Mixing
Why is $mathbb{C}$ over $mathbb{R}$ considered ramified?
$begingroup$
For a number field $K/mathbb{Q}$, we say that a finite place of $Q$ is ramified if there exists a valuation $v_{p_i}$ in $K$ lying over $v_p$ such that it is ramified in the sense of the associated discrete valuation rings. This is the ramification index, denoted $e_{p_i/p}$.
We also have the residue field extension, of degree $f_{p_i/p}$.
With these, we can define $n_{p_i/p}$ as $f_{p_i/p}e_{p_i/p}$, and the $n_{p_i/p}$ behave well, they sum to the degree of the extension and so on.
Now this is all lovely over a finite place, but for infinite places, in the treatment I saw, only the integers $n_{infty_i/infty}$ were defined, as $1$ if the associated extension is real to real, and $2$ if real to complex. These infinite $n_{infty_i/infty}$ behave in the same way as the finite $n_{p_i/p}$.
So then the question is, why do we view $n_{infty_i/infty}=2$ as ramified, eg, from the perspective of CFT. Is there an intuitive explanation of why these ought to be ramified and not just the analogue of a purely $f_{p_i/p}$ extension of finite places?
algebraic-number-theory absolute-value class-field-theory ramification
asked Feb 2 at 11:41
user277182user277182
561212
$endgroup$
add a comment |
$begingroup$
For a number field $K/mathbb{Q}$, we say that a finite place of $Q$ is ramified if there exists a valuation $v_{p_i}$ in $K$ lying over $v_p$ such that it is ramified in the sense of the associated discrete valuation rings. This is the ramification index, denoted $e_{p_i/p}$.
We also have the residue field extension, of degree $f_{p_i/p}$.
With these, we can define $n_{p_i/p}$ as $f_{p_i/p}e_{p_i/p}$, and the $n_{p_i/p}$ behave well, they sum to the degree of the extension and so on.
Now this is all lovely over a finite place, but for infinite places, in the treatment I saw, only the integers $n_{infty_i/infty}$ were defined, as $1$ if the associated extension is real to real, and $2$ if real to complex. These infinite $n_{infty_i/infty}$ behave in the same way as the finite $n_{p_i/p}$.
So then the question is, why do we view $n_{infty_i/infty}=2$ as ramified, eg, from the perspective of CFT. Is there an intuitive explanation of why these ought to be ramified and not just the analogue of a purely $f_{p_i/p}$ extension of finite places?
algebraic-number-theory absolute-value class-field-theory ramification
asked Feb 2 at 11:41
user277182user277182
561212
$endgroup$
1
$begingroup$
See math.stackexchange.com/a/2157935/300700
$endgroup$
– nguyen quang do
Feb 3 at 21:52
add a comment |
$begingroup$
For a number field $K/mathbb{Q}$, we say that a finite place of $Q$ is ramified if there exists a valuation $v_{p_i}$ in $K$ lying over $v_p$ such that it is ramified in the sense of the associated discrete valuation rings. This is the ramification index, denoted $e_{p_i/p}$.
We also have the residue field extension, of degree $f_{p_i/p}$.
With these, we can define $n_{p_i/p}$ as $f_{p_i/p}e_{p_i/p}$, and the $n_{p_i/p}$ behave well, they sum to the degree of the extension and so on.
Now this is all lovely over a finite place, but for infinite places, in the treatment I saw, only the integers $n_{infty_i/infty}$ were defined, as $1$ if the associated extension is real to real, and $2$ if real to complex. These infinite $n_{infty_i/infty}$ behave in the same way as the finite $n_{p_i/p}$.
So then the question is, why do we view $n_{infty_i/infty}=2$ as ramified, eg, from the perspective of CFT. Is there an intuitive explanation of why these ought to be ramified and not just the analogue of a purely $f_{p_i/p}$ extension of finite places?
algebraic-number-theory absolute-value class-field-theory ramification
asked Feb 2 at 11:41
user277182user277182
561212
$endgroup$
For a number field $K/mathbb{Q}$, we say that a finite place of $Q$ is ramified if there exists a valuation $v_{p_i}$ in $K$ lying over $v_p$ such that it is ramified in the sense of the associated discrete valuation rings. This is the ramification index, denoted $e_{p_i/p}$.
We also have the residue field extension, of degree $f_{p_i/p}$.
With these, we can define $n_{p_i/p}$ as $f_{p_i/p}e_{p_i/p}$, and the $n_{p_i/p}$ behave well, they sum to the degree of the extension and so on.
Now this is all lovely over a finite place, but for infinite places, in the treatment I saw, only the integers $n_{infty_i/infty}$ were defined, as $1$ if the associated extension is real to real, and $2$ if real to complex. These infinite $n_{infty_i/infty}$ behave in the same way as the finite $n_{p_i/p}$.
So then the question is, why do we view $n_{infty_i/infty}=2$ as ramified, eg, from the perspective of CFT. Is there an intuitive explanation of why these ought to be ramified and not just the analogue of a purely $f_{p_i/p}$ extension of finite places?
algebraic-number-theory absolute-value class-field-theory ramification
algebraic-number-theory absolute-value class-field-theory ramification
asked Feb 2 at 11:41
user277182user277182
561212
asked Feb 2 at 11:41
user277182user277182
561212
asked Feb 2 at 11:41
user277182user277182
561212
asked Feb 2 at 11:41
user277182user277182
561212
asked Feb 2 at 11:41
user277182user277182
561212
561212
1
$begingroup$
See math.stackexchange.com/a/2157935/300700
$endgroup$
– nguyen quang do
Feb 3 at 21:52
add a comment |
1
$begingroup$
See math.stackexchange.com/a/2157935/300700
$endgroup$
– nguyen quang do
Feb 3 at 21:52
1
1
$begingroup$
See math.stackexchange.com/a/2157935/300700
$endgroup$
– nguyen quang do
Feb 3 at 21:52
$begingroup$
See math.stackexchange.com/a/2157935/300700
$endgroup$
– nguyen quang do
Feb 3 at 21:52
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097210%2fwhy-is-mathbbc-over-mathbbr-considered-ramified%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097210%2fwhy-is-mathbbc-over-mathbbr-considered-ramified%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
See math.stackexchange.com/a/2157935/300700
$endgroup$
– nguyen quang do
Feb 3 at 21:52