Mixing
Example of an infinite simple extension.
$begingroup$
I'm new to Field Theory and I'm looking for an example of an infinite simple extension.
Theorem: The element $alpha$ is algebraic over $F$ if and only if the simple extension $F(alpha)/F$ is finite.
Using the above theorem, I guess that something like $Bbb{Q}({pi})/Bbb{Q}$ is an example of such an extension. But the proof of $pi$ is transcendental is not at all trivial(and I don't think I can understand the proof with my limited knowledge).
So I have got two questions:
Is my example correct?
Are there any methods to construct an infinite simple extension which doesn't require more sophisticated tools?
Also, I would love to see some "trivial" examples(if they exist).
Any help is appreciated.
Thank you.
field-theory extension-field
asked Jan 11 at 7:44
Thomas ShelbyThomas Shelby
2,877421
$endgroup$
add a comment |
$begingroup$
I'm new to Field Theory and I'm looking for an example of an infinite simple extension.
Theorem: The element $alpha$ is algebraic over $F$ if and only if the simple extension $F(alpha)/F$ is finite.
Using the above theorem, I guess that something like $Bbb{Q}({pi})/Bbb{Q}$ is an example of such an extension. But the proof of $pi$ is transcendental is not at all trivial(and I don't think I can understand the proof with my limited knowledge).
So I have got two questions:
Is my example correct?
Are there any methods to construct an infinite simple extension which doesn't require more sophisticated tools?
Also, I would love to see some "trivial" examples(if they exist).
Any help is appreciated.
Thank you.
field-theory extension-field
asked Jan 11 at 7:44
Thomas ShelbyThomas Shelby
2,877421
$endgroup$
1
$begingroup$
Your example is correct. Any transcendental should work, and they should all look the same. You can just let $alpha$ be any transcendental.
$endgroup$
– Gaffney
Jan 11 at 7:57
add a comment |
$begingroup$
I'm new to Field Theory and I'm looking for an example of an infinite simple extension.
Theorem: The element $alpha$ is algebraic over $F$ if and only if the simple extension $F(alpha)/F$ is finite.
Using the above theorem, I guess that something like $Bbb{Q}({pi})/Bbb{Q}$ is an example of such an extension. But the proof of $pi$ is transcendental is not at all trivial(and I don't think I can understand the proof with my limited knowledge).
So I have got two questions:
Is my example correct?
Are there any methods to construct an infinite simple extension which doesn't require more sophisticated tools?
Also, I would love to see some "trivial" examples(if they exist).
Any help is appreciated.
Thank you.
field-theory extension-field
asked Jan 11 at 7:44
Thomas ShelbyThomas Shelby
2,877421
$endgroup$
I'm new to Field Theory and I'm looking for an example of an infinite simple extension.
Theorem: The element $alpha$ is algebraic over $F$ if and only if the simple extension $F(alpha)/F$ is finite.
Using the above theorem, I guess that something like $Bbb{Q}({pi})/Bbb{Q}$ is an example of such an extension. But the proof of $pi$ is transcendental is not at all trivial(and I don't think I can understand the proof with my limited knowledge).
So I have got two questions:
Is my example correct?
Are there any methods to construct an infinite simple extension which doesn't require more sophisticated tools?
Also, I would love to see some "trivial" examples(if they exist).
Any help is appreciated.
Thank you.
field-theory extension-field
field-theory extension-field
asked Jan 11 at 7:44
Thomas ShelbyThomas Shelby
2,877421
asked Jan 11 at 7:44
Thomas ShelbyThomas Shelby
2,877421
asked Jan 11 at 7:44
Thomas ShelbyThomas Shelby
2,877421
asked Jan 11 at 7:44
Thomas ShelbyThomas Shelby
2,877421
asked Jan 11 at 7:44
Thomas ShelbyThomas Shelby
2,877421
2,877421
1
$begingroup$
Your example is correct. Any transcendental should work, and they should all look the same. You can just let $alpha$ be any transcendental.
$endgroup$
– Gaffney
Jan 11 at 7:57
add a comment |
1
$begingroup$
Your example is correct. Any transcendental should work, and they should all look the same. You can just let $alpha$ be any transcendental.
$endgroup$
– Gaffney
Jan 11 at 7:57
1
1
$begingroup$
Your example is correct. Any transcendental should work, and they should all look the same. You can just let $alpha$ be any transcendental.
$endgroup$
– Gaffney
Jan 11 at 7:57
$begingroup$
Your example is correct. Any transcendental should work, and they should all look the same. You can just let $alpha$ be any transcendental.
$endgroup$
– Gaffney
Jan 11 at 7:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes your example is correct. Another example would be $k(X)/k$ where $k(X)$ is the field of fractions of the polynomial ring $k[X]$, and $X$ is clearly transcendental.
answered Jan 11 at 7:58
asdqasdq
1,8211518
$endgroup$
$begingroup$
Thanks a lot. But I cannot understand why $X $ is transcendental and couldn't find any information online. Would you provide a hint or some useful link?
$endgroup$
– Thomas Shelby
Jan 11 at 8:43
2
$begingroup$
An element is algebraic if it is killed by a polynomial, so by definition of $k[X]$ the element $X$ cannot be algebraic.
$endgroup$
– asdq
Jan 11 at 8:47
add a comment |
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$begingroup$
Yes your example is correct. Another example would be $k(X)/k$ where $k(X)$ is the field of fractions of the polynomial ring $k[X]$, and $X$ is clearly transcendental.
answered Jan 11 at 7:58
asdqasdq
1,8211518
$endgroup$
$begingroup$
Thanks a lot. But I cannot understand why $X $ is transcendental and couldn't find any information online. Would you provide a hint or some useful link?
$endgroup$
– Thomas Shelby
Jan 11 at 8:43
2
$begingroup$
An element is algebraic if it is killed by a polynomial, so by definition of $k[X]$ the element $X$ cannot be algebraic.
$endgroup$
– asdq
Jan 11 at 8:47
add a comment |
$begingroup$
Yes your example is correct. Another example would be $k(X)/k$ where $k(X)$ is the field of fractions of the polynomial ring $k[X]$, and $X$ is clearly transcendental.
answered Jan 11 at 7:58
asdqasdq
1,8211518
$endgroup$
$begingroup$
Thanks a lot. But I cannot understand why $X $ is transcendental and couldn't find any information online. Would you provide a hint or some useful link?
$endgroup$
– Thomas Shelby
Jan 11 at 8:43
2
$begingroup$
An element is algebraic if it is killed by a polynomial, so by definition of $k[X]$ the element $X$ cannot be algebraic.
$endgroup$
– asdq
Jan 11 at 8:47
add a comment |
$begingroup$
Yes your example is correct. Another example would be $k(X)/k$ where $k(X)$ is the field of fractions of the polynomial ring $k[X]$, and $X$ is clearly transcendental.
answered Jan 11 at 7:58
asdqasdq
1,8211518
$endgroup$
Yes your example is correct. Another example would be $k(X)/k$ where $k(X)$ is the field of fractions of the polynomial ring $k[X]$, and $X$ is clearly transcendental.
answered Jan 11 at 7:58
asdqasdq
1,8211518
answered Jan 11 at 7:58
asdqasdq
1,8211518
answered Jan 11 at 7:58
asdqasdq
1,8211518
answered Jan 11 at 7:58
asdqasdq
1,8211518
1,8211518
$begingroup$
Thanks a lot. But I cannot understand why $X $ is transcendental and couldn't find any information online. Would you provide a hint or some useful link?
$endgroup$
– Thomas Shelby
Jan 11 at 8:43
2
$begingroup$
An element is algebraic if it is killed by a polynomial, so by definition of $k[X]$ the element $X$ cannot be algebraic.
$endgroup$
– asdq
Jan 11 at 8:47
add a comment |
$begingroup$
Thanks a lot. But I cannot understand why $X $ is transcendental and couldn't find any information online. Would you provide a hint or some useful link?
$endgroup$
– Thomas Shelby
Jan 11 at 8:43
2
$begingroup$
An element is algebraic if it is killed by a polynomial, so by definition of $k[X]$ the element $X$ cannot be algebraic.
$endgroup$
– asdq
Jan 11 at 8:47
$begingroup$
Thanks a lot. But I cannot understand why $X $ is transcendental and couldn't find any information online. Would you provide a hint or some useful link?
$endgroup$
– Thomas Shelby
Jan 11 at 8:43
$begingroup$
Thanks a lot. But I cannot understand why $X $ is transcendental and couldn't find any information online. Would you provide a hint or some useful link?
$endgroup$
– Thomas Shelby
Jan 11 at 8:43
2
2
$begingroup$
An element is algebraic if it is killed by a polynomial, so by definition of $k[X]$ the element $X$ cannot be algebraic.
$endgroup$
– asdq
Jan 11 at 8:47
$begingroup$
An element is algebraic if it is killed by a polynomial, so by definition of $k[X]$ the element $X$ cannot be algebraic.
$endgroup$
– asdq
Jan 11 at 8:47
add a comment |
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$begingroup$
Your example is correct. Any transcendental should work, and they should all look the same. You can just let $alpha$ be any transcendental.
$endgroup$
– Gaffney
Jan 11 at 7:57