Example of a field of functions containing ln(x)
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
10.5k41741
$endgroup$
$begingroup$
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?
$endgroup$
– ChrisWong
Jan 4 at 17:46
$begingroup$
$sin$ 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 $z=i$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 4 at 17:47
add a comment |
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$begingroup$
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
10.5k41741
$endgroup$
$begingroup$
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?
$endgroup$
– ChrisWong
Jan 4 at 17:46
$begingroup$
$sin$ 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 $z=i$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 4 at 17:47
add a comment |
$begingroup$
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
10.5k41741
$endgroup$
$begingroup$
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?
$endgroup$
– ChrisWong
Jan 4 at 17:46
$begingroup$
$sin$ 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 $z=i$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 4 at 17:47
add a comment |
$begingroup$
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
10.5k41741
$endgroup$
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
10.5k41741
answered Jan 4 at 17:31
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 4 at 17:31
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 4 at 17:31
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
10.5k41741
$begingroup$
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?
$endgroup$
– ChrisWong
Jan 4 at 17:46
$begingroup$
$sin$ 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 $z=i$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 4 at 17:47
add a comment |
$begingroup$
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?
$endgroup$
– ChrisWong
Jan 4 at 17:46
$begingroup$
$sin$ 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 $z=i$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 4 at 17:47
$begingroup$
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?
$endgroup$
– ChrisWong
Jan 4 at 17:46
$begingroup$
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?
$endgroup$
– ChrisWong
Jan 4 at 17:46
$begingroup$
$sin$ 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 $z=i$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 4 at 17:47
$begingroup$
$sin$ 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 $z=i$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 4 at 17:47
add a comment |
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