What do we call the algebraic closure of $mathbb{R}(x)$, and how can we describe it?
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$mathbb{R}(x)$ is not algebraically closed, as can easily be seen by considering the equation $xy^2=1$; there is no rational function $r(x)$ with the property that $xr(x)^2=1$. However, everyone knows how to solve this equation for $y$, obtaining
$$y=frac{1}{sqrt{x}}, y=-frac{1}{sqrt{x}}$$
as the two solutions. So the solutions exist, but in order to find them we have to go beyond $mathbb{R}(x)$ to a larger collection of functions.
The algebraic closure of $mathbb{R}(x)$ would contain not only all rational functions, but also roots of all orders, plus some things we don't have names or notation for (solutions of polynomials with coefficients in $mathbb{R}(x)$ of degree 5 or higher that are not solvable by radicals , for example). (The word "functions", here, is used somewhat casually, as most of these things are partial functions at best.) I guess it would have to contain $mathbb{C}$ as well, come to think of it.
What is the algebraic closure called? Does it have a conventional name? I think of it as the "field of algebraic functions" but I don't know if that's standard terminology or if there's another, more commonly-used name. Is there a notation for it? Does it contain any other exotic things, besides what I have already mentioned?
functions field-theory extension-field
add a comment |
up vote
2
down vote
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$mathbb{R}(x)$ is not algebraically closed, as can easily be seen by considering the equation $xy^2=1$; there is no rational function $r(x)$ with the property that $xr(x)^2=1$. However, everyone knows how to solve this equation for $y$, obtaining
$$y=frac{1}{sqrt{x}}, y=-frac{1}{sqrt{x}}$$
as the two solutions. So the solutions exist, but in order to find them we have to go beyond $mathbb{R}(x)$ to a larger collection of functions.
The algebraic closure of $mathbb{R}(x)$ would contain not only all rational functions, but also roots of all orders, plus some things we don't have names or notation for (solutions of polynomials with coefficients in $mathbb{R}(x)$ of degree 5 or higher that are not solvable by radicals , for example). (The word "functions", here, is used somewhat casually, as most of these things are partial functions at best.) I guess it would have to contain $mathbb{C}$ as well, come to think of it.
What is the algebraic closure called? Does it have a conventional name? I think of it as the "field of algebraic functions" but I don't know if that's standard terminology or if there's another, more commonly-used name. Is there a notation for it? Does it contain any other exotic things, besides what I have already mentioned?
functions field-theory extension-field
Possibly related: math.stackexchange.com/questions/2609504, math.stackexchange.com/questions/318690
– Watson
Nov 20 at 8:45
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$mathbb{R}(x)$ is not algebraically closed, as can easily be seen by considering the equation $xy^2=1$; there is no rational function $r(x)$ with the property that $xr(x)^2=1$. However, everyone knows how to solve this equation for $y$, obtaining
$$y=frac{1}{sqrt{x}}, y=-frac{1}{sqrt{x}}$$
as the two solutions. So the solutions exist, but in order to find them we have to go beyond $mathbb{R}(x)$ to a larger collection of functions.
The algebraic closure of $mathbb{R}(x)$ would contain not only all rational functions, but also roots of all orders, plus some things we don't have names or notation for (solutions of polynomials with coefficients in $mathbb{R}(x)$ of degree 5 or higher that are not solvable by radicals , for example). (The word "functions", here, is used somewhat casually, as most of these things are partial functions at best.) I guess it would have to contain $mathbb{C}$ as well, come to think of it.
What is the algebraic closure called? Does it have a conventional name? I think of it as the "field of algebraic functions" but I don't know if that's standard terminology or if there's another, more commonly-used name. Is there a notation for it? Does it contain any other exotic things, besides what I have already mentioned?
functions field-theory extension-field
$mathbb{R}(x)$ is not algebraically closed, as can easily be seen by considering the equation $xy^2=1$; there is no rational function $r(x)$ with the property that $xr(x)^2=1$. However, everyone knows how to solve this equation for $y$, obtaining
$$y=frac{1}{sqrt{x}}, y=-frac{1}{sqrt{x}}$$
as the two solutions. So the solutions exist, but in order to find them we have to go beyond $mathbb{R}(x)$ to a larger collection of functions.
The algebraic closure of $mathbb{R}(x)$ would contain not only all rational functions, but also roots of all orders, plus some things we don't have names or notation for (solutions of polynomials with coefficients in $mathbb{R}(x)$ of degree 5 or higher that are not solvable by radicals , for example). (The word "functions", here, is used somewhat casually, as most of these things are partial functions at best.) I guess it would have to contain $mathbb{C}$ as well, come to think of it.
What is the algebraic closure called? Does it have a conventional name? I think of it as the "field of algebraic functions" but I don't know if that's standard terminology or if there's another, more commonly-used name. Is there a notation for it? Does it contain any other exotic things, besides what I have already mentioned?
functions field-theory extension-field
functions field-theory extension-field
asked Nov 20 at 3:52
mweiss
17.5k23270
17.5k23270
Possibly related: math.stackexchange.com/questions/2609504, math.stackexchange.com/questions/318690
– Watson
Nov 20 at 8:45
add a comment |
Possibly related: math.stackexchange.com/questions/2609504, math.stackexchange.com/questions/318690
– Watson
Nov 20 at 8:45
Possibly related: math.stackexchange.com/questions/2609504, math.stackexchange.com/questions/318690
– Watson
Nov 20 at 8:45
Possibly related: math.stackexchange.com/questions/2609504, math.stackexchange.com/questions/318690
– Watson
Nov 20 at 8:45
add a comment |
2 Answers
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up vote
4
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Let's change to $mathbb{C}(x)$ (we might as well, since certainly the algebraic closure of $mathbb{R}[x]$ will contain a square root of negative one).
Finite extensions of $mathbb{C}(x)$ are in bijection with pairs of a smooth proper algebraic curves $X$ together with a meromorphic function $phi: X to mathbb{P}^1$. The easy direction of the bijection is that a curve together with a $phi$ gives a map of fields $mathbb{C}(x) to mathbb{C}(X)$ (the field of meromorphic functions on $X$) given by pre-composing a function with $phi$. The interesting direction is that the data of a field extension actually determines an algebraic curve (together with a meromorphic function on the curve), so that every field extension of $mathbb{C}(x)$ arises in this way.
This correspondence is such that if $C_1 to C_2$ is a map of curves, then $mathbb{C}(C_2) hookrightarrow mathbb{C}(C_1)$.
Since the algebraic closure is the union of finite extensions, you can therefore think of it as the space of meromorphic functions on the inverse limit of all the algebraic curves; concretely, an element of it is a choice of curve (together with map to $mathbb{P}^1$) and a meromorphic function on that curve. So for instance, $sqrt{x}$ is "really" a function on the pullback of $mathbb{P}^1$ under $x to x^2$.
1
Note that there is not a canonical "inverse limit of all the algebraic curves" since you have to make some non-canonical choices to put them into an inverse system (the maps between them are not unique). Indeed, it is not even clear to me whether you can prove that an algebraic closure of $mathbb{C}(x)$ exists without the axiom of choice.
– Eric Wofsey
Nov 20 at 4:30
2
Ah, you can construct an algebraic closure using Puiseux series. So, you can get a canonical inverse system by taking subfields of the Puiseux series that are finite extensions of $mathbb{C}(x)$. It's not clear to me how one can think of this inverse system geometrically though.
– Eric Wofsey
Nov 20 at 4:45
1
I guess geometrically this corresponds to considering only curves over $mathbb{P}^1$ which are obtained by first taking the $n$th power map $mathbb{P}^1tomathbb{P}^1$ and then taking a branched cover of $mathbb{P}^1$ obtained by analytically continuing some finite collection of germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 5:20
@EricWofsey Well, once you assume choice you get a much nicer (and at the same time much less useful) description of the algebraic closure: It is $mathbb{C}$. This is of course pretty useless since it says nothing about how the original field of rational functions embed into it.
– Tobias Kildetoft
Nov 20 at 7:07
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0
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For $L/mathbb{C}(z)$ a Galois extension, pick a simply connected open set $U subset mathbb{C}$ such that $L$ is a field of meromorphic functions $U to mathbb{C}$.
Let $gamma(t),t in [0,1]$ be a closed curve in the Riemann sphere with $gamma(0)= gamma(1) in U$.
For $f in L$, continue $f$ analytically along $gamma$, assume it is well-defined and that the resulting function is meromorphic on $U$, call it $f^gamma$.
The map $f mapsto f^gamma$ is an element of $G=Gal(L/mathbb{C}(z))$.
Proof : analytic continuation is clearly compatible with products, addition, quotient, so it is a field morphism, and if $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f(z)^j = 0$ then $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f^gamma(z)^j = 0$ so the image lies in $L$, and since the map is inversible, the image is exactly $L$, so the map is $in Aut(L)$, and it leaves $mathbb{C}(z)$ fixed so it is $in G$.
If $f in L$ and $V$ is a simply connected open set then there exists at most $|G|$ different meromorphic functions $V to mathbb{C}$ that are analytic continuations of $f$.
Write $L = mathbb{C}(z,h(z))$ to obtain that for some set of curves $prod_{gamma in Gamma} h^gamma in mathbb{C}(z) implies forall f in L,prod_{gamma in Gamma} f^gamma in mathbb{C}(z)$ and hence $G = Gamma$.
If $f in L$ then any analytic continuation of $f$ is meromorphic everywhere (on the Riemann sphere) except at finitely many points where $f(a+(z-a)^{|G|})$ is meromorphic (if $a = infty$ check that $f(1/z^{|G|})$ is meromorphic at $0$)
For a branch point $a$ of $f$, continue it analytically by rotating $|G|$ times around $a$ to obtain that $a$ isn't a branch point of $f(a+(z-a)^{|G|})$. And $a$ isn't an essential singularity of $f(a+(z-a)^{|G|})$ because $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f(z)^j = 0$, thus $f(a+(z-a)^{|G|})$ is meromorphic at $a$.
Those criterion are a characterization of the algebraic functions : a meromorphic function $f : U to mathbb{C}$ is in $overline{mathbb{C}(z)}$ iff $f$ has finitely many analytic continuation and those are meromorphic everywhere except at finitely many branch points where $f(a+(z-a)^n)$ is meromorphic.
Proof : note that $prod_{gamma in Gamma} (y-f^gamma(z)) in R[y]$ where $R$ is the ring of meromorphic functions on the Riemann sphere, that is $R = mathbb{C}(z)$ and hence $f in overline{mathbb{C}(z)}$.
There is a very major point you are glossing over here. You are describing when a meromorphic function with branch points is algebraic. That doesn't give you a description of an algebraic closure of $mathbb{C}(z)$, though, since you need to construct a field that simultaneously contains all these functions at once (in a suitable sense). This isn't trivial at all, and requires making some sort of coherent choices of branch cuts on all such functions at once.
– Eric Wofsey
Nov 20 at 7:35
As I commented on the other answer, one way to do this is using Puiseux series, which amounts to identifying your functions with the germs at $0$ of $f(z^n)$ for sufficiently large $n$ such that it is meromorphic at $0$. For any fixed $n$, this gives you an actual field because you are just looking at germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 7:51
As an example of the fallacy of just blindly writing things like $f(z) in overline{mathbb{C}(z)}$ without first defining $overline{mathbb{C}(z)}$ as a field somehow, consider $f(z)$ which is a square root function with the negative real axis as a branch cut and $g(z)$ which is a square root function with the positive real axis as a branch cut. There is no field that contains both $f(z)$ and $g(z)$, since they are equal on one half-plane but not equal on the other.
– Eric Wofsey
Nov 20 at 7:56
@EricWofsey Hi, I tried to repair the sloppy parts
– reuns
Nov 21 at 12:38
What is $overline{mathbb{C}(z)}$ in that context. I'd say a ring of functions on the Riemann sphere such that each function is meromorphic on a dense open subset, is the root of some polynomial $in mathbb{C}(z)[y]$, together with a map $m$ given $U$ returning the functions in $overline{mathbb{C}(z)}$ that are meromorphic on $U$ modulo the equivalence relation $f|_U = g|_U$ so that $m(U)$ is a field of algebraic functions.
– reuns
Nov 21 at 14:35
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2 Answers
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2 Answers
2
active
oldest
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active
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active
oldest
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up vote
4
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Let's change to $mathbb{C}(x)$ (we might as well, since certainly the algebraic closure of $mathbb{R}[x]$ will contain a square root of negative one).
Finite extensions of $mathbb{C}(x)$ are in bijection with pairs of a smooth proper algebraic curves $X$ together with a meromorphic function $phi: X to mathbb{P}^1$. The easy direction of the bijection is that a curve together with a $phi$ gives a map of fields $mathbb{C}(x) to mathbb{C}(X)$ (the field of meromorphic functions on $X$) given by pre-composing a function with $phi$. The interesting direction is that the data of a field extension actually determines an algebraic curve (together with a meromorphic function on the curve), so that every field extension of $mathbb{C}(x)$ arises in this way.
This correspondence is such that if $C_1 to C_2$ is a map of curves, then $mathbb{C}(C_2) hookrightarrow mathbb{C}(C_1)$.
Since the algebraic closure is the union of finite extensions, you can therefore think of it as the space of meromorphic functions on the inverse limit of all the algebraic curves; concretely, an element of it is a choice of curve (together with map to $mathbb{P}^1$) and a meromorphic function on that curve. So for instance, $sqrt{x}$ is "really" a function on the pullback of $mathbb{P}^1$ under $x to x^2$.
1
Note that there is not a canonical "inverse limit of all the algebraic curves" since you have to make some non-canonical choices to put them into an inverse system (the maps between them are not unique). Indeed, it is not even clear to me whether you can prove that an algebraic closure of $mathbb{C}(x)$ exists without the axiom of choice.
– Eric Wofsey
Nov 20 at 4:30
2
Ah, you can construct an algebraic closure using Puiseux series. So, you can get a canonical inverse system by taking subfields of the Puiseux series that are finite extensions of $mathbb{C}(x)$. It's not clear to me how one can think of this inverse system geometrically though.
– Eric Wofsey
Nov 20 at 4:45
1
I guess geometrically this corresponds to considering only curves over $mathbb{P}^1$ which are obtained by first taking the $n$th power map $mathbb{P}^1tomathbb{P}^1$ and then taking a branched cover of $mathbb{P}^1$ obtained by analytically continuing some finite collection of germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 5:20
@EricWofsey Well, once you assume choice you get a much nicer (and at the same time much less useful) description of the algebraic closure: It is $mathbb{C}$. This is of course pretty useless since it says nothing about how the original field of rational functions embed into it.
– Tobias Kildetoft
Nov 20 at 7:07
add a comment |
up vote
4
down vote
Let's change to $mathbb{C}(x)$ (we might as well, since certainly the algebraic closure of $mathbb{R}[x]$ will contain a square root of negative one).
Finite extensions of $mathbb{C}(x)$ are in bijection with pairs of a smooth proper algebraic curves $X$ together with a meromorphic function $phi: X to mathbb{P}^1$. The easy direction of the bijection is that a curve together with a $phi$ gives a map of fields $mathbb{C}(x) to mathbb{C}(X)$ (the field of meromorphic functions on $X$) given by pre-composing a function with $phi$. The interesting direction is that the data of a field extension actually determines an algebraic curve (together with a meromorphic function on the curve), so that every field extension of $mathbb{C}(x)$ arises in this way.
This correspondence is such that if $C_1 to C_2$ is a map of curves, then $mathbb{C}(C_2) hookrightarrow mathbb{C}(C_1)$.
Since the algebraic closure is the union of finite extensions, you can therefore think of it as the space of meromorphic functions on the inverse limit of all the algebraic curves; concretely, an element of it is a choice of curve (together with map to $mathbb{P}^1$) and a meromorphic function on that curve. So for instance, $sqrt{x}$ is "really" a function on the pullback of $mathbb{P}^1$ under $x to x^2$.
1
Note that there is not a canonical "inverse limit of all the algebraic curves" since you have to make some non-canonical choices to put them into an inverse system (the maps between them are not unique). Indeed, it is not even clear to me whether you can prove that an algebraic closure of $mathbb{C}(x)$ exists without the axiom of choice.
– Eric Wofsey
Nov 20 at 4:30
2
Ah, you can construct an algebraic closure using Puiseux series. So, you can get a canonical inverse system by taking subfields of the Puiseux series that are finite extensions of $mathbb{C}(x)$. It's not clear to me how one can think of this inverse system geometrically though.
– Eric Wofsey
Nov 20 at 4:45
1
I guess geometrically this corresponds to considering only curves over $mathbb{P}^1$ which are obtained by first taking the $n$th power map $mathbb{P}^1tomathbb{P}^1$ and then taking a branched cover of $mathbb{P}^1$ obtained by analytically continuing some finite collection of germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 5:20
@EricWofsey Well, once you assume choice you get a much nicer (and at the same time much less useful) description of the algebraic closure: It is $mathbb{C}$. This is of course pretty useless since it says nothing about how the original field of rational functions embed into it.
– Tobias Kildetoft
Nov 20 at 7:07
add a comment |
up vote
4
down vote
up vote
4
down vote
Let's change to $mathbb{C}(x)$ (we might as well, since certainly the algebraic closure of $mathbb{R}[x]$ will contain a square root of negative one).
Finite extensions of $mathbb{C}(x)$ are in bijection with pairs of a smooth proper algebraic curves $X$ together with a meromorphic function $phi: X to mathbb{P}^1$. The easy direction of the bijection is that a curve together with a $phi$ gives a map of fields $mathbb{C}(x) to mathbb{C}(X)$ (the field of meromorphic functions on $X$) given by pre-composing a function with $phi$. The interesting direction is that the data of a field extension actually determines an algebraic curve (together with a meromorphic function on the curve), so that every field extension of $mathbb{C}(x)$ arises in this way.
This correspondence is such that if $C_1 to C_2$ is a map of curves, then $mathbb{C}(C_2) hookrightarrow mathbb{C}(C_1)$.
Since the algebraic closure is the union of finite extensions, you can therefore think of it as the space of meromorphic functions on the inverse limit of all the algebraic curves; concretely, an element of it is a choice of curve (together with map to $mathbb{P}^1$) and a meromorphic function on that curve. So for instance, $sqrt{x}$ is "really" a function on the pullback of $mathbb{P}^1$ under $x to x^2$.
Let's change to $mathbb{C}(x)$ (we might as well, since certainly the algebraic closure of $mathbb{R}[x]$ will contain a square root of negative one).
Finite extensions of $mathbb{C}(x)$ are in bijection with pairs of a smooth proper algebraic curves $X$ together with a meromorphic function $phi: X to mathbb{P}^1$. The easy direction of the bijection is that a curve together with a $phi$ gives a map of fields $mathbb{C}(x) to mathbb{C}(X)$ (the field of meromorphic functions on $X$) given by pre-composing a function with $phi$. The interesting direction is that the data of a field extension actually determines an algebraic curve (together with a meromorphic function on the curve), so that every field extension of $mathbb{C}(x)$ arises in this way.
This correspondence is such that if $C_1 to C_2$ is a map of curves, then $mathbb{C}(C_2) hookrightarrow mathbb{C}(C_1)$.
Since the algebraic closure is the union of finite extensions, you can therefore think of it as the space of meromorphic functions on the inverse limit of all the algebraic curves; concretely, an element of it is a choice of curve (together with map to $mathbb{P}^1$) and a meromorphic function on that curve. So for instance, $sqrt{x}$ is "really" a function on the pullback of $mathbb{P}^1$ under $x to x^2$.
answered Nov 20 at 4:12
hunter
14k22437
14k22437
1
Note that there is not a canonical "inverse limit of all the algebraic curves" since you have to make some non-canonical choices to put them into an inverse system (the maps between them are not unique). Indeed, it is not even clear to me whether you can prove that an algebraic closure of $mathbb{C}(x)$ exists without the axiom of choice.
– Eric Wofsey
Nov 20 at 4:30
2
Ah, you can construct an algebraic closure using Puiseux series. So, you can get a canonical inverse system by taking subfields of the Puiseux series that are finite extensions of $mathbb{C}(x)$. It's not clear to me how one can think of this inverse system geometrically though.
– Eric Wofsey
Nov 20 at 4:45
1
I guess geometrically this corresponds to considering only curves over $mathbb{P}^1$ which are obtained by first taking the $n$th power map $mathbb{P}^1tomathbb{P}^1$ and then taking a branched cover of $mathbb{P}^1$ obtained by analytically continuing some finite collection of germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 5:20
@EricWofsey Well, once you assume choice you get a much nicer (and at the same time much less useful) description of the algebraic closure: It is $mathbb{C}$. This is of course pretty useless since it says nothing about how the original field of rational functions embed into it.
– Tobias Kildetoft
Nov 20 at 7:07
add a comment |
1
Note that there is not a canonical "inverse limit of all the algebraic curves" since you have to make some non-canonical choices to put them into an inverse system (the maps between them are not unique). Indeed, it is not even clear to me whether you can prove that an algebraic closure of $mathbb{C}(x)$ exists without the axiom of choice.
– Eric Wofsey
Nov 20 at 4:30
2
Ah, you can construct an algebraic closure using Puiseux series. So, you can get a canonical inverse system by taking subfields of the Puiseux series that are finite extensions of $mathbb{C}(x)$. It's not clear to me how one can think of this inverse system geometrically though.
– Eric Wofsey
Nov 20 at 4:45
1
I guess geometrically this corresponds to considering only curves over $mathbb{P}^1$ which are obtained by first taking the $n$th power map $mathbb{P}^1tomathbb{P}^1$ and then taking a branched cover of $mathbb{P}^1$ obtained by analytically continuing some finite collection of germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 5:20
@EricWofsey Well, once you assume choice you get a much nicer (and at the same time much less useful) description of the algebraic closure: It is $mathbb{C}$. This is of course pretty useless since it says nothing about how the original field of rational functions embed into it.
– Tobias Kildetoft
Nov 20 at 7:07
1
1
Note that there is not a canonical "inverse limit of all the algebraic curves" since you have to make some non-canonical choices to put them into an inverse system (the maps between them are not unique). Indeed, it is not even clear to me whether you can prove that an algebraic closure of $mathbb{C}(x)$ exists without the axiom of choice.
– Eric Wofsey
Nov 20 at 4:30
Note that there is not a canonical "inverse limit of all the algebraic curves" since you have to make some non-canonical choices to put them into an inverse system (the maps between them are not unique). Indeed, it is not even clear to me whether you can prove that an algebraic closure of $mathbb{C}(x)$ exists without the axiom of choice.
– Eric Wofsey
Nov 20 at 4:30
2
2
Ah, you can construct an algebraic closure using Puiseux series. So, you can get a canonical inverse system by taking subfields of the Puiseux series that are finite extensions of $mathbb{C}(x)$. It's not clear to me how one can think of this inverse system geometrically though.
– Eric Wofsey
Nov 20 at 4:45
Ah, you can construct an algebraic closure using Puiseux series. So, you can get a canonical inverse system by taking subfields of the Puiseux series that are finite extensions of $mathbb{C}(x)$. It's not clear to me how one can think of this inverse system geometrically though.
– Eric Wofsey
Nov 20 at 4:45
1
1
I guess geometrically this corresponds to considering only curves over $mathbb{P}^1$ which are obtained by first taking the $n$th power map $mathbb{P}^1tomathbb{P}^1$ and then taking a branched cover of $mathbb{P}^1$ obtained by analytically continuing some finite collection of germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 5:20
I guess geometrically this corresponds to considering only curves over $mathbb{P}^1$ which are obtained by first taking the $n$th power map $mathbb{P}^1tomathbb{P}^1$ and then taking a branched cover of $mathbb{P}^1$ obtained by analytically continuing some finite collection of germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 5:20
@EricWofsey Well, once you assume choice you get a much nicer (and at the same time much less useful) description of the algebraic closure: It is $mathbb{C}$. This is of course pretty useless since it says nothing about how the original field of rational functions embed into it.
– Tobias Kildetoft
Nov 20 at 7:07
@EricWofsey Well, once you assume choice you get a much nicer (and at the same time much less useful) description of the algebraic closure: It is $mathbb{C}$. This is of course pretty useless since it says nothing about how the original field of rational functions embed into it.
– Tobias Kildetoft
Nov 20 at 7:07
add a comment |
up vote
0
down vote
For $L/mathbb{C}(z)$ a Galois extension, pick a simply connected open set $U subset mathbb{C}$ such that $L$ is a field of meromorphic functions $U to mathbb{C}$.
Let $gamma(t),t in [0,1]$ be a closed curve in the Riemann sphere with $gamma(0)= gamma(1) in U$.
For $f in L$, continue $f$ analytically along $gamma$, assume it is well-defined and that the resulting function is meromorphic on $U$, call it $f^gamma$.
The map $f mapsto f^gamma$ is an element of $G=Gal(L/mathbb{C}(z))$.
Proof : analytic continuation is clearly compatible with products, addition, quotient, so it is a field morphism, and if $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f(z)^j = 0$ then $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f^gamma(z)^j = 0$ so the image lies in $L$, and since the map is inversible, the image is exactly $L$, so the map is $in Aut(L)$, and it leaves $mathbb{C}(z)$ fixed so it is $in G$.
If $f in L$ and $V$ is a simply connected open set then there exists at most $|G|$ different meromorphic functions $V to mathbb{C}$ that are analytic continuations of $f$.
Write $L = mathbb{C}(z,h(z))$ to obtain that for some set of curves $prod_{gamma in Gamma} h^gamma in mathbb{C}(z) implies forall f in L,prod_{gamma in Gamma} f^gamma in mathbb{C}(z)$ and hence $G = Gamma$.
If $f in L$ then any analytic continuation of $f$ is meromorphic everywhere (on the Riemann sphere) except at finitely many points where $f(a+(z-a)^{|G|})$ is meromorphic (if $a = infty$ check that $f(1/z^{|G|})$ is meromorphic at $0$)
For a branch point $a$ of $f$, continue it analytically by rotating $|G|$ times around $a$ to obtain that $a$ isn't a branch point of $f(a+(z-a)^{|G|})$. And $a$ isn't an essential singularity of $f(a+(z-a)^{|G|})$ because $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f(z)^j = 0$, thus $f(a+(z-a)^{|G|})$ is meromorphic at $a$.
Those criterion are a characterization of the algebraic functions : a meromorphic function $f : U to mathbb{C}$ is in $overline{mathbb{C}(z)}$ iff $f$ has finitely many analytic continuation and those are meromorphic everywhere except at finitely many branch points where $f(a+(z-a)^n)$ is meromorphic.
Proof : note that $prod_{gamma in Gamma} (y-f^gamma(z)) in R[y]$ where $R$ is the ring of meromorphic functions on the Riemann sphere, that is $R = mathbb{C}(z)$ and hence $f in overline{mathbb{C}(z)}$.
There is a very major point you are glossing over here. You are describing when a meromorphic function with branch points is algebraic. That doesn't give you a description of an algebraic closure of $mathbb{C}(z)$, though, since you need to construct a field that simultaneously contains all these functions at once (in a suitable sense). This isn't trivial at all, and requires making some sort of coherent choices of branch cuts on all such functions at once.
– Eric Wofsey
Nov 20 at 7:35
As I commented on the other answer, one way to do this is using Puiseux series, which amounts to identifying your functions with the germs at $0$ of $f(z^n)$ for sufficiently large $n$ such that it is meromorphic at $0$. For any fixed $n$, this gives you an actual field because you are just looking at germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 7:51
As an example of the fallacy of just blindly writing things like $f(z) in overline{mathbb{C}(z)}$ without first defining $overline{mathbb{C}(z)}$ as a field somehow, consider $f(z)$ which is a square root function with the negative real axis as a branch cut and $g(z)$ which is a square root function with the positive real axis as a branch cut. There is no field that contains both $f(z)$ and $g(z)$, since they are equal on one half-plane but not equal on the other.
– Eric Wofsey
Nov 20 at 7:56
@EricWofsey Hi, I tried to repair the sloppy parts
– reuns
Nov 21 at 12:38
What is $overline{mathbb{C}(z)}$ in that context. I'd say a ring of functions on the Riemann sphere such that each function is meromorphic on a dense open subset, is the root of some polynomial $in mathbb{C}(z)[y]$, together with a map $m$ given $U$ returning the functions in $overline{mathbb{C}(z)}$ that are meromorphic on $U$ modulo the equivalence relation $f|_U = g|_U$ so that $m(U)$ is a field of algebraic functions.
– reuns
Nov 21 at 14:35
|
show 3 more comments
up vote
0
down vote
For $L/mathbb{C}(z)$ a Galois extension, pick a simply connected open set $U subset mathbb{C}$ such that $L$ is a field of meromorphic functions $U to mathbb{C}$.
Let $gamma(t),t in [0,1]$ be a closed curve in the Riemann sphere with $gamma(0)= gamma(1) in U$.
For $f in L$, continue $f$ analytically along $gamma$, assume it is well-defined and that the resulting function is meromorphic on $U$, call it $f^gamma$.
The map $f mapsto f^gamma$ is an element of $G=Gal(L/mathbb{C}(z))$.
Proof : analytic continuation is clearly compatible with products, addition, quotient, so it is a field morphism, and if $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f(z)^j = 0$ then $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f^gamma(z)^j = 0$ so the image lies in $L$, and since the map is inversible, the image is exactly $L$, so the map is $in Aut(L)$, and it leaves $mathbb{C}(z)$ fixed so it is $in G$.
If $f in L$ and $V$ is a simply connected open set then there exists at most $|G|$ different meromorphic functions $V to mathbb{C}$ that are analytic continuations of $f$.
Write $L = mathbb{C}(z,h(z))$ to obtain that for some set of curves $prod_{gamma in Gamma} h^gamma in mathbb{C}(z) implies forall f in L,prod_{gamma in Gamma} f^gamma in mathbb{C}(z)$ and hence $G = Gamma$.
If $f in L$ then any analytic continuation of $f$ is meromorphic everywhere (on the Riemann sphere) except at finitely many points where $f(a+(z-a)^{|G|})$ is meromorphic (if $a = infty$ check that $f(1/z^{|G|})$ is meromorphic at $0$)
For a branch point $a$ of $f$, continue it analytically by rotating $|G|$ times around $a$ to obtain that $a$ isn't a branch point of $f(a+(z-a)^{|G|})$. And $a$ isn't an essential singularity of $f(a+(z-a)^{|G|})$ because $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f(z)^j = 0$, thus $f(a+(z-a)^{|G|})$ is meromorphic at $a$.
Those criterion are a characterization of the algebraic functions : a meromorphic function $f : U to mathbb{C}$ is in $overline{mathbb{C}(z)}$ iff $f$ has finitely many analytic continuation and those are meromorphic everywhere except at finitely many branch points where $f(a+(z-a)^n)$ is meromorphic.
Proof : note that $prod_{gamma in Gamma} (y-f^gamma(z)) in R[y]$ where $R$ is the ring of meromorphic functions on the Riemann sphere, that is $R = mathbb{C}(z)$ and hence $f in overline{mathbb{C}(z)}$.
There is a very major point you are glossing over here. You are describing when a meromorphic function with branch points is algebraic. That doesn't give you a description of an algebraic closure of $mathbb{C}(z)$, though, since you need to construct a field that simultaneously contains all these functions at once (in a suitable sense). This isn't trivial at all, and requires making some sort of coherent choices of branch cuts on all such functions at once.
– Eric Wofsey
Nov 20 at 7:35
As I commented on the other answer, one way to do this is using Puiseux series, which amounts to identifying your functions with the germs at $0$ of $f(z^n)$ for sufficiently large $n$ such that it is meromorphic at $0$. For any fixed $n$, this gives you an actual field because you are just looking at germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 7:51
As an example of the fallacy of just blindly writing things like $f(z) in overline{mathbb{C}(z)}$ without first defining $overline{mathbb{C}(z)}$ as a field somehow, consider $f(z)$ which is a square root function with the negative real axis as a branch cut and $g(z)$ which is a square root function with the positive real axis as a branch cut. There is no field that contains both $f(z)$ and $g(z)$, since they are equal on one half-plane but not equal on the other.
– Eric Wofsey
Nov 20 at 7:56
@EricWofsey Hi, I tried to repair the sloppy parts
– reuns
Nov 21 at 12:38
What is $overline{mathbb{C}(z)}$ in that context. I'd say a ring of functions on the Riemann sphere such that each function is meromorphic on a dense open subset, is the root of some polynomial $in mathbb{C}(z)[y]$, together with a map $m$ given $U$ returning the functions in $overline{mathbb{C}(z)}$ that are meromorphic on $U$ modulo the equivalence relation $f|_U = g|_U$ so that $m(U)$ is a field of algebraic functions.
– reuns
Nov 21 at 14:35
|
show 3 more comments
up vote
0
down vote
up vote
0
down vote
For $L/mathbb{C}(z)$ a Galois extension, pick a simply connected open set $U subset mathbb{C}$ such that $L$ is a field of meromorphic functions $U to mathbb{C}$.
Let $gamma(t),t in [0,1]$ be a closed curve in the Riemann sphere with $gamma(0)= gamma(1) in U$.
For $f in L$, continue $f$ analytically along $gamma$, assume it is well-defined and that the resulting function is meromorphic on $U$, call it $f^gamma$.
The map $f mapsto f^gamma$ is an element of $G=Gal(L/mathbb{C}(z))$.
Proof : analytic continuation is clearly compatible with products, addition, quotient, so it is a field morphism, and if $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f(z)^j = 0$ then $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f^gamma(z)^j = 0$ so the image lies in $L$, and since the map is inversible, the image is exactly $L$, so the map is $in Aut(L)$, and it leaves $mathbb{C}(z)$ fixed so it is $in G$.
If $f in L$ and $V$ is a simply connected open set then there exists at most $|G|$ different meromorphic functions $V to mathbb{C}$ that are analytic continuations of $f$.
Write $L = mathbb{C}(z,h(z))$ to obtain that for some set of curves $prod_{gamma in Gamma} h^gamma in mathbb{C}(z) implies forall f in L,prod_{gamma in Gamma} f^gamma in mathbb{C}(z)$ and hence $G = Gamma$.
If $f in L$ then any analytic continuation of $f$ is meromorphic everywhere (on the Riemann sphere) except at finitely many points where $f(a+(z-a)^{|G|})$ is meromorphic (if $a = infty$ check that $f(1/z^{|G|})$ is meromorphic at $0$)
For a branch point $a$ of $f$, continue it analytically by rotating $|G|$ times around $a$ to obtain that $a$ isn't a branch point of $f(a+(z-a)^{|G|})$. And $a$ isn't an essential singularity of $f(a+(z-a)^{|G|})$ because $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f(z)^j = 0$, thus $f(a+(z-a)^{|G|})$ is meromorphic at $a$.
Those criterion are a characterization of the algebraic functions : a meromorphic function $f : U to mathbb{C}$ is in $overline{mathbb{C}(z)}$ iff $f$ has finitely many analytic continuation and those are meromorphic everywhere except at finitely many branch points where $f(a+(z-a)^n)$ is meromorphic.
Proof : note that $prod_{gamma in Gamma} (y-f^gamma(z)) in R[y]$ where $R$ is the ring of meromorphic functions on the Riemann sphere, that is $R = mathbb{C}(z)$ and hence $f in overline{mathbb{C}(z)}$.
For $L/mathbb{C}(z)$ a Galois extension, pick a simply connected open set $U subset mathbb{C}$ such that $L$ is a field of meromorphic functions $U to mathbb{C}$.
Let $gamma(t),t in [0,1]$ be a closed curve in the Riemann sphere with $gamma(0)= gamma(1) in U$.
For $f in L$, continue $f$ analytically along $gamma$, assume it is well-defined and that the resulting function is meromorphic on $U$, call it $f^gamma$.
The map $f mapsto f^gamma$ is an element of $G=Gal(L/mathbb{C}(z))$.
Proof : analytic continuation is clearly compatible with products, addition, quotient, so it is a field morphism, and if $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f(z)^j = 0$ then $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f^gamma(z)^j = 0$ so the image lies in $L$, and since the map is inversible, the image is exactly $L$, so the map is $in Aut(L)$, and it leaves $mathbb{C}(z)$ fixed so it is $in G$.
If $f in L$ and $V$ is a simply connected open set then there exists at most $|G|$ different meromorphic functions $V to mathbb{C}$ that are analytic continuations of $f$.
Write $L = mathbb{C}(z,h(z))$ to obtain that for some set of curves $prod_{gamma in Gamma} h^gamma in mathbb{C}(z) implies forall f in L,prod_{gamma in Gamma} f^gamma in mathbb{C}(z)$ and hence $G = Gamma$.
If $f in L$ then any analytic continuation of $f$ is meromorphic everywhere (on the Riemann sphere) except at finitely many points where $f(a+(z-a)^{|G|})$ is meromorphic (if $a = infty$ check that $f(1/z^{|G|})$ is meromorphic at $0$)
For a branch point $a$ of $f$, continue it analytically by rotating $|G|$ times around $a$ to obtain that $a$ isn't a branch point of $f(a+(z-a)^{|G|})$. And $a$ isn't an essential singularity of $f(a+(z-a)^{|G|})$ because $sum_{j=0}^J frac{P_j(z)}{Q_j(z)} f(z)^j = 0$, thus $f(a+(z-a)^{|G|})$ is meromorphic at $a$.
Those criterion are a characterization of the algebraic functions : a meromorphic function $f : U to mathbb{C}$ is in $overline{mathbb{C}(z)}$ iff $f$ has finitely many analytic continuation and those are meromorphic everywhere except at finitely many branch points where $f(a+(z-a)^n)$ is meromorphic.
Proof : note that $prod_{gamma in Gamma} (y-f^gamma(z)) in R[y]$ where $R$ is the ring of meromorphic functions on the Riemann sphere, that is $R = mathbb{C}(z)$ and hence $f in overline{mathbb{C}(z)}$.
edited Nov 21 at 12:59
answered Nov 20 at 4:50
reuns
19.3k21046
19.3k21046
There is a very major point you are glossing over here. You are describing when a meromorphic function with branch points is algebraic. That doesn't give you a description of an algebraic closure of $mathbb{C}(z)$, though, since you need to construct a field that simultaneously contains all these functions at once (in a suitable sense). This isn't trivial at all, and requires making some sort of coherent choices of branch cuts on all such functions at once.
– Eric Wofsey
Nov 20 at 7:35
As I commented on the other answer, one way to do this is using Puiseux series, which amounts to identifying your functions with the germs at $0$ of $f(z^n)$ for sufficiently large $n$ such that it is meromorphic at $0$. For any fixed $n$, this gives you an actual field because you are just looking at germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 7:51
As an example of the fallacy of just blindly writing things like $f(z) in overline{mathbb{C}(z)}$ without first defining $overline{mathbb{C}(z)}$ as a field somehow, consider $f(z)$ which is a square root function with the negative real axis as a branch cut and $g(z)$ which is a square root function with the positive real axis as a branch cut. There is no field that contains both $f(z)$ and $g(z)$, since they are equal on one half-plane but not equal on the other.
– Eric Wofsey
Nov 20 at 7:56
@EricWofsey Hi, I tried to repair the sloppy parts
– reuns
Nov 21 at 12:38
What is $overline{mathbb{C}(z)}$ in that context. I'd say a ring of functions on the Riemann sphere such that each function is meromorphic on a dense open subset, is the root of some polynomial $in mathbb{C}(z)[y]$, together with a map $m$ given $U$ returning the functions in $overline{mathbb{C}(z)}$ that are meromorphic on $U$ modulo the equivalence relation $f|_U = g|_U$ so that $m(U)$ is a field of algebraic functions.
– reuns
Nov 21 at 14:35
|
show 3 more comments
There is a very major point you are glossing over here. You are describing when a meromorphic function with branch points is algebraic. That doesn't give you a description of an algebraic closure of $mathbb{C}(z)$, though, since you need to construct a field that simultaneously contains all these functions at once (in a suitable sense). This isn't trivial at all, and requires making some sort of coherent choices of branch cuts on all such functions at once.
– Eric Wofsey
Nov 20 at 7:35
As I commented on the other answer, one way to do this is using Puiseux series, which amounts to identifying your functions with the germs at $0$ of $f(z^n)$ for sufficiently large $n$ such that it is meromorphic at $0$. For any fixed $n$, this gives you an actual field because you are just looking at germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 7:51
As an example of the fallacy of just blindly writing things like $f(z) in overline{mathbb{C}(z)}$ without first defining $overline{mathbb{C}(z)}$ as a field somehow, consider $f(z)$ which is a square root function with the negative real axis as a branch cut and $g(z)$ which is a square root function with the positive real axis as a branch cut. There is no field that contains both $f(z)$ and $g(z)$, since they are equal on one half-plane but not equal on the other.
– Eric Wofsey
Nov 20 at 7:56
@EricWofsey Hi, I tried to repair the sloppy parts
– reuns
Nov 21 at 12:38
What is $overline{mathbb{C}(z)}$ in that context. I'd say a ring of functions on the Riemann sphere such that each function is meromorphic on a dense open subset, is the root of some polynomial $in mathbb{C}(z)[y]$, together with a map $m$ given $U$ returning the functions in $overline{mathbb{C}(z)}$ that are meromorphic on $U$ modulo the equivalence relation $f|_U = g|_U$ so that $m(U)$ is a field of algebraic functions.
– reuns
Nov 21 at 14:35
There is a very major point you are glossing over here. You are describing when a meromorphic function with branch points is algebraic. That doesn't give you a description of an algebraic closure of $mathbb{C}(z)$, though, since you need to construct a field that simultaneously contains all these functions at once (in a suitable sense). This isn't trivial at all, and requires making some sort of coherent choices of branch cuts on all such functions at once.
– Eric Wofsey
Nov 20 at 7:35
There is a very major point you are glossing over here. You are describing when a meromorphic function with branch points is algebraic. That doesn't give you a description of an algebraic closure of $mathbb{C}(z)$, though, since you need to construct a field that simultaneously contains all these functions at once (in a suitable sense). This isn't trivial at all, and requires making some sort of coherent choices of branch cuts on all such functions at once.
– Eric Wofsey
Nov 20 at 7:35
As I commented on the other answer, one way to do this is using Puiseux series, which amounts to identifying your functions with the germs at $0$ of $f(z^n)$ for sufficiently large $n$ such that it is meromorphic at $0$. For any fixed $n$, this gives you an actual field because you are just looking at germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 7:51
As I commented on the other answer, one way to do this is using Puiseux series, which amounts to identifying your functions with the germs at $0$ of $f(z^n)$ for sufficiently large $n$ such that it is meromorphic at $0$. For any fixed $n$, this gives you an actual field because you are just looking at germs of meromorphic functions at $0$.
– Eric Wofsey
Nov 20 at 7:51
As an example of the fallacy of just blindly writing things like $f(z) in overline{mathbb{C}(z)}$ without first defining $overline{mathbb{C}(z)}$ as a field somehow, consider $f(z)$ which is a square root function with the negative real axis as a branch cut and $g(z)$ which is a square root function with the positive real axis as a branch cut. There is no field that contains both $f(z)$ and $g(z)$, since they are equal on one half-plane but not equal on the other.
– Eric Wofsey
Nov 20 at 7:56
As an example of the fallacy of just blindly writing things like $f(z) in overline{mathbb{C}(z)}$ without first defining $overline{mathbb{C}(z)}$ as a field somehow, consider $f(z)$ which is a square root function with the negative real axis as a branch cut and $g(z)$ which is a square root function with the positive real axis as a branch cut. There is no field that contains both $f(z)$ and $g(z)$, since they are equal on one half-plane but not equal on the other.
– Eric Wofsey
Nov 20 at 7:56
@EricWofsey Hi, I tried to repair the sloppy parts
– reuns
Nov 21 at 12:38
@EricWofsey Hi, I tried to repair the sloppy parts
– reuns
Nov 21 at 12:38
What is $overline{mathbb{C}(z)}$ in that context. I'd say a ring of functions on the Riemann sphere such that each function is meromorphic on a dense open subset, is the root of some polynomial $in mathbb{C}(z)[y]$, together with a map $m$ given $U$ returning the functions in $overline{mathbb{C}(z)}$ that are meromorphic on $U$ modulo the equivalence relation $f|_U = g|_U$ so that $m(U)$ is a field of algebraic functions.
– reuns
Nov 21 at 14:35
What is $overline{mathbb{C}(z)}$ in that context. I'd say a ring of functions on the Riemann sphere such that each function is meromorphic on a dense open subset, is the root of some polynomial $in mathbb{C}(z)[y]$, together with a map $m$ given $U$ returning the functions in $overline{mathbb{C}(z)}$ that are meromorphic on $U$ modulo the equivalence relation $f|_U = g|_U$ so that $m(U)$ is a field of algebraic functions.
– reuns
Nov 21 at 14:35
|
show 3 more comments
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Possibly related: math.stackexchange.com/questions/2609504, math.stackexchange.com/questions/318690
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