Show $Delta(theta)mathbb{Z} = Delta(R/mathbb{Z}) implies R = mathbb{Z}[theta]$.
$begingroup$
Let $Delta$ denote the discriminant and $Delta(R/mathbb{Z})$ denote the discriminant ideal of $R$ over $mathbb{Z}$. I'm having trouble proving the following:
Suppose $[mathbb{Q}(theta) : mathbb{Q}] = n$ and $theta in R$ which is the integral closure of $mathbb{Z}$ in $mathbb{Q}(theta)$. Then $Delta(theta)mathbb{Z} = Delta(R/mathbb{Z}) implies R = mathbb{Z}[theta]$.
Attempt: Let $X={x_0, dots, x_{n-1}}$ be a $mathbb{Z}$-basis for $R$. I know from a lemma that $Delta(theta) = det A^2 Delta(X)$, where $A$ is the change of basis matrix between $X$ and ${1, theta, dots, theta^{n-1}}$. So $Delta(X) mid Delta(theta)$, but also, $Delta(theta) mid Delta(X)$ since $Delta(R/mathbb{Z}) = Delta(theta)mathbb{Z}$. This implies that $det A = pm1$.
Now, consider a typical element $y in R$. Then
begin{align*}
y &= sum_{k=0}^{n-1} a_k x_k
\
& = sum_{k=0}^{n-1} sum_{j=0}^{n-1}a_kr_{jk}theta^j
end{align*}
where $(r_{jk}) = A^{-1}$. Since $A^{-1} = frac{1}{det A} C^T = pm C^T$, where $C$ is the cofactor matrix of $A$. Since $C$ has only entries from $mathbb{Z}$ we have that the $r_{jk} in mathbb{Z}$. Thus, $y in mathbb{Z}[theta]$.
Question: Is this right? Also, someone told me that in general $|det A| = [R : mathbb{Z}[theta]]$. I can't see why this would be true. Is there a way to understand this without going into Smith normal form(see here)
?
proof-verification algebraic-number-theory discriminant
asked Dec 21 '18 at 6:43
matt stokesmatt stokes
582310
$endgroup$
add a comment |
$begingroup$
Let $Delta$ denote the discriminant and $Delta(R/mathbb{Z})$ denote the discriminant ideal of $R$ over $mathbb{Z}$. I'm having trouble proving the following:
Suppose $[mathbb{Q}(theta) : mathbb{Q}] = n$ and $theta in R$ which is the integral closure of $mathbb{Z}$ in $mathbb{Q}(theta)$. Then $Delta(theta)mathbb{Z} = Delta(R/mathbb{Z}) implies R = mathbb{Z}[theta]$.
Attempt: Let $X={x_0, dots, x_{n-1}}$ be a $mathbb{Z}$-basis for $R$. I know from a lemma that $Delta(theta) = det A^2 Delta(X)$, where $A$ is the change of basis matrix between $X$ and ${1, theta, dots, theta^{n-1}}$. So $Delta(X) mid Delta(theta)$, but also, $Delta(theta) mid Delta(X)$ since $Delta(R/mathbb{Z}) = Delta(theta)mathbb{Z}$. This implies that $det A = pm1$.
Now, consider a typical element $y in R$. Then
begin{align*}
y &= sum_{k=0}^{n-1} a_k x_k
\
& = sum_{k=0}^{n-1} sum_{j=0}^{n-1}a_kr_{jk}theta^j
end{align*}
where $(r_{jk}) = A^{-1}$. Since $A^{-1} = frac{1}{det A} C^T = pm C^T$, where $C$ is the cofactor matrix of $A$. Since $C$ has only entries from $mathbb{Z}$ we have that the $r_{jk} in mathbb{Z}$. Thus, $y in mathbb{Z}[theta]$.
Question: Is this right? Also, someone told me that in general $|det A| = [R : mathbb{Z}[theta]]$. I can't see why this would be true. Is there a way to understand this without going into Smith normal form(see here)
?
proof-verification algebraic-number-theory discriminant
asked Dec 21 '18 at 6:43
matt stokesmatt stokes
582310
$endgroup$
$begingroup$
What is wrong with using the Smith normal form? Or any other argument leading to the so called stacked bases theorem?
$endgroup$
– Jyrki Lahtonen
Dec 21 '18 at 8:11
$begingroup$
I've never seen it before, but I guess it wouldn't hurt to learn something new. I was just wondering if there is a more elementary way to see this.
$endgroup$
– matt stokes
Dec 21 '18 at 14:55
add a comment |
$begingroup$
Let $Delta$ denote the discriminant and $Delta(R/mathbb{Z})$ denote the discriminant ideal of $R$ over $mathbb{Z}$. I'm having trouble proving the following:
Suppose $[mathbb{Q}(theta) : mathbb{Q}] = n$ and $theta in R$ which is the integral closure of $mathbb{Z}$ in $mathbb{Q}(theta)$. Then $Delta(theta)mathbb{Z} = Delta(R/mathbb{Z}) implies R = mathbb{Z}[theta]$.
Attempt: Let $X={x_0, dots, x_{n-1}}$ be a $mathbb{Z}$-basis for $R$. I know from a lemma that $Delta(theta) = det A^2 Delta(X)$, where $A$ is the change of basis matrix between $X$ and ${1, theta, dots, theta^{n-1}}$. So $Delta(X) mid Delta(theta)$, but also, $Delta(theta) mid Delta(X)$ since $Delta(R/mathbb{Z}) = Delta(theta)mathbb{Z}$. This implies that $det A = pm1$.
Now, consider a typical element $y in R$. Then
begin{align*}
y &= sum_{k=0}^{n-1} a_k x_k
\
& = sum_{k=0}^{n-1} sum_{j=0}^{n-1}a_kr_{jk}theta^j
end{align*}
where $(r_{jk}) = A^{-1}$. Since $A^{-1} = frac{1}{det A} C^T = pm C^T$, where $C$ is the cofactor matrix of $A$. Since $C$ has only entries from $mathbb{Z}$ we have that the $r_{jk} in mathbb{Z}$. Thus, $y in mathbb{Z}[theta]$.
Question: Is this right? Also, someone told me that in general $|det A| = [R : mathbb{Z}[theta]]$. I can't see why this would be true. Is there a way to understand this without going into Smith normal form(see here)
?
proof-verification algebraic-number-theory discriminant
asked Dec 21 '18 at 6:43
matt stokesmatt stokes
582310
$endgroup$
Let $Delta$ denote the discriminant and $Delta(R/mathbb{Z})$ denote the discriminant ideal of $R$ over $mathbb{Z}$. I'm having trouble proving the following:
Suppose $[mathbb{Q}(theta) : mathbb{Q}] = n$ and $theta in R$ which is the integral closure of $mathbb{Z}$ in $mathbb{Q}(theta)$. Then $Delta(theta)mathbb{Z} = Delta(R/mathbb{Z}) implies R = mathbb{Z}[theta]$.
Attempt: Let $X={x_0, dots, x_{n-1}}$ be a $mathbb{Z}$-basis for $R$. I know from a lemma that $Delta(theta) = det A^2 Delta(X)$, where $A$ is the change of basis matrix between $X$ and ${1, theta, dots, theta^{n-1}}$. So $Delta(X) mid Delta(theta)$, but also, $Delta(theta) mid Delta(X)$ since $Delta(R/mathbb{Z}) = Delta(theta)mathbb{Z}$. This implies that $det A = pm1$.
Now, consider a typical element $y in R$. Then
begin{align*}
y &= sum_{k=0}^{n-1} a_k x_k
\
& = sum_{k=0}^{n-1} sum_{j=0}^{n-1}a_kr_{jk}theta^j
end{align*}
where $(r_{jk}) = A^{-1}$. Since $A^{-1} = frac{1}{det A} C^T = pm C^T$, where $C$ is the cofactor matrix of $A$. Since $C$ has only entries from $mathbb{Z}$ we have that the $r_{jk} in mathbb{Z}$. Thus, $y in mathbb{Z}[theta]$.
Question: Is this right? Also, someone told me that in general $|det A| = [R : mathbb{Z}[theta]]$. I can't see why this would be true. Is there a way to understand this without going into Smith normal form(see here)
?
proof-verification algebraic-number-theory discriminant
proof-verification algebraic-number-theory discriminant
asked Dec 21 '18 at 6:43
matt stokesmatt stokes
582310
asked Dec 21 '18 at 6:43
matt stokesmatt stokes
582310
edited Dec 21 '18 at 6:56
matt stokes
asked Dec 21 '18 at 6:43
matt stokesmatt stokes
582310
asked Dec 21 '18 at 6:43
matt stokesmatt stokes
582310
asked Dec 21 '18 at 6:43
matt stokesmatt stokes
582310
582310
$begingroup$
What is wrong with using the Smith normal form? Or any other argument leading to the so called stacked bases theorem?
$endgroup$
– Jyrki Lahtonen
Dec 21 '18 at 8:11
$begingroup$
I've never seen it before, but I guess it wouldn't hurt to learn something new. I was just wondering if there is a more elementary way to see this.
$endgroup$
– matt stokes
Dec 21 '18 at 14:55
add a comment |
$begingroup$
What is wrong with using the Smith normal form? Or any other argument leading to the so called stacked bases theorem?
$endgroup$
– Jyrki Lahtonen
Dec 21 '18 at 8:11
$begingroup$
I've never seen it before, but I guess it wouldn't hurt to learn something new. I was just wondering if there is a more elementary way to see this.
$endgroup$
– matt stokes
Dec 21 '18 at 14:55
$begingroup$
What is wrong with using the Smith normal form? Or any other argument leading to the so called stacked bases theorem?
$endgroup$
– Jyrki Lahtonen
Dec 21 '18 at 8:11
$begingroup$
What is wrong with using the Smith normal form? Or any other argument leading to the so called stacked bases theorem?
$endgroup$
– Jyrki Lahtonen
Dec 21 '18 at 8:11
$begingroup$
I've never seen it before, but I guess it wouldn't hurt to learn something new. I was just wondering if there is a more elementary way to see this.
$endgroup$
– matt stokes
Dec 21 '18 at 14:55
$begingroup$
I've never seen it before, but I guess it wouldn't hurt to learn something new. I was just wondering if there is a more elementary way to see this.
$endgroup$
– matt stokes
Dec 21 '18 at 14:55
add a comment |
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
});
}
});
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%2f3048244%2fshow-delta-theta-mathbbz-deltar-mathbbz-implies-r-mathbbz-t%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%2f3048244%2fshow-delta-theta-mathbbz-deltar-mathbbz-implies-r-mathbbz-t%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
$begingroup$
What is wrong with using the Smith normal form? Or any other argument leading to the so called stacked bases theorem?
$endgroup$
– Jyrki Lahtonen
Dec 21 '18 at 8:11
$begingroup$
I've never seen it before, but I guess it wouldn't hurt to learn something new. I was just wondering if there is a more elementary way to see this.
$endgroup$
– matt stokes
Dec 21 '18 at 14:55