Algebraic integers in a quadratic number field
I want to show that an element in $alpha$ in a quadratic number field, $mathbb{Q}(d)$, is an algebraic integer if the Norm and the Trace of $alpha$ are in $mathbb{Z}.$
algebraic-number-theory
asked Dec 5 '18 at 17:50
pullofthemoonpullofthemoon
535
add a comment |
I want to show that an element in $alpha$ in a quadratic number field, $mathbb{Q}(d)$, is an algebraic integer if the Norm and the Trace of $alpha$ are in $mathbb{Z}.$
algebraic-number-theory
asked Dec 5 '18 at 17:50
pullofthemoonpullofthemoon
535
What have you tried and where are you stuck?
– rogerl
Dec 5 '18 at 18:05
add a comment |
I want to show that an element in $alpha$ in a quadratic number field, $mathbb{Q}(d)$, is an algebraic integer if the Norm and the Trace of $alpha$ are in $mathbb{Z}.$
algebraic-number-theory
asked Dec 5 '18 at 17:50
pullofthemoonpullofthemoon
535
I want to show that an element in $alpha$ in a quadratic number field, $mathbb{Q}(d)$, is an algebraic integer if the Norm and the Trace of $alpha$ are in $mathbb{Z}.$
algebraic-number-theory
algebraic-number-theory
asked Dec 5 '18 at 17:50
pullofthemoonpullofthemoon
535
asked Dec 5 '18 at 17:50
pullofthemoonpullofthemoon
535
asked Dec 5 '18 at 17:50
pullofthemoonpullofthemoon
535
asked Dec 5 '18 at 17:50
pullofthemoonpullofthemoon
535
asked Dec 5 '18 at 17:50
pullofthemoonpullofthemoon
535
535
What have you tried and where are you stuck?
– rogerl
Dec 5 '18 at 18:05
add a comment |
What have you tried and where are you stuck?
– rogerl
Dec 5 '18 at 18:05
What have you tried and where are you stuck?
– rogerl
Dec 5 '18 at 18:05
What have you tried and where are you stuck?
– rogerl
Dec 5 '18 at 18:05
add a comment |
1 Answer
1
active
oldest
votes
Let $alphainmathbb{Q}(sqrt d)$ be an algebraic integer; By definition, it is the root of a polynomial $f(x)inmathbb{Z}[x]$. One can show that in fact, the minimal polynomial of $alpha$ over $mathbb{Q}$, is with integral coefficients
(please tell me if you want me to elaborate on this point). Note that if $alpha=a+bsqrt d$, then $alpha$ is a root of $x^2-2ax+a^2-db^2inmathbb{Q}[x]$ and that this is in fact its minimal polynomial, as there is no polynomial of degree 1 over $mathbb{Q}$ has $alpha$ as a root (otherwise, $alpha$ would be a rational number, and this case is easy to handle with). But then it follows that $2a,a^2-db^2inmathbb{Z}$, and these are the trace and norm of $alpha$, so we're finished.
Conversely, write $alpha=a+bsqrt d$. Then $tr_{mathbb{Q}(sqrt d)/mathbb{Q}}(alpha)=2a$ and $Norm_{mathbb{Q}(sqrt d)/mathbb{Q}}(alpha)=a^2-db^2$. If they're both in $mathbb{Z}$, then $f(x)=x^2-2ax+(a^2-db^2)inmathbb{Z}[x]$. Now by Vietta's formula (or by substitution, whatever you like...) we get that $f(alpha)=0$ and so $alpha$ is algebraic.
Conceptually, $alpha$ is an algebraic integer, iff its minimal polynomial over $mathbb{Q}$ is a polynomial with integer coefficients. In the case of a quadratic extension, the coefficients of the minimal polynomial (except for the leading coefficient, which is 1) are precisely the norm and the trace of $alpha$.
answered Dec 7 '18 at 12:26
MadarbMadarb
346111
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%2f3027408%2falgebraic-integers-in-a-quadratic-number-field%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $alphainmathbb{Q}(sqrt d)$ be an algebraic integer; By definition, it is the root of a polynomial $f(x)inmathbb{Z}[x]$. One can show that in fact, the minimal polynomial of $alpha$ over $mathbb{Q}$, is with integral coefficients
(please tell me if you want me to elaborate on this point). Note that if $alpha=a+bsqrt d$, then $alpha$ is a root of $x^2-2ax+a^2-db^2inmathbb{Q}[x]$ and that this is in fact its minimal polynomial, as there is no polynomial of degree 1 over $mathbb{Q}$ has $alpha$ as a root (otherwise, $alpha$ would be a rational number, and this case is easy to handle with). But then it follows that $2a,a^2-db^2inmathbb{Z}$, and these are the trace and norm of $alpha$, so we're finished.
Conversely, write $alpha=a+bsqrt d$. Then $tr_{mathbb{Q}(sqrt d)/mathbb{Q}}(alpha)=2a$ and $Norm_{mathbb{Q}(sqrt d)/mathbb{Q}}(alpha)=a^2-db^2$. If they're both in $mathbb{Z}$, then $f(x)=x^2-2ax+(a^2-db^2)inmathbb{Z}[x]$. Now by Vietta's formula (or by substitution, whatever you like...) we get that $f(alpha)=0$ and so $alpha$ is algebraic.
Conceptually, $alpha$ is an algebraic integer, iff its minimal polynomial over $mathbb{Q}$ is a polynomial with integer coefficients. In the case of a quadratic extension, the coefficients of the minimal polynomial (except for the leading coefficient, which is 1) are precisely the norm and the trace of $alpha$.
answered Dec 7 '18 at 12:26
MadarbMadarb
346111
add a comment |
Let $alphainmathbb{Q}(sqrt d)$ be an algebraic integer; By definition, it is the root of a polynomial $f(x)inmathbb{Z}[x]$. One can show that in fact, the minimal polynomial of $alpha$ over $mathbb{Q}$, is with integral coefficients
(please tell me if you want me to elaborate on this point). Note that if $alpha=a+bsqrt d$, then $alpha$ is a root of $x^2-2ax+a^2-db^2inmathbb{Q}[x]$ and that this is in fact its minimal polynomial, as there is no polynomial of degree 1 over $mathbb{Q}$ has $alpha$ as a root (otherwise, $alpha$ would be a rational number, and this case is easy to handle with). But then it follows that $2a,a^2-db^2inmathbb{Z}$, and these are the trace and norm of $alpha$, so we're finished.
Conversely, write $alpha=a+bsqrt d$. Then $tr_{mathbb{Q}(sqrt d)/mathbb{Q}}(alpha)=2a$ and $Norm_{mathbb{Q}(sqrt d)/mathbb{Q}}(alpha)=a^2-db^2$. If they're both in $mathbb{Z}$, then $f(x)=x^2-2ax+(a^2-db^2)inmathbb{Z}[x]$. Now by Vietta's formula (or by substitution, whatever you like...) we get that $f(alpha)=0$ and so $alpha$ is algebraic.
Conceptually, $alpha$ is an algebraic integer, iff its minimal polynomial over $mathbb{Q}$ is a polynomial with integer coefficients. In the case of a quadratic extension, the coefficients of the minimal polynomial (except for the leading coefficient, which is 1) are precisely the norm and the trace of $alpha$.
answered Dec 7 '18 at 12:26
MadarbMadarb
346111
add a comment |
Let $alphainmathbb{Q}(sqrt d)$ be an algebraic integer; By definition, it is the root of a polynomial $f(x)inmathbb{Z}[x]$. One can show that in fact, the minimal polynomial of $alpha$ over $mathbb{Q}$, is with integral coefficients
(please tell me if you want me to elaborate on this point). Note that if $alpha=a+bsqrt d$, then $alpha$ is a root of $x^2-2ax+a^2-db^2inmathbb{Q}[x]$ and that this is in fact its minimal polynomial, as there is no polynomial of degree 1 over $mathbb{Q}$ has $alpha$ as a root (otherwise, $alpha$ would be a rational number, and this case is easy to handle with). But then it follows that $2a,a^2-db^2inmathbb{Z}$, and these are the trace and norm of $alpha$, so we're finished.
Conversely, write $alpha=a+bsqrt d$. Then $tr_{mathbb{Q}(sqrt d)/mathbb{Q}}(alpha)=2a$ and $Norm_{mathbb{Q}(sqrt d)/mathbb{Q}}(alpha)=a^2-db^2$. If they're both in $mathbb{Z}$, then $f(x)=x^2-2ax+(a^2-db^2)inmathbb{Z}[x]$. Now by Vietta's formula (or by substitution, whatever you like...) we get that $f(alpha)=0$ and so $alpha$ is algebraic.
Conceptually, $alpha$ is an algebraic integer, iff its minimal polynomial over $mathbb{Q}$ is a polynomial with integer coefficients. In the case of a quadratic extension, the coefficients of the minimal polynomial (except for the leading coefficient, which is 1) are precisely the norm and the trace of $alpha$.
answered Dec 7 '18 at 12:26
MadarbMadarb
346111
Let $alphainmathbb{Q}(sqrt d)$ be an algebraic integer; By definition, it is the root of a polynomial $f(x)inmathbb{Z}[x]$. One can show that in fact, the minimal polynomial of $alpha$ over $mathbb{Q}$, is with integral coefficients
(please tell me if you want me to elaborate on this point). Note that if $alpha=a+bsqrt d$, then $alpha$ is a root of $x^2-2ax+a^2-db^2inmathbb{Q}[x]$ and that this is in fact its minimal polynomial, as there is no polynomial of degree 1 over $mathbb{Q}$ has $alpha$ as a root (otherwise, $alpha$ would be a rational number, and this case is easy to handle with). But then it follows that $2a,a^2-db^2inmathbb{Z}$, and these are the trace and norm of $alpha$, so we're finished.
Conversely, write $alpha=a+bsqrt d$. Then $tr_{mathbb{Q}(sqrt d)/mathbb{Q}}(alpha)=2a$ and $Norm_{mathbb{Q}(sqrt d)/mathbb{Q}}(alpha)=a^2-db^2$. If they're both in $mathbb{Z}$, then $f(x)=x^2-2ax+(a^2-db^2)inmathbb{Z}[x]$. Now by Vietta's formula (or by substitution, whatever you like...) we get that $f(alpha)=0$ and so $alpha$ is algebraic.
Conceptually, $alpha$ is an algebraic integer, iff its minimal polynomial over $mathbb{Q}$ is a polynomial with integer coefficients. In the case of a quadratic extension, the coefficients of the minimal polynomial (except for the leading coefficient, which is 1) are precisely the norm and the trace of $alpha$.
answered Dec 7 '18 at 12:26
MadarbMadarb
346111
answered Dec 7 '18 at 12:26
MadarbMadarb
346111
answered Dec 7 '18 at 12:26
MadarbMadarb
346111
answered Dec 7 '18 at 12:26
MadarbMadarb
346111
346111
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%2f3027408%2falgebraic-integers-in-a-quadratic-number-field%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
What have you tried and where are you stuck?
– rogerl
Dec 5 '18 at 18:05