Why does the minimal polynomial of α divide all polynomials for which α is a root?
$begingroup$
Suppose α ∈ E with E being a field extension of F. Let S be the set of all polynomials in F[x] for which α is a root. Prove that the minimal polynomial of α over F divides every polynomial in S.
field-theory minimal-polynomials
asked Jun 14 '14 at 3:28
user156724user156724
163
$endgroup$
add a comment |
$begingroup$
Suppose α ∈ E with E being a field extension of F. Let S be the set of all polynomials in F[x] for which α is a root. Prove that the minimal polynomial of α over F divides every polynomial in S.
field-theory minimal-polynomials
asked Jun 14 '14 at 3:28
user156724user156724
163
$endgroup$
4
$begingroup$
Hint: the remainder upon division by $m(x)$ has degree $<deg m$.
$endgroup$
– blue
Jun 14 '14 at 3:31
add a comment |
$begingroup$
Suppose α ∈ E with E being a field extension of F. Let S be the set of all polynomials in F[x] for which α is a root. Prove that the minimal polynomial of α over F divides every polynomial in S.
field-theory minimal-polynomials
asked Jun 14 '14 at 3:28
user156724user156724
163
$endgroup$
Suppose α ∈ E with E being a field extension of F. Let S be the set of all polynomials in F[x] for which α is a root. Prove that the minimal polynomial of α over F divides every polynomial in S.
field-theory minimal-polynomials
field-theory minimal-polynomials
asked Jun 14 '14 at 3:28
user156724user156724
163
asked Jun 14 '14 at 3:28
user156724user156724
163
asked Jun 14 '14 at 3:28
user156724user156724
163
asked Jun 14 '14 at 3:28
user156724user156724
163
asked Jun 14 '14 at 3:28
user156724user156724
163
163
4
$begingroup$
Hint: the remainder upon division by $m(x)$ has degree $<deg m$.
$endgroup$
– blue
Jun 14 '14 at 3:31
add a comment |
4
$begingroup$
Hint: the remainder upon division by $m(x)$ has degree $<deg m$.
$endgroup$
– blue
Jun 14 '14 at 3:31
4
4
$begingroup$
Hint: the remainder upon division by $m(x)$ has degree $<deg m$.
$endgroup$
– blue
Jun 14 '14 at 3:31
$begingroup$
Hint: the remainder upon division by $m(x)$ has degree $<deg m$.
$endgroup$
– blue
Jun 14 '14 at 3:31
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Both these answers are much more complicated than is necessary. Simply note that by the division algorithm, we can always write
$$f(alpha) = q(alpha)m(alpha) + r(alpha)$$
for some $q in F[x]$ and $f in S$, with $m in S$ being the minimal polynomial. Since $alpha$ is a root of $f$, we either have $r=0$, as desired, or $r(alpha) = 0$. But, since the degree of $r$ is smaller than the degree of $m$, this would violate the minimality of $m$, so we must have $r=0$.
answered Mar 14 '18 at 4:16
Elliot GorokhovskyElliot Gorokhovsky
1,03511024
$endgroup$
add a comment |
$begingroup$
Consider the evaluation homomorphism $ev_alpha:F[x] rightarrow F[alpha]$ defined as follows: $g(x) mapsto g(alpha)$. If you haven't seen this before, you'll want to write up a quick proof that this is indeed a homomorphism.
As with any homomorphism, its kernel is an ideal (in $F[x]$), and it will be the set of all polynomials with $alpha$ as a root. Since $F$ is a field, then $F[x]$ is a principal ideal domain. That is, every ideal in $F[x]$ is generated by a single element. For completeness, you'll want to prove this fact as well. Therefore, $ker(ev_alpha)$ must be generated by a single element.
Now you just need to convince yourself that that element must be the minimal polynomial with $alpha$ as a root. Finally, we can use this to conclude that the minimal polynomial will divide every polynomial with $alpha$ as a root.
answered Jun 14 '14 at 4:08
Kaj HansenKaj Hansen
27.8k43980
$endgroup$
add a comment |
$begingroup$
Hint $ S $ is an ideal of $,F[x],,$ which, having a Euclidean algorithm, is a PID. Hence the ideal $,S,$ is principal, generated by any element of minimal degree (= gcd of all elements of $S).$
Or, if ideals are unfamiliar, note that $,S,$ is closed under gcd. Indeed, by Bezout, $,gcd(f,g) = a f + b g,$ so if $,alpha,$ is a root of $,f,g,$ then it is a root of their gcd. Thus any nonzero element $,fin S,$ of minimal degree divides every $,gin S,,$ for else their gcd $in S,$ and has smaller degree than $,f,$ (or, equivalently, $,0neq gbmod f = g - fhin S,$ has smaller degree than $,f,,$ contra minimality of $,f)$.
answered Jun 14 '14 at 3:31
Bill DubuqueBill Dubuque
214k29197660
$endgroup$
add a comment |
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%2f833535%2fwhy-does-the-minimal-polynomial-of-%25ce%25b1-divide-all-polynomials-for-which-%25ce%25b1-is-a-roo%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Both these answers are much more complicated than is necessary. Simply note that by the division algorithm, we can always write
$$f(alpha) = q(alpha)m(alpha) + r(alpha)$$
for some $q in F[x]$ and $f in S$, with $m in S$ being the minimal polynomial. Since $alpha$ is a root of $f$, we either have $r=0$, as desired, or $r(alpha) = 0$. But, since the degree of $r$ is smaller than the degree of $m$, this would violate the minimality of $m$, so we must have $r=0$.
answered Mar 14 '18 at 4:16
Elliot GorokhovskyElliot Gorokhovsky
1,03511024
$endgroup$
add a comment |
$begingroup$
Both these answers are much more complicated than is necessary. Simply note that by the division algorithm, we can always write
$$f(alpha) = q(alpha)m(alpha) + r(alpha)$$
for some $q in F[x]$ and $f in S$, with $m in S$ being the minimal polynomial. Since $alpha$ is a root of $f$, we either have $r=0$, as desired, or $r(alpha) = 0$. But, since the degree of $r$ is smaller than the degree of $m$, this would violate the minimality of $m$, so we must have $r=0$.
answered Mar 14 '18 at 4:16
Elliot GorokhovskyElliot Gorokhovsky
1,03511024
$endgroup$
add a comment |
$begingroup$
Both these answers are much more complicated than is necessary. Simply note that by the division algorithm, we can always write
$$f(alpha) = q(alpha)m(alpha) + r(alpha)$$
for some $q in F[x]$ and $f in S$, with $m in S$ being the minimal polynomial. Since $alpha$ is a root of $f$, we either have $r=0$, as desired, or $r(alpha) = 0$. But, since the degree of $r$ is smaller than the degree of $m$, this would violate the minimality of $m$, so we must have $r=0$.
answered Mar 14 '18 at 4:16
Elliot GorokhovskyElliot Gorokhovsky
1,03511024
$endgroup$
Both these answers are much more complicated than is necessary. Simply note that by the division algorithm, we can always write
$$f(alpha) = q(alpha)m(alpha) + r(alpha)$$
for some $q in F[x]$ and $f in S$, with $m in S$ being the minimal polynomial. Since $alpha$ is a root of $f$, we either have $r=0$, as desired, or $r(alpha) = 0$. But, since the degree of $r$ is smaller than the degree of $m$, this would violate the minimality of $m$, so we must have $r=0$.
answered Mar 14 '18 at 4:16
Elliot GorokhovskyElliot Gorokhovsky
1,03511024
answered Mar 14 '18 at 4:16
Elliot GorokhovskyElliot Gorokhovsky
1,03511024
answered Mar 14 '18 at 4:16
Elliot GorokhovskyElliot Gorokhovsky
1,03511024
answered Mar 14 '18 at 4:16
Elliot GorokhovskyElliot Gorokhovsky
1,03511024
1,03511024
add a comment |
add a comment |
$begingroup$
Consider the evaluation homomorphism $ev_alpha:F[x] rightarrow F[alpha]$ defined as follows: $g(x) mapsto g(alpha)$. If you haven't seen this before, you'll want to write up a quick proof that this is indeed a homomorphism.
As with any homomorphism, its kernel is an ideal (in $F[x]$), and it will be the set of all polynomials with $alpha$ as a root. Since $F$ is a field, then $F[x]$ is a principal ideal domain. That is, every ideal in $F[x]$ is generated by a single element. For completeness, you'll want to prove this fact as well. Therefore, $ker(ev_alpha)$ must be generated by a single element.
Now you just need to convince yourself that that element must be the minimal polynomial with $alpha$ as a root. Finally, we can use this to conclude that the minimal polynomial will divide every polynomial with $alpha$ as a root.
answered Jun 14 '14 at 4:08
Kaj HansenKaj Hansen
27.8k43980
$endgroup$
add a comment |
$begingroup$
Consider the evaluation homomorphism $ev_alpha:F[x] rightarrow F[alpha]$ defined as follows: $g(x) mapsto g(alpha)$. If you haven't seen this before, you'll want to write up a quick proof that this is indeed a homomorphism.
As with any homomorphism, its kernel is an ideal (in $F[x]$), and it will be the set of all polynomials with $alpha$ as a root. Since $F$ is a field, then $F[x]$ is a principal ideal domain. That is, every ideal in $F[x]$ is generated by a single element. For completeness, you'll want to prove this fact as well. Therefore, $ker(ev_alpha)$ must be generated by a single element.
Now you just need to convince yourself that that element must be the minimal polynomial with $alpha$ as a root. Finally, we can use this to conclude that the minimal polynomial will divide every polynomial with $alpha$ as a root.
answered Jun 14 '14 at 4:08
Kaj HansenKaj Hansen
27.8k43980
$endgroup$
add a comment |
$begingroup$
Consider the evaluation homomorphism $ev_alpha:F[x] rightarrow F[alpha]$ defined as follows: $g(x) mapsto g(alpha)$. If you haven't seen this before, you'll want to write up a quick proof that this is indeed a homomorphism.
As with any homomorphism, its kernel is an ideal (in $F[x]$), and it will be the set of all polynomials with $alpha$ as a root. Since $F$ is a field, then $F[x]$ is a principal ideal domain. That is, every ideal in $F[x]$ is generated by a single element. For completeness, you'll want to prove this fact as well. Therefore, $ker(ev_alpha)$ must be generated by a single element.
Now you just need to convince yourself that that element must be the minimal polynomial with $alpha$ as a root. Finally, we can use this to conclude that the minimal polynomial will divide every polynomial with $alpha$ as a root.
answered Jun 14 '14 at 4:08
Kaj HansenKaj Hansen
27.8k43980
$endgroup$
Consider the evaluation homomorphism $ev_alpha:F[x] rightarrow F[alpha]$ defined as follows: $g(x) mapsto g(alpha)$. If you haven't seen this before, you'll want to write up a quick proof that this is indeed a homomorphism.
As with any homomorphism, its kernel is an ideal (in $F[x]$), and it will be the set of all polynomials with $alpha$ as a root. Since $F$ is a field, then $F[x]$ is a principal ideal domain. That is, every ideal in $F[x]$ is generated by a single element. For completeness, you'll want to prove this fact as well. Therefore, $ker(ev_alpha)$ must be generated by a single element.
Now you just need to convince yourself that that element must be the minimal polynomial with $alpha$ as a root. Finally, we can use this to conclude that the minimal polynomial will divide every polynomial with $alpha$ as a root.
answered Jun 14 '14 at 4:08
Kaj HansenKaj Hansen
27.8k43980
edited Jun 14 '14 at 4:17
answered Jun 14 '14 at 4:08
Kaj HansenKaj Hansen
27.8k43980
answered Jun 14 '14 at 4:08
Kaj HansenKaj Hansen
27.8k43980
answered Jun 14 '14 at 4:08
Kaj HansenKaj Hansen
27.8k43980
27.8k43980
add a comment |
add a comment |
$begingroup$
Hint $ S $ is an ideal of $,F[x],,$ which, having a Euclidean algorithm, is a PID. Hence the ideal $,S,$ is principal, generated by any element of minimal degree (= gcd of all elements of $S).$
Or, if ideals are unfamiliar, note that $,S,$ is closed under gcd. Indeed, by Bezout, $,gcd(f,g) = a f + b g,$ so if $,alpha,$ is a root of $,f,g,$ then it is a root of their gcd. Thus any nonzero element $,fin S,$ of minimal degree divides every $,gin S,,$ for else their gcd $in S,$ and has smaller degree than $,f,$ (or, equivalently, $,0neq gbmod f = g - fhin S,$ has smaller degree than $,f,,$ contra minimality of $,f)$.
answered Jun 14 '14 at 3:31
Bill DubuqueBill Dubuque
214k29197660
$endgroup$
add a comment |
$begingroup$
Hint $ S $ is an ideal of $,F[x],,$ which, having a Euclidean algorithm, is a PID. Hence the ideal $,S,$ is principal, generated by any element of minimal degree (= gcd of all elements of $S).$
Or, if ideals are unfamiliar, note that $,S,$ is closed under gcd. Indeed, by Bezout, $,gcd(f,g) = a f + b g,$ so if $,alpha,$ is a root of $,f,g,$ then it is a root of their gcd. Thus any nonzero element $,fin S,$ of minimal degree divides every $,gin S,,$ for else their gcd $in S,$ and has smaller degree than $,f,$ (or, equivalently, $,0neq gbmod f = g - fhin S,$ has smaller degree than $,f,,$ contra minimality of $,f)$.
answered Jun 14 '14 at 3:31
Bill DubuqueBill Dubuque
214k29197660
$endgroup$
add a comment |
$begingroup$
Hint $ S $ is an ideal of $,F[x],,$ which, having a Euclidean algorithm, is a PID. Hence the ideal $,S,$ is principal, generated by any element of minimal degree (= gcd of all elements of $S).$
Or, if ideals are unfamiliar, note that $,S,$ is closed under gcd. Indeed, by Bezout, $,gcd(f,g) = a f + b g,$ so if $,alpha,$ is a root of $,f,g,$ then it is a root of their gcd. Thus any nonzero element $,fin S,$ of minimal degree divides every $,gin S,,$ for else their gcd $in S,$ and has smaller degree than $,f,$ (or, equivalently, $,0neq gbmod f = g - fhin S,$ has smaller degree than $,f,,$ contra minimality of $,f)$.
answered Jun 14 '14 at 3:31
Bill DubuqueBill Dubuque
214k29197660
$endgroup$
Hint $ S $ is an ideal of $,F[x],,$ which, having a Euclidean algorithm, is a PID. Hence the ideal $,S,$ is principal, generated by any element of minimal degree (= gcd of all elements of $S).$
Or, if ideals are unfamiliar, note that $,S,$ is closed under gcd. Indeed, by Bezout, $,gcd(f,g) = a f + b g,$ so if $,alpha,$ is a root of $,f,g,$ then it is a root of their gcd. Thus any nonzero element $,fin S,$ of minimal degree divides every $,gin S,,$ for else their gcd $in S,$ and has smaller degree than $,f,$ (or, equivalently, $,0neq gbmod f = g - fhin S,$ has smaller degree than $,f,,$ contra minimality of $,f)$.
answered Jun 14 '14 at 3:31
Bill DubuqueBill Dubuque
214k29197660
edited Dec 23 '18 at 18:23
answered Jun 14 '14 at 3:31
Bill DubuqueBill Dubuque
214k29197660
answered Jun 14 '14 at 3:31
Bill DubuqueBill Dubuque
214k29197660
answered Jun 14 '14 at 3:31
Bill DubuqueBill Dubuque
214k29197660
214k29197660
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%2f833535%2fwhy-does-the-minimal-polynomial-of-%25ce%25b1-divide-all-polynomials-for-which-%25ce%25b1-is-a-roo%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
4
$begingroup$
Hint: the remainder upon division by $m(x)$ has degree $<deg m$.
$endgroup$
– blue
Jun 14 '14 at 3:31