Ordered subfields of $mathbb{Q}_p$
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I recently read about real ordered fields. Using real closures, I figured out that for each algebraically closed field $C$ of characteristic $0$, there exists a real closed subfield $Rsubseteq C$ such that the extension is algebraic (and $[C:R]=2$). This follows very easily using real closures and Zorn's lemma.
Applying this to the field $overline{mathbb{Q}}_p$, it follows that there exists an ordered subfield $Rsubseteq mathbb{Q}_p$ such that the extension is algebraic. The field $mathbb{Q}_p$ cannot be ordered itself (as $-1$ is a sum of squares), it follows that $Rsubseteq mathbb{Q}_p$ is a strict extension. However, $mathbb{Q}_p$ has no nontrivial endomorphisms, so the extension cannot be normal.
Are there any explicit constructions for $R$? And if the extension is finite, what would be a generator for $mathbb{Q}_p$?
abstract-algebra p-adic-number-theory ordered-fields
asked Dec 9 '18 at 21:35
MarMar
786411
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add a comment |
$begingroup$
I recently read about real ordered fields. Using real closures, I figured out that for each algebraically closed field $C$ of characteristic $0$, there exists a real closed subfield $Rsubseteq C$ such that the extension is algebraic (and $[C:R]=2$). This follows very easily using real closures and Zorn's lemma.
Applying this to the field $overline{mathbb{Q}}_p$, it follows that there exists an ordered subfield $Rsubseteq mathbb{Q}_p$ such that the extension is algebraic. The field $mathbb{Q}_p$ cannot be ordered itself (as $-1$ is a sum of squares), it follows that $Rsubseteq mathbb{Q}_p$ is a strict extension. However, $mathbb{Q}_p$ has no nontrivial endomorphisms, so the extension cannot be normal.
Are there any explicit constructions for $R$? And if the extension is finite, what would be a generator for $mathbb{Q}_p$?
abstract-algebra p-adic-number-theory ordered-fields
asked Dec 9 '18 at 21:35
MarMar
786411
$endgroup$
2
$begingroup$
Here's a terrible construction: pick any isomorphism from $overline{mathbb{Q}}_p$ to $mathbb{C}$, then consider the image of $mathbb{R}$.
$endgroup$
– Qiaochu Yuan
Dec 9 '18 at 21:44
4
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You’ve made (extremely) fundamental use of Axiom of Choice here, I think. You should expect that there is no explicit construction of the type you ask for. More generally speaking, it seems to me that you’re asking to do something radically unnatural. If it yields something, more power to you. But I don’t think it will.
$endgroup$
– Lubin
Dec 10 '18 at 2:08
add a comment |
$begingroup$
I recently read about real ordered fields. Using real closures, I figured out that for each algebraically closed field $C$ of characteristic $0$, there exists a real closed subfield $Rsubseteq C$ such that the extension is algebraic (and $[C:R]=2$). This follows very easily using real closures and Zorn's lemma.
Applying this to the field $overline{mathbb{Q}}_p$, it follows that there exists an ordered subfield $Rsubseteq mathbb{Q}_p$ such that the extension is algebraic. The field $mathbb{Q}_p$ cannot be ordered itself (as $-1$ is a sum of squares), it follows that $Rsubseteq mathbb{Q}_p$ is a strict extension. However, $mathbb{Q}_p$ has no nontrivial endomorphisms, so the extension cannot be normal.
Are there any explicit constructions for $R$? And if the extension is finite, what would be a generator for $mathbb{Q}_p$?
abstract-algebra p-adic-number-theory ordered-fields
asked Dec 9 '18 at 21:35
MarMar
786411
$endgroup$
I recently read about real ordered fields. Using real closures, I figured out that for each algebraically closed field $C$ of characteristic $0$, there exists a real closed subfield $Rsubseteq C$ such that the extension is algebraic (and $[C:R]=2$). This follows very easily using real closures and Zorn's lemma.
Applying this to the field $overline{mathbb{Q}}_p$, it follows that there exists an ordered subfield $Rsubseteq mathbb{Q}_p$ such that the extension is algebraic. The field $mathbb{Q}_p$ cannot be ordered itself (as $-1$ is a sum of squares), it follows that $Rsubseteq mathbb{Q}_p$ is a strict extension. However, $mathbb{Q}_p$ has no nontrivial endomorphisms, so the extension cannot be normal.
Are there any explicit constructions for $R$? And if the extension is finite, what would be a generator for $mathbb{Q}_p$?
abstract-algebra p-adic-number-theory ordered-fields
abstract-algebra p-adic-number-theory ordered-fields
asked Dec 9 '18 at 21:35
MarMar
786411
asked Dec 9 '18 at 21:35
MarMar
786411
asked Dec 9 '18 at 21:35
MarMar
786411
asked Dec 9 '18 at 21:35
MarMar
786411
asked Dec 9 '18 at 21:35
MarMar
786411
786411
2
$begingroup$
Here's a terrible construction: pick any isomorphism from $overline{mathbb{Q}}_p$ to $mathbb{C}$, then consider the image of $mathbb{R}$.
$endgroup$
– Qiaochu Yuan
Dec 9 '18 at 21:44
4
$begingroup$
You’ve made (extremely) fundamental use of Axiom of Choice here, I think. You should expect that there is no explicit construction of the type you ask for. More generally speaking, it seems to me that you’re asking to do something radically unnatural. If it yields something, more power to you. But I don’t think it will.
$endgroup$
– Lubin
Dec 10 '18 at 2:08
add a comment |
2
$begingroup$
Here's a terrible construction: pick any isomorphism from $overline{mathbb{Q}}_p$ to $mathbb{C}$, then consider the image of $mathbb{R}$.
$endgroup$
– Qiaochu Yuan
Dec 9 '18 at 21:44
4
$begingroup$
You’ve made (extremely) fundamental use of Axiom of Choice here, I think. You should expect that there is no explicit construction of the type you ask for. More generally speaking, it seems to me that you’re asking to do something radically unnatural. If it yields something, more power to you. But I don’t think it will.
$endgroup$
– Lubin
Dec 10 '18 at 2:08
2
2
$begingroup$
Here's a terrible construction: pick any isomorphism from $overline{mathbb{Q}}_p$ to $mathbb{C}$, then consider the image of $mathbb{R}$.
$endgroup$
– Qiaochu Yuan
Dec 9 '18 at 21:44
$begingroup$
Here's a terrible construction: pick any isomorphism from $overline{mathbb{Q}}_p$ to $mathbb{C}$, then consider the image of $mathbb{R}$.
$endgroup$
– Qiaochu Yuan
Dec 9 '18 at 21:44
4
4
$begingroup$
You’ve made (extremely) fundamental use of Axiom of Choice here, I think. You should expect that there is no explicit construction of the type you ask for. More generally speaking, it seems to me that you’re asking to do something radically unnatural. If it yields something, more power to you. But I don’t think it will.
$endgroup$
– Lubin
Dec 10 '18 at 2:08
$begingroup$
You’ve made (extremely) fundamental use of Axiom of Choice here, I think. You should expect that there is no explicit construction of the type you ask for. More generally speaking, it seems to me that you’re asking to do something radically unnatural. If it yields something, more power to you. But I don’t think it will.
$endgroup$
– Lubin
Dec 10 '18 at 2:08
add a comment |
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$begingroup$
Here's a terrible construction: pick any isomorphism from $overline{mathbb{Q}}_p$ to $mathbb{C}$, then consider the image of $mathbb{R}$.
$endgroup$
– Qiaochu Yuan
Dec 9 '18 at 21:44
4
$begingroup$
You’ve made (extremely) fundamental use of Axiom of Choice here, I think. You should expect that there is no explicit construction of the type you ask for. More generally speaking, it seems to me that you’re asking to do something radically unnatural. If it yields something, more power to you. But I don’t think it will.
$endgroup$
– Lubin
Dec 10 '18 at 2:08