Mixing
What does $(a,b)_{zeta}$ correspond to in $mathrm{Br}(mathbb{Q}_p)=mathbb{Q}/mathbb{Z}$
$begingroup$
Let $p$ be a prime number, let $mathbb{Q}_p$ be the local field, by Hensel's lemma, we know it has $p-1$-th roots of unity, let $zeta$ be a fixed primitive $p-1$-th root of unity in $mathbb{Q}_p$.
Let $a,binmathbb{Q}_p^*$, let $(a,b)_{zeta}$ be the cyclic algebra defined by $mathbb{Q}_plangle x,y|x^{p-1}=a,y^{p-1}=b, xy=zeta yxrangle$.
Is there an explicit way to tell which class $[(a,b)_zeta]inmathrm{Br}(mathbb{Q}_p)=mathbb{Q}/mathbb{Z}$ is? (For example, when $a,binmathbb{Q}$, how does the two rational numbers determine an element in $mathbb{Q}/mathbb{Z}$?)
class-field-theory division-algebras
asked Nov 13 '18 at 20:04
QixiaoQixiao
2,8621628
$endgroup$
|
show 1 more comment
$begingroup$
Let $p$ be a prime number, let $mathbb{Q}_p$ be the local field, by Hensel's lemma, we know it has $p-1$-th roots of unity, let $zeta$ be a fixed primitive $p-1$-th root of unity in $mathbb{Q}_p$.
Let $a,binmathbb{Q}_p^*$, let $(a,b)_{zeta}$ be the cyclic algebra defined by $mathbb{Q}_plangle x,y|x^{p-1}=a,y^{p-1}=b, xy=zeta yxrangle$.
Is there an explicit way to tell which class $[(a,b)_zeta]inmathrm{Br}(mathbb{Q}_p)=mathbb{Q}/mathbb{Z}$ is? (For example, when $a,binmathbb{Q}$, how does the two rational numbers determine an element in $mathbb{Q}/mathbb{Z}$?)
class-field-theory division-algebras
asked Nov 13 '18 at 20:04
QixiaoQixiao
2,8621628
$endgroup$
$begingroup$
en.wikipedia.org/wiki/Hilbert_symbol
$endgroup$
– Qiaochu Yuan
Nov 13 '18 at 22:49
$begingroup$
@QiaochuYuan Thanks! But for example how could one calculate $(2,3)_5$ explicitly?
$endgroup$
– Qixiao
Nov 14 '18 at 17:15
$begingroup$
@QiaochuYuan Sorry this may be a stupid question, but I feel confused: from the wikipedia description, the Hilbert symbol is expressed by Legendre symbols, $(a,b)_p=(-1)^{alphabetaepsilon(p)}(frac{u}{p})^{beta}(frac{v}{p})^{alpha}$, but it seems this expression only take $pm1$, but $Br(mathbb{Q}_p)=mathbb{Q}/mathbb{Z}$.
$endgroup$
– Qixiao
Nov 14 '18 at 17:37
$begingroup$
That section is about the quadratic Hilbert symbol, which corresponds to $2$-torsion in the Brauer group. Scroll down further for the general Hilbert symbol
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 19:56
$begingroup$
@QiaochuYuan Thanks! But is the Artin symbol easier to calculate explicitly? For example we have two integers $2,3inmathbb{Z}$ and a $4$-th root of unity $zeta$ in $mathbb{Q}_5$, is there an explicit way to determine what the class of $(2,3)_{zeta}$ in $mathbb{Q}/mathbb{Z}$ is?(As for quadratic symbols, this can be determined by solution of a degenerate conic)
$endgroup$
– Qixiao
Nov 14 '18 at 20:12
|
show 1 more comment
$begingroup$
Let $p$ be a prime number, let $mathbb{Q}_p$ be the local field, by Hensel's lemma, we know it has $p-1$-th roots of unity, let $zeta$ be a fixed primitive $p-1$-th root of unity in $mathbb{Q}_p$.
Let $a,binmathbb{Q}_p^*$, let $(a,b)_{zeta}$ be the cyclic algebra defined by $mathbb{Q}_plangle x,y|x^{p-1}=a,y^{p-1}=b, xy=zeta yxrangle$.
Is there an explicit way to tell which class $[(a,b)_zeta]inmathrm{Br}(mathbb{Q}_p)=mathbb{Q}/mathbb{Z}$ is? (For example, when $a,binmathbb{Q}$, how does the two rational numbers determine an element in $mathbb{Q}/mathbb{Z}$?)
class-field-theory division-algebras
asked Nov 13 '18 at 20:04
QixiaoQixiao
2,8621628
$endgroup$
Let $p$ be a prime number, let $mathbb{Q}_p$ be the local field, by Hensel's lemma, we know it has $p-1$-th roots of unity, let $zeta$ be a fixed primitive $p-1$-th root of unity in $mathbb{Q}_p$.
Let $a,binmathbb{Q}_p^*$, let $(a,b)_{zeta}$ be the cyclic algebra defined by $mathbb{Q}_plangle x,y|x^{p-1}=a,y^{p-1}=b, xy=zeta yxrangle$.
Is there an explicit way to tell which class $[(a,b)_zeta]inmathrm{Br}(mathbb{Q}_p)=mathbb{Q}/mathbb{Z}$ is? (For example, when $a,binmathbb{Q}$, how does the two rational numbers determine an element in $mathbb{Q}/mathbb{Z}$?)
class-field-theory division-algebras
class-field-theory division-algebras
asked Nov 13 '18 at 20:04
QixiaoQixiao
2,8621628
asked Nov 13 '18 at 20:04
QixiaoQixiao
2,8621628
asked Nov 13 '18 at 20:04
QixiaoQixiao
2,8621628
asked Nov 13 '18 at 20:04
QixiaoQixiao
2,8621628
asked Nov 13 '18 at 20:04
QixiaoQixiao
2,8621628
2,8621628
$begingroup$
en.wikipedia.org/wiki/Hilbert_symbol
$endgroup$
– Qiaochu Yuan
Nov 13 '18 at 22:49
$begingroup$
@QiaochuYuan Thanks! But for example how could one calculate $(2,3)_5$ explicitly?
$endgroup$
– Qixiao
Nov 14 '18 at 17:15
$begingroup$
@QiaochuYuan Sorry this may be a stupid question, but I feel confused: from the wikipedia description, the Hilbert symbol is expressed by Legendre symbols, $(a,b)_p=(-1)^{alphabetaepsilon(p)}(frac{u}{p})^{beta}(frac{v}{p})^{alpha}$, but it seems this expression only take $pm1$, but $Br(mathbb{Q}_p)=mathbb{Q}/mathbb{Z}$.
$endgroup$
– Qixiao
Nov 14 '18 at 17:37
$begingroup$
That section is about the quadratic Hilbert symbol, which corresponds to $2$-torsion in the Brauer group. Scroll down further for the general Hilbert symbol
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 19:56
$begingroup$
@QiaochuYuan Thanks! But is the Artin symbol easier to calculate explicitly? For example we have two integers $2,3inmathbb{Z}$ and a $4$-th root of unity $zeta$ in $mathbb{Q}_5$, is there an explicit way to determine what the class of $(2,3)_{zeta}$ in $mathbb{Q}/mathbb{Z}$ is?(As for quadratic symbols, this can be determined by solution of a degenerate conic)
$endgroup$
– Qixiao
Nov 14 '18 at 20:12
|
show 1 more comment
$begingroup$
en.wikipedia.org/wiki/Hilbert_symbol
$endgroup$
– Qiaochu Yuan
Nov 13 '18 at 22:49
$begingroup$
@QiaochuYuan Thanks! But for example how could one calculate $(2,3)_5$ explicitly?
$endgroup$
– Qixiao
Nov 14 '18 at 17:15
$begingroup$
@QiaochuYuan Sorry this may be a stupid question, but I feel confused: from the wikipedia description, the Hilbert symbol is expressed by Legendre symbols, $(a,b)_p=(-1)^{alphabetaepsilon(p)}(frac{u}{p})^{beta}(frac{v}{p})^{alpha}$, but it seems this expression only take $pm1$, but $Br(mathbb{Q}_p)=mathbb{Q}/mathbb{Z}$.
$endgroup$
– Qixiao
Nov 14 '18 at 17:37
$begingroup$
That section is about the quadratic Hilbert symbol, which corresponds to $2$-torsion in the Brauer group. Scroll down further for the general Hilbert symbol
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 19:56
$begingroup$
@QiaochuYuan Thanks! But is the Artin symbol easier to calculate explicitly? For example we have two integers $2,3inmathbb{Z}$ and a $4$-th root of unity $zeta$ in $mathbb{Q}_5$, is there an explicit way to determine what the class of $(2,3)_{zeta}$ in $mathbb{Q}/mathbb{Z}$ is?(As for quadratic symbols, this can be determined by solution of a degenerate conic)
$endgroup$
– Qixiao
Nov 14 '18 at 20:12
$begingroup$
en.wikipedia.org/wiki/Hilbert_symbol
$endgroup$
– Qiaochu Yuan
Nov 13 '18 at 22:49
$begingroup$
en.wikipedia.org/wiki/Hilbert_symbol
$endgroup$
– Qiaochu Yuan
Nov 13 '18 at 22:49
$begingroup$
@QiaochuYuan Thanks! But for example how could one calculate $(2,3)_5$ explicitly?
$endgroup$
– Qixiao
Nov 14 '18 at 17:15
$begingroup$
@QiaochuYuan Thanks! But for example how could one calculate $(2,3)_5$ explicitly?
$endgroup$
– Qixiao
Nov 14 '18 at 17:15
$begingroup$
@QiaochuYuan Sorry this may be a stupid question, but I feel confused: from the wikipedia description, the Hilbert symbol is expressed by Legendre symbols, $(a,b)_p=(-1)^{alphabetaepsilon(p)}(frac{u}{p})^{beta}(frac{v}{p})^{alpha}$, but it seems this expression only take $pm1$, but $Br(mathbb{Q}_p)=mathbb{Q}/mathbb{Z}$.
$endgroup$
– Qixiao
Nov 14 '18 at 17:37
$begingroup$
@QiaochuYuan Sorry this may be a stupid question, but I feel confused: from the wikipedia description, the Hilbert symbol is expressed by Legendre symbols, $(a,b)_p=(-1)^{alphabetaepsilon(p)}(frac{u}{p})^{beta}(frac{v}{p})^{alpha}$, but it seems this expression only take $pm1$, but $Br(mathbb{Q}_p)=mathbb{Q}/mathbb{Z}$.
$endgroup$
– Qixiao
Nov 14 '18 at 17:37
$begingroup$
That section is about the quadratic Hilbert symbol, which corresponds to $2$-torsion in the Brauer group. Scroll down further for the general Hilbert symbol
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 19:56
$begingroup$
That section is about the quadratic Hilbert symbol, which corresponds to $2$-torsion in the Brauer group. Scroll down further for the general Hilbert symbol
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 19:56
$begingroup$
@QiaochuYuan Thanks! But is the Artin symbol easier to calculate explicitly? For example we have two integers $2,3inmathbb{Z}$ and a $4$-th root of unity $zeta$ in $mathbb{Q}_5$, is there an explicit way to determine what the class of $(2,3)_{zeta}$ in $mathbb{Q}/mathbb{Z}$ is?(As for quadratic symbols, this can be determined by solution of a degenerate conic)
$endgroup$
– Qixiao
Nov 14 '18 at 20:12
$begingroup$
@QiaochuYuan Thanks! But is the Artin symbol easier to calculate explicitly? For example we have two integers $2,3inmathbb{Z}$ and a $4$-th root of unity $zeta$ in $mathbb{Q}_5$, is there an explicit way to determine what the class of $(2,3)_{zeta}$ in $mathbb{Q}/mathbb{Z}$ is?(As for quadratic symbols, this can be determined by solution of a degenerate conic)
$endgroup$
– Qixiao
Nov 14 '18 at 20:12
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
I should clarify going into this that I don't know any class field theory, so I might say some silly things. The number you're trying to compute is the Hasse invariant of the cyclic algebra, and it should also correspond to a Hilbert symbol as in the comments. I found a nice formula for a special case of the Hilbert symbol in Section 3.3 of these notes (Lemma 3.22). Specialized to this case it gives
$$(a, b)_{xi} = omega left( (-1)^{nu(a) nu(b)} frac{b^{nu(a)}}{a^{nu(b)}} bmod p right)$$
where $nu : mathbb{Q}_p to mathbb{Z}$ is the valuation and $omega : mathbb{F}_p^{times} to mu_{p-1}(mathbb{Q}_p)$ is the Teichmüller character.
You asked in the comments about $p = 5, a = 2, b = 3$. In this case $nu(a) = nu(b) = 0$ so the Hilbert symbol is trivial. In order for it to be nontrivial at least one of $nu(a)$ and $nu(b)$ must be nonzero. For example, sticking to $p = 5$, if we take $a = 2, b = 5$ then we get
$$(1, 5)_{xi} = omega left( frac{1}{2} bmod 5 right) = omega(3)$$
which is the unique $4^{th}$ root of unity in $mathbb{Z}_5$ congruent to $3 bmod 5$, and in particular primitive. Unfortunately I don't know exactly how this is identified with an element of $mathbb{Q}/mathbb{Z}$, although it must be either $frac{1}{4}$ or $frac{3}{4}$.
answered Nov 14 '18 at 20:52
Qiaochu YuanQiaochu Yuan
281k32593938
$endgroup$
$begingroup$
I've tried to decide the last question in my answer, please check if you agree.
$endgroup$
– Torsten Schoeneberg
Jan 26 at 6:47
add a comment |
$begingroup$
Qiaochu Yuan's answer gives good insight into computing the Hilbert symbol here, giving the result as an element of $mu_{p-1}(Bbb Q_p)$; but it avoids the machinery of cyclic (division) algebras which are sort of the original setting to define the Hasse invariant as element of $Br(Bbb Q_p) simeq Bbb Q/Bbb Z$. I want to amend that. It turns out that in the example at the end of his answer, whether the result is $frac14$ or $frac34$ depends on (some conventions, and:) which primitive $(p-1)$-th root of unity $zeta$ you choose in the original question.
References: I learned a lot of this stuff from R. Pierce: Associative Algebras (GTM 88, Springer, 1982) and I. Reiner: Maximal Orders (LMS Monographs 5, Academic Press, 1975) some years ago, although I don't have them at hand by now, so I cannot even guarantee that I follow the exact same conventions.
Namely, let us restrict to $p neq 2$ for simplicity, and to the basic interesting case $nu(a) = 0, nu(b) =1$, i.e. $ain Bbb Z_p^times simeq mu_{p-1} times (1+Bbb Z_p)$. Since elements in the second factor have $(p-1)$-th roots, we can easily restrict to the case $a in mu_{p-1}$; however, for your definition of the cyclic algebra to work, I think we should further restrict to the case that $a$ is a primitive $(p-1)$-th root of unity. Namely, now one would usually define the cyclic algebra as follows: Let $r in overline{Bbb Q_p}$ satisfy $r^{p-1}=a$. The extension $L= Bbb Q_p(r) vert Bbb Q_p$ is the unique unramified extension of degree $p-1$; its Galois group is cyclic and has a distinct generator, namely the (lift of the) Frobenius, let's call it $sigma$. Then the cyclic algebra you define can be realised as a subalgebra of $M_{(p-1) times (p-1)}(L)$, generated by
$x:= pmatrix{r&0& &\
0&zeta^{-1} r& &\
&& ddots &0\
&&0&zeta^{2-p}r}$
and $y:= pmatrix{0&1&0& &\
0&0&1 &&\
&&ddots& ddots &\
&&& &1\
b&&&0&0}.$
Note that we have $x^{p-1}=a$ and $y^{p-1}=b$ as well as $yxy^{-1}= zeta^{-1}x$, as demanded. One checks that this is a division algebra, which contains as a subfield the diagonal matrices
$$pmatrix{z&0& &\
0&tau(z)& &\
&& ddots &0\
&&0&tau^{p-1}(z)}$$
where $z in L$ and $tau in Gal(Lvert Bbb Q_p)$ is the automorphism induced by $rmapsto zeta^{-1}r.$ We identify this subfield with $L$. This division algebra, which in this case corresponds exactly to your definition of $(a,b)_zeta$, is usually denoted by something like $(Lvert Bbb Q_p, tau, b)$.
Now the original definition of the Hasse invariant goes like this: One can extend the $p$-adic valuation $nu$ to a valuation of the division algebra. Note that e.g. $nu(x) = 0$ and $nu(y) = dfrac{1}{p-1}$. In the division algebra, there exist elements $gamma$ such that the Frobenius $sigma$ on $L$ is induced by conjugation with $gamma$, i.e.
$$gamma z gamma^{-1} = sigma(z) text{ for all } z in L . (*)$$
Then the Hasse invariant is (well-)defined as $nu(gamma) + Bbb Z in Bbb Q/ Bbb Z$.
A quick calculation shows that $sigma(r) = ar$. Now find $1le k le p-1$ such that $tau^k = sigma$, or in other words, $zeta^{-k} = a$. Then one checks that $gamma := y^k$ satisfies $(*)$ and hence
$$(a,b)_zeta = dfrac{k}{n} + Bbb Z in Bbb Q / Bbb Z.$$
In the example $a=2, b=p=5$ of the other answer, it depends: There are two primitive $4$-th roots of unity in $Bbb Q_5$, namely $omega(2)$ and $omega(3)$ ($omega$ denoting the Teichmüller map). With your definitions we have
$$(2,5)_{omega(2)} = dfrac{3}{4} + Bbb Z$$
$$(2,5)_{omega(3)} = dfrac{1}{4} + Bbb Z$$
Unravelling many restrictions we made:
For $a in Bbb Z_p^times, b in p Bbb Z_p$, we have $$(a,b)_{zeta} = dfrac{k}{p-1} + Bbb Z in Bbb Q/Bbb Z simeq Br(Bbb Q_p)$$
where $zeta^{-k} equiv a$ mod $p$.
And from there one easily generalises to (consistent with the formula in Qiaochu's answer):
For $a in Bbb Z_p^times, b in Bbb Q_p$, we have $$(a,b)_{zeta} = nu(b)dfrac{k}{p-1} + Bbb Z in Bbb Q/Bbb Z simeq Br(Bbb Q_p)$$
where $zeta^{-k} equiv a$ mod $p$.
(The only class field theory that goes into this case is basic Kummer theory, relating $sigma$ and $tau$ and $mu_{p-1}$.)
Note that if one rewrote conjugation in $(*)$ as $gamma^{-1}zgamma$, everything would be multiplied by $(-1)$ in $Bbb Q/Bbb Z$ which in Qiaochu's example would exactly flip the two values $frac14$ and $frac34$. So in a way, it's really a convention which one is which. However, if one looks at the full Brauer group instead of just the $p-1$-torsion part as here, one indeed has to fix such a convention.
answered Jan 26 at 6:46
Torsten SchoenebergTorsten Schoeneberg
4,3312834
$endgroup$
add a comment |
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
});
}
});
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%2f2997243%2fwhat-does-a-b-zeta-correspond-to-in-mathrmbr-mathbbq-p-mathbbq%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I should clarify going into this that I don't know any class field theory, so I might say some silly things. The number you're trying to compute is the Hasse invariant of the cyclic algebra, and it should also correspond to a Hilbert symbol as in the comments. I found a nice formula for a special case of the Hilbert symbol in Section 3.3 of these notes (Lemma 3.22). Specialized to this case it gives
$$(a, b)_{xi} = omega left( (-1)^{nu(a) nu(b)} frac{b^{nu(a)}}{a^{nu(b)}} bmod p right)$$
where $nu : mathbb{Q}_p to mathbb{Z}$ is the valuation and $omega : mathbb{F}_p^{times} to mu_{p-1}(mathbb{Q}_p)$ is the Teichmüller character.
You asked in the comments about $p = 5, a = 2, b = 3$. In this case $nu(a) = nu(b) = 0$ so the Hilbert symbol is trivial. In order for it to be nontrivial at least one of $nu(a)$ and $nu(b)$ must be nonzero. For example, sticking to $p = 5$, if we take $a = 2, b = 5$ then we get
$$(1, 5)_{xi} = omega left( frac{1}{2} bmod 5 right) = omega(3)$$
which is the unique $4^{th}$ root of unity in $mathbb{Z}_5$ congruent to $3 bmod 5$, and in particular primitive. Unfortunately I don't know exactly how this is identified with an element of $mathbb{Q}/mathbb{Z}$, although it must be either $frac{1}{4}$ or $frac{3}{4}$.
answered Nov 14 '18 at 20:52
Qiaochu YuanQiaochu Yuan
281k32593938
$endgroup$
$begingroup$
I've tried to decide the last question in my answer, please check if you agree.
$endgroup$
– Torsten Schoeneberg
Jan 26 at 6:47
add a comment |
$begingroup$
I should clarify going into this that I don't know any class field theory, so I might say some silly things. The number you're trying to compute is the Hasse invariant of the cyclic algebra, and it should also correspond to a Hilbert symbol as in the comments. I found a nice formula for a special case of the Hilbert symbol in Section 3.3 of these notes (Lemma 3.22). Specialized to this case it gives
$$(a, b)_{xi} = omega left( (-1)^{nu(a) nu(b)} frac{b^{nu(a)}}{a^{nu(b)}} bmod p right)$$
where $nu : mathbb{Q}_p to mathbb{Z}$ is the valuation and $omega : mathbb{F}_p^{times} to mu_{p-1}(mathbb{Q}_p)$ is the Teichmüller character.
You asked in the comments about $p = 5, a = 2, b = 3$. In this case $nu(a) = nu(b) = 0$ so the Hilbert symbol is trivial. In order for it to be nontrivial at least one of $nu(a)$ and $nu(b)$ must be nonzero. For example, sticking to $p = 5$, if we take $a = 2, b = 5$ then we get
$$(1, 5)_{xi} = omega left( frac{1}{2} bmod 5 right) = omega(3)$$
which is the unique $4^{th}$ root of unity in $mathbb{Z}_5$ congruent to $3 bmod 5$, and in particular primitive. Unfortunately I don't know exactly how this is identified with an element of $mathbb{Q}/mathbb{Z}$, although it must be either $frac{1}{4}$ or $frac{3}{4}$.
answered Nov 14 '18 at 20:52
Qiaochu YuanQiaochu Yuan
281k32593938
$endgroup$
$begingroup$
I've tried to decide the last question in my answer, please check if you agree.
$endgroup$
– Torsten Schoeneberg
Jan 26 at 6:47
add a comment |
$begingroup$
I should clarify going into this that I don't know any class field theory, so I might say some silly things. The number you're trying to compute is the Hasse invariant of the cyclic algebra, and it should also correspond to a Hilbert symbol as in the comments. I found a nice formula for a special case of the Hilbert symbol in Section 3.3 of these notes (Lemma 3.22). Specialized to this case it gives
$$(a, b)_{xi} = omega left( (-1)^{nu(a) nu(b)} frac{b^{nu(a)}}{a^{nu(b)}} bmod p right)$$
where $nu : mathbb{Q}_p to mathbb{Z}$ is the valuation and $omega : mathbb{F}_p^{times} to mu_{p-1}(mathbb{Q}_p)$ is the Teichmüller character.
You asked in the comments about $p = 5, a = 2, b = 3$. In this case $nu(a) = nu(b) = 0$ so the Hilbert symbol is trivial. In order for it to be nontrivial at least one of $nu(a)$ and $nu(b)$ must be nonzero. For example, sticking to $p = 5$, if we take $a = 2, b = 5$ then we get
$$(1, 5)_{xi} = omega left( frac{1}{2} bmod 5 right) = omega(3)$$
which is the unique $4^{th}$ root of unity in $mathbb{Z}_5$ congruent to $3 bmod 5$, and in particular primitive. Unfortunately I don't know exactly how this is identified with an element of $mathbb{Q}/mathbb{Z}$, although it must be either $frac{1}{4}$ or $frac{3}{4}$.
answered Nov 14 '18 at 20:52
Qiaochu YuanQiaochu Yuan
281k32593938
$endgroup$
I should clarify going into this that I don't know any class field theory, so I might say some silly things. The number you're trying to compute is the Hasse invariant of the cyclic algebra, and it should also correspond to a Hilbert symbol as in the comments. I found a nice formula for a special case of the Hilbert symbol in Section 3.3 of these notes (Lemma 3.22). Specialized to this case it gives
$$(a, b)_{xi} = omega left( (-1)^{nu(a) nu(b)} frac{b^{nu(a)}}{a^{nu(b)}} bmod p right)$$
where $nu : mathbb{Q}_p to mathbb{Z}$ is the valuation and $omega : mathbb{F}_p^{times} to mu_{p-1}(mathbb{Q}_p)$ is the Teichmüller character.
You asked in the comments about $p = 5, a = 2, b = 3$. In this case $nu(a) = nu(b) = 0$ so the Hilbert symbol is trivial. In order for it to be nontrivial at least one of $nu(a)$ and $nu(b)$ must be nonzero. For example, sticking to $p = 5$, if we take $a = 2, b = 5$ then we get
$$(1, 5)_{xi} = omega left( frac{1}{2} bmod 5 right) = omega(3)$$
which is the unique $4^{th}$ root of unity in $mathbb{Z}_5$ congruent to $3 bmod 5$, and in particular primitive. Unfortunately I don't know exactly how this is identified with an element of $mathbb{Q}/mathbb{Z}$, although it must be either $frac{1}{4}$ or $frac{3}{4}$.
answered Nov 14 '18 at 20:52
Qiaochu YuanQiaochu Yuan
281k32593938
edited Nov 14 '18 at 22:15
answered Nov 14 '18 at 20:52
Qiaochu YuanQiaochu Yuan
281k32593938
answered Nov 14 '18 at 20:52
Qiaochu YuanQiaochu Yuan
281k32593938
answered Nov 14 '18 at 20:52
Qiaochu YuanQiaochu Yuan
281k32593938
281k32593938
$begingroup$
I've tried to decide the last question in my answer, please check if you agree.
$endgroup$
– Torsten Schoeneberg
Jan 26 at 6:47
add a comment |
$begingroup$
I've tried to decide the last question in my answer, please check if you agree.
$endgroup$
– Torsten Schoeneberg
Jan 26 at 6:47
$begingroup$
I've tried to decide the last question in my answer, please check if you agree.
$endgroup$
– Torsten Schoeneberg
Jan 26 at 6:47
$begingroup$
I've tried to decide the last question in my answer, please check if you agree.
$endgroup$
– Torsten Schoeneberg
Jan 26 at 6:47
add a comment |
$begingroup$
Qiaochu Yuan's answer gives good insight into computing the Hilbert symbol here, giving the result as an element of $mu_{p-1}(Bbb Q_p)$; but it avoids the machinery of cyclic (division) algebras which are sort of the original setting to define the Hasse invariant as element of $Br(Bbb Q_p) simeq Bbb Q/Bbb Z$. I want to amend that. It turns out that in the example at the end of his answer, whether the result is $frac14$ or $frac34$ depends on (some conventions, and:) which primitive $(p-1)$-th root of unity $zeta$ you choose in the original question.
References: I learned a lot of this stuff from R. Pierce: Associative Algebras (GTM 88, Springer, 1982) and I. Reiner: Maximal Orders (LMS Monographs 5, Academic Press, 1975) some years ago, although I don't have them at hand by now, so I cannot even guarantee that I follow the exact same conventions.
Namely, let us restrict to $p neq 2$ for simplicity, and to the basic interesting case $nu(a) = 0, nu(b) =1$, i.e. $ain Bbb Z_p^times simeq mu_{p-1} times (1+Bbb Z_p)$. Since elements in the second factor have $(p-1)$-th roots, we can easily restrict to the case $a in mu_{p-1}$; however, for your definition of the cyclic algebra to work, I think we should further restrict to the case that $a$ is a primitive $(p-1)$-th root of unity. Namely, now one would usually define the cyclic algebra as follows: Let $r in overline{Bbb Q_p}$ satisfy $r^{p-1}=a$. The extension $L= Bbb Q_p(r) vert Bbb Q_p$ is the unique unramified extension of degree $p-1$; its Galois group is cyclic and has a distinct generator, namely the (lift of the) Frobenius, let's call it $sigma$. Then the cyclic algebra you define can be realised as a subalgebra of $M_{(p-1) times (p-1)}(L)$, generated by
$x:= pmatrix{r&0& &\
0&zeta^{-1} r& &\
&& ddots &0\
&&0&zeta^{2-p}r}$
and $y:= pmatrix{0&1&0& &\
0&0&1 &&\
&&ddots& ddots &\
&&& &1\
b&&&0&0}.$
Note that we have $x^{p-1}=a$ and $y^{p-1}=b$ as well as $yxy^{-1}= zeta^{-1}x$, as demanded. One checks that this is a division algebra, which contains as a subfield the diagonal matrices
$$pmatrix{z&0& &\
0&tau(z)& &\
&& ddots &0\
&&0&tau^{p-1}(z)}$$
where $z in L$ and $tau in Gal(Lvert Bbb Q_p)$ is the automorphism induced by $rmapsto zeta^{-1}r.$ We identify this subfield with $L$. This division algebra, which in this case corresponds exactly to your definition of $(a,b)_zeta$, is usually denoted by something like $(Lvert Bbb Q_p, tau, b)$.
Now the original definition of the Hasse invariant goes like this: One can extend the $p$-adic valuation $nu$ to a valuation of the division algebra. Note that e.g. $nu(x) = 0$ and $nu(y) = dfrac{1}{p-1}$. In the division algebra, there exist elements $gamma$ such that the Frobenius $sigma$ on $L$ is induced by conjugation with $gamma$, i.e.
$$gamma z gamma^{-1} = sigma(z) text{ for all } z in L . (*)$$
Then the Hasse invariant is (well-)defined as $nu(gamma) + Bbb Z in Bbb Q/ Bbb Z$.
A quick calculation shows that $sigma(r) = ar$. Now find $1le k le p-1$ such that $tau^k = sigma$, or in other words, $zeta^{-k} = a$. Then one checks that $gamma := y^k$ satisfies $(*)$ and hence
$$(a,b)_zeta = dfrac{k}{n} + Bbb Z in Bbb Q / Bbb Z.$$
In the example $a=2, b=p=5$ of the other answer, it depends: There are two primitive $4$-th roots of unity in $Bbb Q_5$, namely $omega(2)$ and $omega(3)$ ($omega$ denoting the Teichmüller map). With your definitions we have
$$(2,5)_{omega(2)} = dfrac{3}{4} + Bbb Z$$
$$(2,5)_{omega(3)} = dfrac{1}{4} + Bbb Z$$
Unravelling many restrictions we made:
For $a in Bbb Z_p^times, b in p Bbb Z_p$, we have $$(a,b)_{zeta} = dfrac{k}{p-1} + Bbb Z in Bbb Q/Bbb Z simeq Br(Bbb Q_p)$$
where $zeta^{-k} equiv a$ mod $p$.
And from there one easily generalises to (consistent with the formula in Qiaochu's answer):
For $a in Bbb Z_p^times, b in Bbb Q_p$, we have $$(a,b)_{zeta} = nu(b)dfrac{k}{p-1} + Bbb Z in Bbb Q/Bbb Z simeq Br(Bbb Q_p)$$
where $zeta^{-k} equiv a$ mod $p$.
(The only class field theory that goes into this case is basic Kummer theory, relating $sigma$ and $tau$ and $mu_{p-1}$.)
Note that if one rewrote conjugation in $(*)$ as $gamma^{-1}zgamma$, everything would be multiplied by $(-1)$ in $Bbb Q/Bbb Z$ which in Qiaochu's example would exactly flip the two values $frac14$ and $frac34$. So in a way, it's really a convention which one is which. However, if one looks at the full Brauer group instead of just the $p-1$-torsion part as here, one indeed has to fix such a convention.
answered Jan 26 at 6:46
Torsten SchoenebergTorsten Schoeneberg
4,3312834
$endgroup$
add a comment |
$begingroup$
Qiaochu Yuan's answer gives good insight into computing the Hilbert symbol here, giving the result as an element of $mu_{p-1}(Bbb Q_p)$; but it avoids the machinery of cyclic (division) algebras which are sort of the original setting to define the Hasse invariant as element of $Br(Bbb Q_p) simeq Bbb Q/Bbb Z$. I want to amend that. It turns out that in the example at the end of his answer, whether the result is $frac14$ or $frac34$ depends on (some conventions, and:) which primitive $(p-1)$-th root of unity $zeta$ you choose in the original question.
References: I learned a lot of this stuff from R. Pierce: Associative Algebras (GTM 88, Springer, 1982) and I. Reiner: Maximal Orders (LMS Monographs 5, Academic Press, 1975) some years ago, although I don't have them at hand by now, so I cannot even guarantee that I follow the exact same conventions.
Namely, let us restrict to $p neq 2$ for simplicity, and to the basic interesting case $nu(a) = 0, nu(b) =1$, i.e. $ain Bbb Z_p^times simeq mu_{p-1} times (1+Bbb Z_p)$. Since elements in the second factor have $(p-1)$-th roots, we can easily restrict to the case $a in mu_{p-1}$; however, for your definition of the cyclic algebra to work, I think we should further restrict to the case that $a$ is a primitive $(p-1)$-th root of unity. Namely, now one would usually define the cyclic algebra as follows: Let $r in overline{Bbb Q_p}$ satisfy $r^{p-1}=a$. The extension $L= Bbb Q_p(r) vert Bbb Q_p$ is the unique unramified extension of degree $p-1$; its Galois group is cyclic and has a distinct generator, namely the (lift of the) Frobenius, let's call it $sigma$. Then the cyclic algebra you define can be realised as a subalgebra of $M_{(p-1) times (p-1)}(L)$, generated by
$x:= pmatrix{r&0& &\
0&zeta^{-1} r& &\
&& ddots &0\
&&0&zeta^{2-p}r}$
and $y:= pmatrix{0&1&0& &\
0&0&1 &&\
&&ddots& ddots &\
&&& &1\
b&&&0&0}.$
Note that we have $x^{p-1}=a$ and $y^{p-1}=b$ as well as $yxy^{-1}= zeta^{-1}x$, as demanded. One checks that this is a division algebra, which contains as a subfield the diagonal matrices
$$pmatrix{z&0& &\
0&tau(z)& &\
&& ddots &0\
&&0&tau^{p-1}(z)}$$
where $z in L$ and $tau in Gal(Lvert Bbb Q_p)$ is the automorphism induced by $rmapsto zeta^{-1}r.$ We identify this subfield with $L$. This division algebra, which in this case corresponds exactly to your definition of $(a,b)_zeta$, is usually denoted by something like $(Lvert Bbb Q_p, tau, b)$.
Now the original definition of the Hasse invariant goes like this: One can extend the $p$-adic valuation $nu$ to a valuation of the division algebra. Note that e.g. $nu(x) = 0$ and $nu(y) = dfrac{1}{p-1}$. In the division algebra, there exist elements $gamma$ such that the Frobenius $sigma$ on $L$ is induced by conjugation with $gamma$, i.e.
$$gamma z gamma^{-1} = sigma(z) text{ for all } z in L . (*)$$
Then the Hasse invariant is (well-)defined as $nu(gamma) + Bbb Z in Bbb Q/ Bbb Z$.
A quick calculation shows that $sigma(r) = ar$. Now find $1le k le p-1$ such that $tau^k = sigma$, or in other words, $zeta^{-k} = a$. Then one checks that $gamma := y^k$ satisfies $(*)$ and hence
$$(a,b)_zeta = dfrac{k}{n} + Bbb Z in Bbb Q / Bbb Z.$$
In the example $a=2, b=p=5$ of the other answer, it depends: There are two primitive $4$-th roots of unity in $Bbb Q_5$, namely $omega(2)$ and $omega(3)$ ($omega$ denoting the Teichmüller map). With your definitions we have
$$(2,5)_{omega(2)} = dfrac{3}{4} + Bbb Z$$
$$(2,5)_{omega(3)} = dfrac{1}{4} + Bbb Z$$
Unravelling many restrictions we made:
For $a in Bbb Z_p^times, b in p Bbb Z_p$, we have $$(a,b)_{zeta} = dfrac{k}{p-1} + Bbb Z in Bbb Q/Bbb Z simeq Br(Bbb Q_p)$$
where $zeta^{-k} equiv a$ mod $p$.
And from there one easily generalises to (consistent with the formula in Qiaochu's answer):
For $a in Bbb Z_p^times, b in Bbb Q_p$, we have $$(a,b)_{zeta} = nu(b)dfrac{k}{p-1} + Bbb Z in Bbb Q/Bbb Z simeq Br(Bbb Q_p)$$
where $zeta^{-k} equiv a$ mod $p$.
(The only class field theory that goes into this case is basic Kummer theory, relating $sigma$ and $tau$ and $mu_{p-1}$.)
Note that if one rewrote conjugation in $(*)$ as $gamma^{-1}zgamma$, everything would be multiplied by $(-1)$ in $Bbb Q/Bbb Z$ which in Qiaochu's example would exactly flip the two values $frac14$ and $frac34$. So in a way, it's really a convention which one is which. However, if one looks at the full Brauer group instead of just the $p-1$-torsion part as here, one indeed has to fix such a convention.
answered Jan 26 at 6:46
Torsten SchoenebergTorsten Schoeneberg
4,3312834
$endgroup$
add a comment |
$begingroup$
Qiaochu Yuan's answer gives good insight into computing the Hilbert symbol here, giving the result as an element of $mu_{p-1}(Bbb Q_p)$; but it avoids the machinery of cyclic (division) algebras which are sort of the original setting to define the Hasse invariant as element of $Br(Bbb Q_p) simeq Bbb Q/Bbb Z$. I want to amend that. It turns out that in the example at the end of his answer, whether the result is $frac14$ or $frac34$ depends on (some conventions, and:) which primitive $(p-1)$-th root of unity $zeta$ you choose in the original question.
References: I learned a lot of this stuff from R. Pierce: Associative Algebras (GTM 88, Springer, 1982) and I. Reiner: Maximal Orders (LMS Monographs 5, Academic Press, 1975) some years ago, although I don't have them at hand by now, so I cannot even guarantee that I follow the exact same conventions.
Namely, let us restrict to $p neq 2$ for simplicity, and to the basic interesting case $nu(a) = 0, nu(b) =1$, i.e. $ain Bbb Z_p^times simeq mu_{p-1} times (1+Bbb Z_p)$. Since elements in the second factor have $(p-1)$-th roots, we can easily restrict to the case $a in mu_{p-1}$; however, for your definition of the cyclic algebra to work, I think we should further restrict to the case that $a$ is a primitive $(p-1)$-th root of unity. Namely, now one would usually define the cyclic algebra as follows: Let $r in overline{Bbb Q_p}$ satisfy $r^{p-1}=a$. The extension $L= Bbb Q_p(r) vert Bbb Q_p$ is the unique unramified extension of degree $p-1$; its Galois group is cyclic and has a distinct generator, namely the (lift of the) Frobenius, let's call it $sigma$. Then the cyclic algebra you define can be realised as a subalgebra of $M_{(p-1) times (p-1)}(L)$, generated by
$x:= pmatrix{r&0& &\
0&zeta^{-1} r& &\
&& ddots &0\
&&0&zeta^{2-p}r}$
and $y:= pmatrix{0&1&0& &\
0&0&1 &&\
&&ddots& ddots &\
&&& &1\
b&&&0&0}.$
Note that we have $x^{p-1}=a$ and $y^{p-1}=b$ as well as $yxy^{-1}= zeta^{-1}x$, as demanded. One checks that this is a division algebra, which contains as a subfield the diagonal matrices
$$pmatrix{z&0& &\
0&tau(z)& &\
&& ddots &0\
&&0&tau^{p-1}(z)}$$
where $z in L$ and $tau in Gal(Lvert Bbb Q_p)$ is the automorphism induced by $rmapsto zeta^{-1}r.$ We identify this subfield with $L$. This division algebra, which in this case corresponds exactly to your definition of $(a,b)_zeta$, is usually denoted by something like $(Lvert Bbb Q_p, tau, b)$.
Now the original definition of the Hasse invariant goes like this: One can extend the $p$-adic valuation $nu$ to a valuation of the division algebra. Note that e.g. $nu(x) = 0$ and $nu(y) = dfrac{1}{p-1}$. In the division algebra, there exist elements $gamma$ such that the Frobenius $sigma$ on $L$ is induced by conjugation with $gamma$, i.e.
$$gamma z gamma^{-1} = sigma(z) text{ for all } z in L . (*)$$
Then the Hasse invariant is (well-)defined as $nu(gamma) + Bbb Z in Bbb Q/ Bbb Z$.
A quick calculation shows that $sigma(r) = ar$. Now find $1le k le p-1$ such that $tau^k = sigma$, or in other words, $zeta^{-k} = a$. Then one checks that $gamma := y^k$ satisfies $(*)$ and hence
$$(a,b)_zeta = dfrac{k}{n} + Bbb Z in Bbb Q / Bbb Z.$$
In the example $a=2, b=p=5$ of the other answer, it depends: There are two primitive $4$-th roots of unity in $Bbb Q_5$, namely $omega(2)$ and $omega(3)$ ($omega$ denoting the Teichmüller map). With your definitions we have
$$(2,5)_{omega(2)} = dfrac{3}{4} + Bbb Z$$
$$(2,5)_{omega(3)} = dfrac{1}{4} + Bbb Z$$
Unravelling many restrictions we made:
For $a in Bbb Z_p^times, b in p Bbb Z_p$, we have $$(a,b)_{zeta} = dfrac{k}{p-1} + Bbb Z in Bbb Q/Bbb Z simeq Br(Bbb Q_p)$$
where $zeta^{-k} equiv a$ mod $p$.
And from there one easily generalises to (consistent with the formula in Qiaochu's answer):
For $a in Bbb Z_p^times, b in Bbb Q_p$, we have $$(a,b)_{zeta} = nu(b)dfrac{k}{p-1} + Bbb Z in Bbb Q/Bbb Z simeq Br(Bbb Q_p)$$
where $zeta^{-k} equiv a$ mod $p$.
(The only class field theory that goes into this case is basic Kummer theory, relating $sigma$ and $tau$ and $mu_{p-1}$.)
Note that if one rewrote conjugation in $(*)$ as $gamma^{-1}zgamma$, everything would be multiplied by $(-1)$ in $Bbb Q/Bbb Z$ which in Qiaochu's example would exactly flip the two values $frac14$ and $frac34$. So in a way, it's really a convention which one is which. However, if one looks at the full Brauer group instead of just the $p-1$-torsion part as here, one indeed has to fix such a convention.
answered Jan 26 at 6:46
Torsten SchoenebergTorsten Schoeneberg
4,3312834
$endgroup$
Qiaochu Yuan's answer gives good insight into computing the Hilbert symbol here, giving the result as an element of $mu_{p-1}(Bbb Q_p)$; but it avoids the machinery of cyclic (division) algebras which are sort of the original setting to define the Hasse invariant as element of $Br(Bbb Q_p) simeq Bbb Q/Bbb Z$. I want to amend that. It turns out that in the example at the end of his answer, whether the result is $frac14$ or $frac34$ depends on (some conventions, and:) which primitive $(p-1)$-th root of unity $zeta$ you choose in the original question.
References: I learned a lot of this stuff from R. Pierce: Associative Algebras (GTM 88, Springer, 1982) and I. Reiner: Maximal Orders (LMS Monographs 5, Academic Press, 1975) some years ago, although I don't have them at hand by now, so I cannot even guarantee that I follow the exact same conventions.
Namely, let us restrict to $p neq 2$ for simplicity, and to the basic interesting case $nu(a) = 0, nu(b) =1$, i.e. $ain Bbb Z_p^times simeq mu_{p-1} times (1+Bbb Z_p)$. Since elements in the second factor have $(p-1)$-th roots, we can easily restrict to the case $a in mu_{p-1}$; however, for your definition of the cyclic algebra to work, I think we should further restrict to the case that $a$ is a primitive $(p-1)$-th root of unity. Namely, now one would usually define the cyclic algebra as follows: Let $r in overline{Bbb Q_p}$ satisfy $r^{p-1}=a$. The extension $L= Bbb Q_p(r) vert Bbb Q_p$ is the unique unramified extension of degree $p-1$; its Galois group is cyclic and has a distinct generator, namely the (lift of the) Frobenius, let's call it $sigma$. Then the cyclic algebra you define can be realised as a subalgebra of $M_{(p-1) times (p-1)}(L)$, generated by
$x:= pmatrix{r&0& &\
0&zeta^{-1} r& &\
&& ddots &0\
&&0&zeta^{2-p}r}$
and $y:= pmatrix{0&1&0& &\
0&0&1 &&\
&&ddots& ddots &\
&&& &1\
b&&&0&0}.$
Note that we have $x^{p-1}=a$ and $y^{p-1}=b$ as well as $yxy^{-1}= zeta^{-1}x$, as demanded. One checks that this is a division algebra, which contains as a subfield the diagonal matrices
$$pmatrix{z&0& &\
0&tau(z)& &\
&& ddots &0\
&&0&tau^{p-1}(z)}$$
where $z in L$ and $tau in Gal(Lvert Bbb Q_p)$ is the automorphism induced by $rmapsto zeta^{-1}r.$ We identify this subfield with $L$. This division algebra, which in this case corresponds exactly to your definition of $(a,b)_zeta$, is usually denoted by something like $(Lvert Bbb Q_p, tau, b)$.
Now the original definition of the Hasse invariant goes like this: One can extend the $p$-adic valuation $nu$ to a valuation of the division algebra. Note that e.g. $nu(x) = 0$ and $nu(y) = dfrac{1}{p-1}$. In the division algebra, there exist elements $gamma$ such that the Frobenius $sigma$ on $L$ is induced by conjugation with $gamma$, i.e.
$$gamma z gamma^{-1} = sigma(z) text{ for all } z in L . (*)$$
Then the Hasse invariant is (well-)defined as $nu(gamma) + Bbb Z in Bbb Q/ Bbb Z$.
A quick calculation shows that $sigma(r) = ar$. Now find $1le k le p-1$ such that $tau^k = sigma$, or in other words, $zeta^{-k} = a$. Then one checks that $gamma := y^k$ satisfies $(*)$ and hence
$$(a,b)_zeta = dfrac{k}{n} + Bbb Z in Bbb Q / Bbb Z.$$
In the example $a=2, b=p=5$ of the other answer, it depends: There are two primitive $4$-th roots of unity in $Bbb Q_5$, namely $omega(2)$ and $omega(3)$ ($omega$ denoting the Teichmüller map). With your definitions we have
$$(2,5)_{omega(2)} = dfrac{3}{4} + Bbb Z$$
$$(2,5)_{omega(3)} = dfrac{1}{4} + Bbb Z$$
Unravelling many restrictions we made:
For $a in Bbb Z_p^times, b in p Bbb Z_p$, we have $$(a,b)_{zeta} = dfrac{k}{p-1} + Bbb Z in Bbb Q/Bbb Z simeq Br(Bbb Q_p)$$
where $zeta^{-k} equiv a$ mod $p$.
And from there one easily generalises to (consistent with the formula in Qiaochu's answer):
For $a in Bbb Z_p^times, b in Bbb Q_p$, we have $$(a,b)_{zeta} = nu(b)dfrac{k}{p-1} + Bbb Z in Bbb Q/Bbb Z simeq Br(Bbb Q_p)$$
where $zeta^{-k} equiv a$ mod $p$.
(The only class field theory that goes into this case is basic Kummer theory, relating $sigma$ and $tau$ and $mu_{p-1}$.)
Note that if one rewrote conjugation in $(*)$ as $gamma^{-1}zgamma$, everything would be multiplied by $(-1)$ in $Bbb Q/Bbb Z$ which in Qiaochu's example would exactly flip the two values $frac14$ and $frac34$. So in a way, it's really a convention which one is which. However, if one looks at the full Brauer group instead of just the $p-1$-torsion part as here, one indeed has to fix such a convention.
answered Jan 26 at 6:46
Torsten SchoenebergTorsten Schoeneberg
4,3312834
edited Jan 26 at 17:25
answered Jan 26 at 6:46
Torsten SchoenebergTorsten Schoeneberg
4,3312834
answered Jan 26 at 6:46
Torsten SchoenebergTorsten Schoeneberg
4,3312834
answered Jan 26 at 6:46
Torsten SchoenebergTorsten Schoeneberg
4,3312834
4,3312834
add a comment |
add a comment |
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%2f2997243%2fwhat-does-a-b-zeta-correspond-to-in-mathrmbr-mathbbq-p-mathbbq%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
$begingroup$
en.wikipedia.org/wiki/Hilbert_symbol
$endgroup$
– Qiaochu Yuan
Nov 13 '18 at 22:49
$begingroup$
@QiaochuYuan Thanks! But for example how could one calculate $(2,3)_5$ explicitly?
$endgroup$
– Qixiao
Nov 14 '18 at 17:15
$begingroup$
@QiaochuYuan Sorry this may be a stupid question, but I feel confused: from the wikipedia description, the Hilbert symbol is expressed by Legendre symbols, $(a,b)_p=(-1)^{alphabetaepsilon(p)}(frac{u}{p})^{beta}(frac{v}{p})^{alpha}$, but it seems this expression only take $pm1$, but $Br(mathbb{Q}_p)=mathbb{Q}/mathbb{Z}$.
$endgroup$
– Qixiao
Nov 14 '18 at 17:37
$begingroup$
That section is about the quadratic Hilbert symbol, which corresponds to $2$-torsion in the Brauer group. Scroll down further for the general Hilbert symbol
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 19:56
$begingroup$
@QiaochuYuan Thanks! But is the Artin symbol easier to calculate explicitly? For example we have two integers $2,3inmathbb{Z}$ and a $4$-th root of unity $zeta$ in $mathbb{Q}_5$, is there an explicit way to determine what the class of $(2,3)_{zeta}$ in $mathbb{Q}/mathbb{Z}$ is?(As for quadratic symbols, this can be determined by solution of a degenerate conic)
$endgroup$
– Qixiao
Nov 14 '18 at 20:12