What is the condoctor of $ mathbb Q(sqrt 2 )/mathbb Q?$ $4mathbb Z or 8 mathbb Z?$












0














By execise 6.8 of Childress's Algebraic Number Theory, $mathbb f(mathbb Q(sqrt 2)/ mathbb Q)=8mathbb Z$. But considerring the norm of the local field about the place corespondent with 2, it should be $4mathbb Z$ and there is contraction to the ording theorem.






algebraic-number-theory







share|cite|improve this question







New contributor




lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    Certainly $m=8$ is the smallest integer such that $Bbb{Q}(sqrt2)subseteqBbb{Q}(zeta_m)$. What does the ording theorem say. I couldn't find it.
    – Jyrki Lahtonen
    11 hours ago












  • Also, it seems to me that $1+8Bbb{Z}$ is contained in the image of the norm map, but $1+4Bbb{Z}$ certainly is not: $x^2-2y^2notequiv5pmod8$. So the local conductor is $3$ resulting in a factor $2^3$. What result suggests that it should $4$ only?
    – Jyrki Lahtonen
    11 hours ago












  • Yes, you are right. By the lemme 5.3 of Childress's Class Field Theory, I get that. And the ordering theorem is on P136. It seems that I am wrong about the image of the norm map.
    – lorry
    10 hours ago










  • I don't have Childress, that's why I asked you to quote the theorem. Anyway, integers congruent to $1$ modulo $8$ have square roots in $Bbb{Q}_2$, so they are trivially included in the image of the norm.
    – Jyrki Lahtonen
    8 hours ago
















0














By execise 6.8 of Childress's Algebraic Number Theory, $mathbb f(mathbb Q(sqrt 2)/ mathbb Q)=8mathbb Z$. But considerring the norm of the local field about the place corespondent with 2, it should be $4mathbb Z$ and there is contraction to the ording theorem.






algebraic-number-theory







share|cite|improve this question







New contributor




lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    Certainly $m=8$ is the smallest integer such that $Bbb{Q}(sqrt2)subseteqBbb{Q}(zeta_m)$. What does the ording theorem say. I couldn't find it.
    – Jyrki Lahtonen
    11 hours ago












  • Also, it seems to me that $1+8Bbb{Z}$ is contained in the image of the norm map, but $1+4Bbb{Z}$ certainly is not: $x^2-2y^2notequiv5pmod8$. So the local conductor is $3$ resulting in a factor $2^3$. What result suggests that it should $4$ only?
    – Jyrki Lahtonen
    11 hours ago












  • Yes, you are right. By the lemme 5.3 of Childress's Class Field Theory, I get that. And the ordering theorem is on P136. It seems that I am wrong about the image of the norm map.
    – lorry
    10 hours ago










  • I don't have Childress, that's why I asked you to quote the theorem. Anyway, integers congruent to $1$ modulo $8$ have square roots in $Bbb{Q}_2$, so they are trivially included in the image of the norm.
    – Jyrki Lahtonen
    8 hours ago














0












0








0







By execise 6.8 of Childress's Algebraic Number Theory, $mathbb f(mathbb Q(sqrt 2)/ mathbb Q)=8mathbb Z$. But considerring the norm of the local field about the place corespondent with 2, it should be $4mathbb Z$ and there is contraction to the ording theorem.






algebraic-number-theory







share|cite|improve this question







New contributor




lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











By execise 6.8 of Childress's Algebraic Number Theory, $mathbb f(mathbb Q(sqrt 2)/ mathbb Q)=8mathbb Z$. But considerring the norm of the local field about the place corespondent with 2, it should be $4mathbb Z$ and there is contraction to the ording theorem.






algebraic-number-theory




algebraic-number-theory






share|cite|improve this question







New contributor




lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 12 hours ago









lorry

1




1




New contributor




lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    Certainly $m=8$ is the smallest integer such that $Bbb{Q}(sqrt2)subseteqBbb{Q}(zeta_m)$. What does the ording theorem say. I couldn't find it.
    – Jyrki Lahtonen
    11 hours ago












  • Also, it seems to me that $1+8Bbb{Z}$ is contained in the image of the norm map, but $1+4Bbb{Z}$ certainly is not: $x^2-2y^2notequiv5pmod8$. So the local conductor is $3$ resulting in a factor $2^3$. What result suggests that it should $4$ only?
    – Jyrki Lahtonen
    11 hours ago












  • Yes, you are right. By the lemme 5.3 of Childress's Class Field Theory, I get that. And the ordering theorem is on P136. It seems that I am wrong about the image of the norm map.
    – lorry
    10 hours ago










  • I don't have Childress, that's why I asked you to quote the theorem. Anyway, integers congruent to $1$ modulo $8$ have square roots in $Bbb{Q}_2$, so they are trivially included in the image of the norm.
    – Jyrki Lahtonen
    8 hours ago














  • 1




    Certainly $m=8$ is the smallest integer such that $Bbb{Q}(sqrt2)subseteqBbb{Q}(zeta_m)$. What does the ording theorem say. I couldn't find it.
    – Jyrki Lahtonen
    11 hours ago












  • Also, it seems to me that $1+8Bbb{Z}$ is contained in the image of the norm map, but $1+4Bbb{Z}$ certainly is not: $x^2-2y^2notequiv5pmod8$. So the local conductor is $3$ resulting in a factor $2^3$. What result suggests that it should $4$ only?
    – Jyrki Lahtonen
    11 hours ago












  • Yes, you are right. By the lemme 5.3 of Childress's Class Field Theory, I get that. And the ordering theorem is on P136. It seems that I am wrong about the image of the norm map.
    – lorry
    10 hours ago










  • I don't have Childress, that's why I asked you to quote the theorem. Anyway, integers congruent to $1$ modulo $8$ have square roots in $Bbb{Q}_2$, so they are trivially included in the image of the norm.
    – Jyrki Lahtonen
    8 hours ago








1




1




Certainly $m=8$ is the smallest integer such that $Bbb{Q}(sqrt2)subseteqBbb{Q}(zeta_m)$. What does the ording theorem say. I couldn't find it.
– Jyrki Lahtonen
11 hours ago






Certainly $m=8$ is the smallest integer such that $Bbb{Q}(sqrt2)subseteqBbb{Q}(zeta_m)$. What does the ording theorem say. I couldn't find it.
– Jyrki Lahtonen
11 hours ago














Also, it seems to me that $1+8Bbb{Z}$ is contained in the image of the norm map, but $1+4Bbb{Z}$ certainly is not: $x^2-2y^2notequiv5pmod8$. So the local conductor is $3$ resulting in a factor $2^3$. What result suggests that it should $4$ only?
– Jyrki Lahtonen
11 hours ago






Also, it seems to me that $1+8Bbb{Z}$ is contained in the image of the norm map, but $1+4Bbb{Z}$ certainly is not: $x^2-2y^2notequiv5pmod8$. So the local conductor is $3$ resulting in a factor $2^3$. What result suggests that it should $4$ only?
– Jyrki Lahtonen
11 hours ago














Yes, you are right. By the lemme 5.3 of Childress's Class Field Theory, I get that. And the ordering theorem is on P136. It seems that I am wrong about the image of the norm map.
– lorry
10 hours ago




Yes, you are right. By the lemme 5.3 of Childress's Class Field Theory, I get that. And the ordering theorem is on P136. It seems that I am wrong about the image of the norm map.
– lorry
10 hours ago












I don't have Childress, that's why I asked you to quote the theorem. Anyway, integers congruent to $1$ modulo $8$ have square roots in $Bbb{Q}_2$, so they are trivially included in the image of the norm.
– Jyrki Lahtonen
8 hours ago




I don't have Childress, that's why I asked you to quote the theorem. Anyway, integers congruent to $1$ modulo $8$ have square roots in $Bbb{Q}_2$, so they are trivially included in the image of the norm.
– Jyrki Lahtonen
8 hours ago










0






active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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
});


}
});






lorry is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060308%2fwhat-is-the-condoctor-of-mathbb-q-sqrt-2-mathbb-q-4-mathbb-z-or-8%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








lorry is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















lorry is a new contributor. Be nice, and check out our Code of Conduct.













lorry is a new contributor. Be nice, and check out our Code of Conduct.












lorry is a new contributor. Be nice, and check out our Code of Conduct.
















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060308%2fwhat-is-the-condoctor-of-mathbb-q-sqrt-2-mathbb-q-4-mathbb-z-or-8%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

An IMO inspired problem

Image
14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
Read more

Management

Image
Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
Read more

Has there ever been an instance of an active nuclear power plant within or near a war zone?

Image
36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
Read more