mathbb{Q}, x mapsto x + 1$.
$endgroup$
– Adam Higgins
Jan 7 at 1:01




$begingroup$
I'm not sure I entirely understand your question, but as it reads to me right now, the following is a very simple counterexample : $F: mathbb{Q} rightarrow mathbb{Q}, x mapsto x + 1$.
$endgroup$
– Adam Higgins
Jan 7 at 1:01












$begingroup$
I think you mean to say that you need this to be true for all possible such functions.
$endgroup$
– jgon
Jan 7 at 1:03




$begingroup$
I think you mean to say that you need this to be true for all possible such functions.
$endgroup$
– jgon
Jan 7 at 1:03












$begingroup$
What you have shown here is that if $mathbb{F}$ is algebraically closed field, and $F : mathbb{F} rightarrow mathbb{F}$ is a function (well, partial function) that is given by a combination of the field operations, then it is surjective.
$endgroup$
– Adam Higgins
Jan 7 at 1:04




$begingroup$
What you have shown here is that if $mathbb{F}$ is algebraically closed field, and $F : mathbb{F} rightarrow mathbb{F}$ is a function (well, partial function) that is given by a combination of the field operations, then it is surjective.
$endgroup$
– Adam Higgins
Jan 7 at 1:04












$begingroup$
@jgon your right, I meant every possible function. Is this a valid definition of an algebraically closed field?
$endgroup$
– SFSplastic
Jan 7 at 1:33




$begingroup$
@jgon your right, I meant every possible function. Is this a valid definition of an algebraically closed field?
$endgroup$
– SFSplastic
Jan 7 at 1:33












$begingroup$
Well, if you remove inversion from the list of allowed operations, then it should be, yes. The problem with inversion is that e.g. $1/x$ is not surjective, since $1/xne 0$ for any $x$.
$endgroup$
– jgon
Jan 7 at 1:58






$begingroup$
Well, if you remove inversion from the list of allowed operations, then it should be, yes. The problem with inversion is that e.g. $1/x$ is not surjective, since $1/xne 0$ for any $x$.
$endgroup$
– jgon
Jan 7 at 1:58












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