Example of a field of functions containing ln(x)
Up until now, I've mainly worked with the polynomial ring $mathbb{R}[x_1,...,x_n]$ or the corresponding field of fractions. But I can't think of an example of a field of functions that contain non-polynomial functions. What examples are there of a field of functions that contain the set of real rational functions and the natural logarithm function $ln(x)$?
Thanks!
commutative-algebra field-theory extension-field
asked Jan 4 at 17:23
ChrisWongChrisWong
1508
add a comment |
Up until now, I've mainly worked with the polynomial ring $mathbb{R}[x_1,...,x_n]$ or the corresponding field of fractions. But I can't think of an example of a field of functions that contain non-polynomial functions. What examples are there of a field of functions that contain the set of real rational functions and the natural logarithm function $ln(x)$?
Thanks!
commutative-algebra field-theory extension-field
asked Jan 4 at 17:23
ChrisWongChrisWong
1508
add a comment |
Up until now, I've mainly worked with the polynomial ring $mathbb{R}[x_1,...,x_n]$ or the corresponding field of fractions. But I can't think of an example of a field of functions that contain non-polynomial functions. What examples are there of a field of functions that contain the set of real rational functions and the natural logarithm function $ln(x)$?
Thanks!
commutative-algebra field-theory extension-field
asked Jan 4 at 17:23
ChrisWongChrisWong
1508
Up until now, I've mainly worked with the polynomial ring $mathbb{R}[x_1,...,x_n]$ or the corresponding field of fractions. But I can't think of an example of a field of functions that contain non-polynomial functions. What examples are there of a field of functions that contain the set of real rational functions and the natural logarithm function $ln(x)$?
Thanks!
commutative-algebra field-theory extension-field
commutative-algebra field-theory extension-field
asked Jan 4 at 17:23
ChrisWongChrisWong
1508
asked Jan 4 at 17:23
ChrisWongChrisWong
1508
asked Jan 4 at 17:23
ChrisWongChrisWong
1508
asked Jan 4 at 17:23
ChrisWongChrisWong
1508
asked Jan 4 at 17:23
ChrisWongChrisWong
1508
1508
add a comment |
add a comment |
1 Answer
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You can definitely study the field of functions $mathbb{R}(x_1,ldots, x_n)=text{Frac}(mathbb{R}[x_1,ldots, x_n]).$ This is the space of rational functions in the variables $x_1,ldots, x_n$. More abstractly, for any integral domain $A$, one can study $A[x_1,ldots, x_n]$ and $A(x_1,ldots, x_n)$.
As far as an example of a field containing $log$ or something like that, we can take $Omega=mathbb{C}setminus [0,infty)$ and study
$$mathcal{M}(Omega)={text{meromorphic functions on}:Omega}.$$
Among these functions is $log(z)$ where we use the branch cut $text{arg}(z)in (0,2pi)$. It can be shown that the ring $mathcal{M}(Omega)$ is actually a field, containing $log$.
answered Jan 4 at 17:31
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
That makes sense, would the field of meromorphic functions also include sin(z) and cos(z)? (I haven't taken complex analysis yet so my knowledge of this stuff is a little shaky). Is the set of holomorphic functions on $Omega$ also a field?
– ChrisWong
Jan 4 at 17:46
$sin$ and $cos$ would also be included. The holomorphic functions are not a field, because if you take $1/(z-i)$ (for instance), we can see that it isn't holomorphic: it has a singularity at $z=i$.
– Antonios-Alexandros Robotis
Jan 4 at 17:47
add a comment |
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You can definitely study the field of functions $mathbb{R}(x_1,ldots, x_n)=text{Frac}(mathbb{R}[x_1,ldots, x_n]).$ This is the space of rational functions in the variables $x_1,ldots, x_n$. More abstractly, for any integral domain $A$, one can study $A[x_1,ldots, x_n]$ and $A(x_1,ldots, x_n)$.
As far as an example of a field containing $log$ or something like that, we can take $Omega=mathbb{C}setminus [0,infty)$ and study
$$mathcal{M}(Omega)={text{meromorphic functions on}:Omega}.$$
Among these functions is $log(z)$ where we use the branch cut $text{arg}(z)in (0,2pi)$. It can be shown that the ring $mathcal{M}(Omega)$ is actually a field, containing $log$.
answered Jan 4 at 17:31
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
That makes sense, would the field of meromorphic functions also include sin(z) and cos(z)? (I haven't taken complex analysis yet so my knowledge of this stuff is a little shaky). Is the set of holomorphic functions on $Omega$ also a field?
– ChrisWong
Jan 4 at 17:46
$sin$ and $cos$ would also be included. The holomorphic functions are not a field, because if you take $1/(z-i)$ (for instance), we can see that it isn't holomorphic: it has a singularity at $z=i$.
– Antonios-Alexandros Robotis
Jan 4 at 17:47
add a comment |
You can definitely study the field of functions $mathbb{R}(x_1,ldots, x_n)=text{Frac}(mathbb{R}[x_1,ldots, x_n]).$ This is the space of rational functions in the variables $x_1,ldots, x_n$. More abstractly, for any integral domain $A$, one can study $A[x_1,ldots, x_n]$ and $A(x_1,ldots, x_n)$.
As far as an example of a field containing $log$ or something like that, we can take $Omega=mathbb{C}setminus [0,infty)$ and study
$$mathcal{M}(Omega)={text{meromorphic functions on}:Omega}.$$
Among these functions is $log(z)$ where we use the branch cut $text{arg}(z)in (0,2pi)$. It can be shown that the ring $mathcal{M}(Omega)$ is actually a field, containing $log$.
answered Jan 4 at 17:31
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
That makes sense, would the field of meromorphic functions also include sin(z) and cos(z)? (I haven't taken complex analysis yet so my knowledge of this stuff is a little shaky). Is the set of holomorphic functions on $Omega$ also a field?
– ChrisWong
Jan 4 at 17:46
$sin$ and $cos$ would also be included. The holomorphic functions are not a field, because if you take $1/(z-i)$ (for instance), we can see that it isn't holomorphic: it has a singularity at $z=i$.
– Antonios-Alexandros Robotis
Jan 4 at 17:47
add a comment |
You can definitely study the field of functions $mathbb{R}(x_1,ldots, x_n)=text{Frac}(mathbb{R}[x_1,ldots, x_n]).$ This is the space of rational functions in the variables $x_1,ldots, x_n$. More abstractly, for any integral domain $A$, one can study $A[x_1,ldots, x_n]$ and $A(x_1,ldots, x_n)$.
As far as an example of a field containing $log$ or something like that, we can take $Omega=mathbb{C}setminus [0,infty)$ and study
$$mathcal{M}(Omega)={text{meromorphic functions on}:Omega}.$$
Among these functions is $log(z)$ where we use the branch cut $text{arg}(z)in (0,2pi)$. It can be shown that the ring $mathcal{M}(Omega)$ is actually a field, containing $log$.
answered Jan 4 at 17:31
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
You can definitely study the field of functions $mathbb{R}(x_1,ldots, x_n)=text{Frac}(mathbb{R}[x_1,ldots, x_n]).$ This is the space of rational functions in the variables $x_1,ldots, x_n$. More abstractly, for any integral domain $A$, one can study $A[x_1,ldots, x_n]$ and $A(x_1,ldots, x_n)$.
As far as an example of a field containing $log$ or something like that, we can take $Omega=mathbb{C}setminus [0,infty)$ and study
$$mathcal{M}(Omega)={text{meromorphic functions on}:Omega}.$$
Among these functions is $log(z)$ where we use the branch cut $text{arg}(z)in (0,2pi)$. It can be shown that the ring $mathcal{M}(Omega)$ is actually a field, containing $log$.
answered Jan 4 at 17:31
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
answered Jan 4 at 17:31
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
answered Jan 4 at 17:31
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
answered Jan 4 at 17:31
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
9,71741640
That makes sense, would the field of meromorphic functions also include sin(z) and cos(z)? (I haven't taken complex analysis yet so my knowledge of this stuff is a little shaky). Is the set of holomorphic functions on $Omega$ also a field?
– ChrisWong
Jan 4 at 17:46
$sin$ and $cos$ would also be included. The holomorphic functions are not a field, because if you take $1/(z-i)$ (for instance), we can see that it isn't holomorphic: it has a singularity at $z=i$.
– Antonios-Alexandros Robotis
Jan 4 at 17:47
add a comment |
That makes sense, would the field of meromorphic functions also include sin(z) and cos(z)? (I haven't taken complex analysis yet so my knowledge of this stuff is a little shaky). Is the set of holomorphic functions on $Omega$ also a field?
– ChrisWong
Jan 4 at 17:46
$sin$ and $cos$ would also be included. The holomorphic functions are not a field, because if you take $1/(z-i)$ (for instance), we can see that it isn't holomorphic: it has a singularity at $z=i$.
– Antonios-Alexandros Robotis
Jan 4 at 17:47
That makes sense, would the field of meromorphic functions also include sin(z) and cos(z)? (I haven't taken complex analysis yet so my knowledge of this stuff is a little shaky). Is the set of holomorphic functions on $Omega$ also a field?
– ChrisWong
Jan 4 at 17:46
That makes sense, would the field of meromorphic functions also include sin(z) and cos(z)? (I haven't taken complex analysis yet so my knowledge of this stuff is a little shaky). Is the set of holomorphic functions on $Omega$ also a field?
– ChrisWong
Jan 4 at 17:46
$sin$ and $cos$ would also be included. The holomorphic functions are not a field, because if you take $1/(z-i)$ (for instance), we can see that it isn't holomorphic: it has a singularity at $z=i$.
– Antonios-Alexandros Robotis
Jan 4 at 17:47
$sin$ and $cos$ would also be included. The holomorphic functions are not a field, because if you take $1/(z-i)$ (for instance), we can see that it isn't holomorphic: it has a singularity at $z=i$.
– Antonios-Alexandros Robotis
Jan 4 at 17:47
add a comment |
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