Agebraically closed field equivalent definition
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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
asked Jan 7 at 0:56
SFSplasticSFSplastic
1
$endgroup$
add a comment |
$begingroup$
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?
abstract-algebra field-theory definition
asked Jan 7 at 0:56
SFSplasticSFSplastic
1
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
abstract-algebra field-theory definition
asked Jan 7 at 0:56
SFSplasticSFSplastic
1
$endgroup$
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
abstract-algebra field-theory definition
asked Jan 7 at 0:56
SFSplasticSFSplastic
1
asked Jan 7 at 0:56
SFSplasticSFSplastic
1
asked Jan 7 at 0:56
SFSplasticSFSplastic
1
asked Jan 7 at 0:56
SFSplasticSFSplastic
1
asked Jan 7 at 0:56
SFSplasticSFSplastic
1
1
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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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