intermediate field $mathbb{F_{p}} subset mathbb{F_{p^{n}}} subset overline{mathbb{F_{p}}}$
$begingroup$
I have trouble solving the following problems:
Let $p$ be a prime number and $K$ a field with characteristic $p$. Let $overline{K} supset K$ be the algebraic closure of $K$.
i) Let $n>0$ be a natural number. Prove that the polynomial $f(X)=X^{p^{n}}-X$ has exactly $p^{n}$ different roots in $overline{K}$.
ii) Now consider $K=mathbb{F_{p}}$ embedded into the algebraic closure $overline{ mathbb{F_{p}}}supset mathbb{F_{p}}$. Define $mathbb{F_{p^{n}}}:={xin overline{ mathbb{F_{p}}}| x^{p^{n}}=x}$. Show that $mathbb{F_{p}} subset mathbb{F_{p^{n}}} subset overline{mathbb{F_{p}}}$ is an intermediate field.
iii) Prove that $mathbb{F_{p}}subset mathbb{F_{p^{n}}}$ is a normal field extension.
i) I know that I can write $X^{p^{n}}-X$ as $X(X^{p^{n-1}}-1)$ and then I have to find the remaining $p^{n-1}$ roots but I'm not sure how.
Thanks for any help in advance.
Clarification of used definitons:
The characteristic of a ring is the smallest number such that $1+...+1=0$.
Let $K,M$ be fields, the ring homomorphism $a:K to M$ is called embedding.
Let $Ksubset L$ be a field extension of the field $K$. An intermediate field is a field $M$ such that $Ksubset M subset L$ and $M$ is a subfiled of $L$.
Let $K$ be a field and $K subset overline{K}$ be the algebraic closure. The intermediate field $K subset L subsetoverline{K}$ is called normal if the embedding $b: L to overline{K}$ satisfies $b(L) subset L$.
abstract-algebra
asked Dec 6 '18 at 12:21
ManwellManwell
114
$endgroup$
add a comment |
$begingroup$
I have trouble solving the following problems:
Let $p$ be a prime number and $K$ a field with characteristic $p$. Let $overline{K} supset K$ be the algebraic closure of $K$.
i) Let $n>0$ be a natural number. Prove that the polynomial $f(X)=X^{p^{n}}-X$ has exactly $p^{n}$ different roots in $overline{K}$.
ii) Now consider $K=mathbb{F_{p}}$ embedded into the algebraic closure $overline{ mathbb{F_{p}}}supset mathbb{F_{p}}$. Define $mathbb{F_{p^{n}}}:={xin overline{ mathbb{F_{p}}}| x^{p^{n}}=x}$. Show that $mathbb{F_{p}} subset mathbb{F_{p^{n}}} subset overline{mathbb{F_{p}}}$ is an intermediate field.
iii) Prove that $mathbb{F_{p}}subset mathbb{F_{p^{n}}}$ is a normal field extension.
i) I know that I can write $X^{p^{n}}-X$ as $X(X^{p^{n-1}}-1)$ and then I have to find the remaining $p^{n-1}$ roots but I'm not sure how.
Thanks for any help in advance.
Clarification of used definitons:
The characteristic of a ring is the smallest number such that $1+...+1=0$.
Let $K,M$ be fields, the ring homomorphism $a:K to M$ is called embedding.
Let $Ksubset L$ be a field extension of the field $K$. An intermediate field is a field $M$ such that $Ksubset M subset L$ and $M$ is a subfiled of $L$.
Let $K$ be a field and $K subset overline{K}$ be the algebraic closure. The intermediate field $K subset L subsetoverline{K}$ is called normal if the embedding $b: L to overline{K}$ satisfies $b(L) subset L$.
abstract-algebra
asked Dec 6 '18 at 12:21
ManwellManwell
114
$endgroup$
add a comment |
$begingroup$
I have trouble solving the following problems:
Let $p$ be a prime number and $K$ a field with characteristic $p$. Let $overline{K} supset K$ be the algebraic closure of $K$.
i) Let $n>0$ be a natural number. Prove that the polynomial $f(X)=X^{p^{n}}-X$ has exactly $p^{n}$ different roots in $overline{K}$.
ii) Now consider $K=mathbb{F_{p}}$ embedded into the algebraic closure $overline{ mathbb{F_{p}}}supset mathbb{F_{p}}$. Define $mathbb{F_{p^{n}}}:={xin overline{ mathbb{F_{p}}}| x^{p^{n}}=x}$. Show that $mathbb{F_{p}} subset mathbb{F_{p^{n}}} subset overline{mathbb{F_{p}}}$ is an intermediate field.
iii) Prove that $mathbb{F_{p}}subset mathbb{F_{p^{n}}}$ is a normal field extension.
i) I know that I can write $X^{p^{n}}-X$ as $X(X^{p^{n-1}}-1)$ and then I have to find the remaining $p^{n-1}$ roots but I'm not sure how.
Thanks for any help in advance.
Clarification of used definitons:
The characteristic of a ring is the smallest number such that $1+...+1=0$.
Let $K,M$ be fields, the ring homomorphism $a:K to M$ is called embedding.
Let $Ksubset L$ be a field extension of the field $K$. An intermediate field is a field $M$ such that $Ksubset M subset L$ and $M$ is a subfiled of $L$.
Let $K$ be a field and $K subset overline{K}$ be the algebraic closure. The intermediate field $K subset L subsetoverline{K}$ is called normal if the embedding $b: L to overline{K}$ satisfies $b(L) subset L$.
abstract-algebra
asked Dec 6 '18 at 12:21
ManwellManwell
114
$endgroup$
I have trouble solving the following problems:
Let $p$ be a prime number and $K$ a field with characteristic $p$. Let $overline{K} supset K$ be the algebraic closure of $K$.
i) Let $n>0$ be a natural number. Prove that the polynomial $f(X)=X^{p^{n}}-X$ has exactly $p^{n}$ different roots in $overline{K}$.
ii) Now consider $K=mathbb{F_{p}}$ embedded into the algebraic closure $overline{ mathbb{F_{p}}}supset mathbb{F_{p}}$. Define $mathbb{F_{p^{n}}}:={xin overline{ mathbb{F_{p}}}| x^{p^{n}}=x}$. Show that $mathbb{F_{p}} subset mathbb{F_{p^{n}}} subset overline{mathbb{F_{p}}}$ is an intermediate field.
iii) Prove that $mathbb{F_{p}}subset mathbb{F_{p^{n}}}$ is a normal field extension.
i) I know that I can write $X^{p^{n}}-X$ as $X(X^{p^{n-1}}-1)$ and then I have to find the remaining $p^{n-1}$ roots but I'm not sure how.
Thanks for any help in advance.
Clarification of used definitons:
The characteristic of a ring is the smallest number such that $1+...+1=0$.
Let $K,M$ be fields, the ring homomorphism $a:K to M$ is called embedding.
Let $Ksubset L$ be a field extension of the field $K$. An intermediate field is a field $M$ such that $Ksubset M subset L$ and $M$ is a subfiled of $L$.
Let $K$ be a field and $K subset overline{K}$ be the algebraic closure. The intermediate field $K subset L subsetoverline{K}$ is called normal if the embedding $b: L to overline{K}$ satisfies $b(L) subset L$.
abstract-algebra
abstract-algebra
asked Dec 6 '18 at 12:21
ManwellManwell
114
asked Dec 6 '18 at 12:21
ManwellManwell
114
edited Dec 6 '18 at 22:22
Manwell
asked Dec 6 '18 at 12:21
ManwellManwell
114
asked Dec 6 '18 at 12:21
ManwellManwell
114
asked Dec 6 '18 at 12:21
ManwellManwell
114
114
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
i) The derivative of $X^{p^n}-X$ is $-1$, which is coprime to $X^{p^n}-X$. So $X^{p^n}-X$ has no repeated roots.
ii) You need to prove that $Bbb{F}_{p^n}$ is a field containing $Bbb{F}_p$. Since $x^p-x=0$ for all $xin Bbb{F}_p$, $Bbb{F}_psubseteq Bbb{F}_{p^n}$. To show that $Bbb{F}_{p^n}$ is closed under addition, prove the identity $$(X+Y)^{p^n}=X^{p^n}+Y^{p^n}.$$ It is clear that $Bbb{F}_{p^n}$ is closed under multiplication and taking inverses (of non-zero elements).
iii) If an irreducible polynomial $p(X)inBbb{F}_p[X]$ has a root in $Bbb{F}_{p^n}$, then $p(X)$ is not coprime to $X^{p^n}-X$, so $p(X)$ must divide $X^{p^n}-X$. Therefore, all roots of $p(X)$ are in $Bbb{F}_{p^n}$, and this means $Bbb{F}_{p^n}$ is a normal extension of $Bbb{F}_p$.
$endgroup$
$begingroup$
Thank you. Why is the derivative -1? And I think we proved that identity after introducing the Frobenius endomorphism in the lecture.
$endgroup$
– Manwell
Dec 6 '18 at 14:12
$begingroup$
The derivative of $X^{p^n}-X$ is $p^nX^{p^n-1}-1$, but $p=0$ here.
$endgroup$
– user593746
Dec 6 '18 at 14:28
$begingroup$
Thanks. Why is $p=0$ all the sudden? It's supposed to be a prime number. Why are we considering the derivative to begin with?
$endgroup$
– Manwell
Dec 6 '18 at 14:34
$begingroup$
Because your field has characteristic $p$, we have $p=0$ in this field.
$endgroup$
– user593746
Dec 6 '18 at 14:36
$begingroup$
About the derivative, read this pdf.
$endgroup$
– user593746
Dec 6 '18 at 14:37
|
show 1 more 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%2f3028433%2fintermediate-field-mathbbf-p-subset-mathbbf-pn-subset-overline%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
$begingroup$
i) The derivative of $X^{p^n}-X$ is $-1$, which is coprime to $X^{p^n}-X$. So $X^{p^n}-X$ has no repeated roots.
ii) You need to prove that $Bbb{F}_{p^n}$ is a field containing $Bbb{F}_p$. Since $x^p-x=0$ for all $xin Bbb{F}_p$, $Bbb{F}_psubseteq Bbb{F}_{p^n}$. To show that $Bbb{F}_{p^n}$ is closed under addition, prove the identity $$(X+Y)^{p^n}=X^{p^n}+Y^{p^n}.$$ It is clear that $Bbb{F}_{p^n}$ is closed under multiplication and taking inverses (of non-zero elements).
iii) If an irreducible polynomial $p(X)inBbb{F}_p[X]$ has a root in $Bbb{F}_{p^n}$, then $p(X)$ is not coprime to $X^{p^n}-X$, so $p(X)$ must divide $X^{p^n}-X$. Therefore, all roots of $p(X)$ are in $Bbb{F}_{p^n}$, and this means $Bbb{F}_{p^n}$ is a normal extension of $Bbb{F}_p$.
$endgroup$
$begingroup$
Thank you. Why is the derivative -1? And I think we proved that identity after introducing the Frobenius endomorphism in the lecture.
$endgroup$
– Manwell
Dec 6 '18 at 14:12
$begingroup$
The derivative of $X^{p^n}-X$ is $p^nX^{p^n-1}-1$, but $p=0$ here.
$endgroup$
– user593746
Dec 6 '18 at 14:28
$begingroup$
Thanks. Why is $p=0$ all the sudden? It's supposed to be a prime number. Why are we considering the derivative to begin with?
$endgroup$
– Manwell
Dec 6 '18 at 14:34
$begingroup$
Because your field has characteristic $p$, we have $p=0$ in this field.
$endgroup$
– user593746
Dec 6 '18 at 14:36
$begingroup$
About the derivative, read this pdf.
$endgroup$
– user593746
Dec 6 '18 at 14:37
|
show 1 more comment
$begingroup$
i) The derivative of $X^{p^n}-X$ is $-1$, which is coprime to $X^{p^n}-X$. So $X^{p^n}-X$ has no repeated roots.
ii) You need to prove that $Bbb{F}_{p^n}$ is a field containing $Bbb{F}_p$. Since $x^p-x=0$ for all $xin Bbb{F}_p$, $Bbb{F}_psubseteq Bbb{F}_{p^n}$. To show that $Bbb{F}_{p^n}$ is closed under addition, prove the identity $$(X+Y)^{p^n}=X^{p^n}+Y^{p^n}.$$ It is clear that $Bbb{F}_{p^n}$ is closed under multiplication and taking inverses (of non-zero elements).
iii) If an irreducible polynomial $p(X)inBbb{F}_p[X]$ has a root in $Bbb{F}_{p^n}$, then $p(X)$ is not coprime to $X^{p^n}-X$, so $p(X)$ must divide $X^{p^n}-X$. Therefore, all roots of $p(X)$ are in $Bbb{F}_{p^n}$, and this means $Bbb{F}_{p^n}$ is a normal extension of $Bbb{F}_p$.
$endgroup$
$begingroup$
Thank you. Why is the derivative -1? And I think we proved that identity after introducing the Frobenius endomorphism in the lecture.
$endgroup$
– Manwell
Dec 6 '18 at 14:12
$begingroup$
The derivative of $X^{p^n}-X$ is $p^nX^{p^n-1}-1$, but $p=0$ here.
$endgroup$
– user593746
Dec 6 '18 at 14:28
$begingroup$
Thanks. Why is $p=0$ all the sudden? It's supposed to be a prime number. Why are we considering the derivative to begin with?
$endgroup$
– Manwell
Dec 6 '18 at 14:34
$begingroup$
Because your field has characteristic $p$, we have $p=0$ in this field.
$endgroup$
– user593746
Dec 6 '18 at 14:36
$begingroup$
About the derivative, read this pdf.
$endgroup$
– user593746
Dec 6 '18 at 14:37
|
show 1 more comment
$begingroup$
i) The derivative of $X^{p^n}-X$ is $-1$, which is coprime to $X^{p^n}-X$. So $X^{p^n}-X$ has no repeated roots.
ii) You need to prove that $Bbb{F}_{p^n}$ is a field containing $Bbb{F}_p$. Since $x^p-x=0$ for all $xin Bbb{F}_p$, $Bbb{F}_psubseteq Bbb{F}_{p^n}$. To show that $Bbb{F}_{p^n}$ is closed under addition, prove the identity $$(X+Y)^{p^n}=X^{p^n}+Y^{p^n}.$$ It is clear that $Bbb{F}_{p^n}$ is closed under multiplication and taking inverses (of non-zero elements).
iii) If an irreducible polynomial $p(X)inBbb{F}_p[X]$ has a root in $Bbb{F}_{p^n}$, then $p(X)$ is not coprime to $X^{p^n}-X$, so $p(X)$ must divide $X^{p^n}-X$. Therefore, all roots of $p(X)$ are in $Bbb{F}_{p^n}$, and this means $Bbb{F}_{p^n}$ is a normal extension of $Bbb{F}_p$.
$endgroup$
i) The derivative of $X^{p^n}-X$ is $-1$, which is coprime to $X^{p^n}-X$. So $X^{p^n}-X$ has no repeated roots.
ii) You need to prove that $Bbb{F}_{p^n}$ is a field containing $Bbb{F}_p$. Since $x^p-x=0$ for all $xin Bbb{F}_p$, $Bbb{F}_psubseteq Bbb{F}_{p^n}$. To show that $Bbb{F}_{p^n}$ is closed under addition, prove the identity $$(X+Y)^{p^n}=X^{p^n}+Y^{p^n}.$$ It is clear that $Bbb{F}_{p^n}$ is closed under multiplication and taking inverses (of non-zero elements).
iii) If an irreducible polynomial $p(X)inBbb{F}_p[X]$ has a root in $Bbb{F}_{p^n}$, then $p(X)$ is not coprime to $X^{p^n}-X$, so $p(X)$ must divide $X^{p^n}-X$. Therefore, all roots of $p(X)$ are in $Bbb{F}_{p^n}$, and this means $Bbb{F}_{p^n}$ is a normal extension of $Bbb{F}_p$.
answered Dec 6 '18 at 13:43
user593746
$begingroup$
Thank you. Why is the derivative -1? And I think we proved that identity after introducing the Frobenius endomorphism in the lecture.
$endgroup$
– Manwell
Dec 6 '18 at 14:12
$begingroup$
The derivative of $X^{p^n}-X$ is $p^nX^{p^n-1}-1$, but $p=0$ here.
$endgroup$
– user593746
Dec 6 '18 at 14:28
$begingroup$
Thanks. Why is $p=0$ all the sudden? It's supposed to be a prime number. Why are we considering the derivative to begin with?
$endgroup$
– Manwell
Dec 6 '18 at 14:34
$begingroup$
Because your field has characteristic $p$, we have $p=0$ in this field.
$endgroup$
– user593746
Dec 6 '18 at 14:36
$begingroup$
About the derivative, read this pdf.
$endgroup$
– user593746
Dec 6 '18 at 14:37
|
show 1 more comment
$begingroup$
Thank you. Why is the derivative -1? And I think we proved that identity after introducing the Frobenius endomorphism in the lecture.
$endgroup$
– Manwell
Dec 6 '18 at 14:12
$begingroup$
The derivative of $X^{p^n}-X$ is $p^nX^{p^n-1}-1$, but $p=0$ here.
$endgroup$
– user593746
Dec 6 '18 at 14:28
$begingroup$
Thanks. Why is $p=0$ all the sudden? It's supposed to be a prime number. Why are we considering the derivative to begin with?
$endgroup$
– Manwell
Dec 6 '18 at 14:34
$begingroup$
Because your field has characteristic $p$, we have $p=0$ in this field.
$endgroup$
– user593746
Dec 6 '18 at 14:36
$begingroup$
About the derivative, read this pdf.
$endgroup$
– user593746
Dec 6 '18 at 14:37
$begingroup$
Thank you. Why is the derivative -1? And I think we proved that identity after introducing the Frobenius endomorphism in the lecture.
$endgroup$
– Manwell
Dec 6 '18 at 14:12
$begingroup$
Thank you. Why is the derivative -1? And I think we proved that identity after introducing the Frobenius endomorphism in the lecture.
$endgroup$
– Manwell
Dec 6 '18 at 14:12
$begingroup$
The derivative of $X^{p^n}-X$ is $p^nX^{p^n-1}-1$, but $p=0$ here.
$endgroup$
– user593746
Dec 6 '18 at 14:28
$begingroup$
The derivative of $X^{p^n}-X$ is $p^nX^{p^n-1}-1$, but $p=0$ here.
$endgroup$
– user593746
Dec 6 '18 at 14:28
$begingroup$
Thanks. Why is $p=0$ all the sudden? It's supposed to be a prime number. Why are we considering the derivative to begin with?
$endgroup$
– Manwell
Dec 6 '18 at 14:34
$begingroup$
Thanks. Why is $p=0$ all the sudden? It's supposed to be a prime number. Why are we considering the derivative to begin with?
$endgroup$
– Manwell
Dec 6 '18 at 14:34
$begingroup$
Because your field has characteristic $p$, we have $p=0$ in this field.
$endgroup$
– user593746
Dec 6 '18 at 14:36
$begingroup$
Because your field has characteristic $p$, we have $p=0$ in this field.
$endgroup$
– user593746
Dec 6 '18 at 14:36
$begingroup$
About the derivative, read this pdf.
$endgroup$
– user593746
Dec 6 '18 at 14:37
$begingroup$
About the derivative, read this pdf.
$endgroup$
– user593746
Dec 6 '18 at 14:37
|
show 1 more 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%2f3028433%2fintermediate-field-mathbbf-p-subset-mathbbf-pn-subset-overline%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
