Strange proof on the irreducibility of $Psi(x)$, the cyclotomic polynomial?












4














Some lemma ahead:




Lemma Let $E/F$ be the field extension. Then all F-automorphisms defines a permutation of roots of any polynomials in $F$






Proof For any root $alpha$ of any polynomial $p(x) in F[x]$, it is straightforward that
$$ p(sigma(alpha)) = sigma(p(alpha)) = 0$$
for any $sigma in text{Gal}(E/F)$.
Therefore $sigma(alpha)$ is indeed a root of $p(x)$.




The proof of irreducibility of $Psi_n(x)$ over $mathbb{Q}[x]$:
$$Psi_n(x) := prod_{(i,n)=1}(x - zeta^i_n)$$
It is proven $Psi_n(x) in mathbb{Z}[x]$.
Consider the group of nth roots of unity, cyclic and isomorphic to $mathbb{Z}_n$.



First note that any $mathbb{Q}$-automorphism must permute the group of nth roots of unity (actually defines an automorphism of that group). It is easy to check any automorphism on $mathbb{Z}_n$ can and can only map a unit to another. Likewise any $mathbb{Q}$-automorphism must map a primitive to another primitive, and we can identify any $mathbb{Q}$-automorphism: $zeta_n mapsto zeta_n^r, (n, r) = 1$ solely by r.



By the lemma, either none of the elements of the primitive nth roots of unity is a root of a polynomial, or all of them. In other words, a minimum polynomial for any $zeta^i_n$ must have degree more than $Psi_n$, therefore must be $Psi_n$. By the definition of minimum polynomial, $Psi_n$ is irreducible.



Is this proof correct?






abstract-algebra galois-theory







share|cite|improve this question




















  • 2




    Until you know the cyclotomic polynomial is urreducible you don't know that every $zeta_n mapsto zeta_n^r, (n, r) = 1$ is an automorphism.
    – ancientmathematician
    Dec 29 '18 at 9:24






  • 2




    What @ancientmathematician says. A census of $Bbb{Q}$-automorphisms is equivalent to irreducibility. Otherwise you can't know that all choices of $r$ are actually automorphisms. For a different way of looking at the weak link in your argument I suggest the following thought experiment. What would change if you look at it over a base field other than $Bbb{Q}$? Remember that over $Bbb{Q}(i)$ the eighth cyclotomic polynomial factors $x^4+1=(x^2+i)(x^2-i)$. And over $Bbb{Q}(sqrt5)$ the fifth cyclotomic polynomial factors. See my comments here
    – Jyrki Lahtonen
    Dec 29 '18 at 9:44










  • @ancientmathematician I still don't get this... Why cannot I state $zeta_n mapsto zeta_n^r, (n, r) = 1$ as an automorphism? Observe the group of nth roots of unity is isomorphic to $Z_n$, therefore every automorphism of $Z_n$ must define a $mathbb{Q}$-automorphism of $mathbb{Q}(zeta_n)$?
    – Astrick Harren
    Dec 29 '18 at 10:31








  • 4




    Group automorphisms are not necessarily field automorphisms. Look at what @JyrkiLahtonen has pointed you to.
    – ancientmathematician
    Dec 29 '18 at 10:34






  • 1




    Your proof does not seem to use the fact that the base field $F$ is $Bbb Q$.
    – Lubin
    Jan 2 at 1:32
















4














Some lemma ahead:




Lemma Let $E/F$ be the field extension. Then all F-automorphisms defines a permutation of roots of any polynomials in $F$






Proof For any root $alpha$ of any polynomial $p(x) in F[x]$, it is straightforward that
$$ p(sigma(alpha)) = sigma(p(alpha)) = 0$$
for any $sigma in text{Gal}(E/F)$.
Therefore $sigma(alpha)$ is indeed a root of $p(x)$.




The proof of irreducibility of $Psi_n(x)$ over $mathbb{Q}[x]$:
$$Psi_n(x) := prod_{(i,n)=1}(x - zeta^i_n)$$
It is proven $Psi_n(x) in mathbb{Z}[x]$.
Consider the group of nth roots of unity, cyclic and isomorphic to $mathbb{Z}_n$.



First note that any $mathbb{Q}$-automorphism must permute the group of nth roots of unity (actually defines an automorphism of that group). It is easy to check any automorphism on $mathbb{Z}_n$ can and can only map a unit to another. Likewise any $mathbb{Q}$-automorphism must map a primitive to another primitive, and we can identify any $mathbb{Q}$-automorphism: $zeta_n mapsto zeta_n^r, (n, r) = 1$ solely by r.



By the lemma, either none of the elements of the primitive nth roots of unity is a root of a polynomial, or all of them. In other words, a minimum polynomial for any $zeta^i_n$ must have degree more than $Psi_n$, therefore must be $Psi_n$. By the definition of minimum polynomial, $Psi_n$ is irreducible.



Is this proof correct?






abstract-algebra galois-theory







share|cite|improve this question




















  • 2




    Until you know the cyclotomic polynomial is urreducible you don't know that every $zeta_n mapsto zeta_n^r, (n, r) = 1$ is an automorphism.
    – ancientmathematician
    Dec 29 '18 at 9:24






  • 2




    What @ancientmathematician says. A census of $Bbb{Q}$-automorphisms is equivalent to irreducibility. Otherwise you can't know that all choices of $r$ are actually automorphisms. For a different way of looking at the weak link in your argument I suggest the following thought experiment. What would change if you look at it over a base field other than $Bbb{Q}$? Remember that over $Bbb{Q}(i)$ the eighth cyclotomic polynomial factors $x^4+1=(x^2+i)(x^2-i)$. And over $Bbb{Q}(sqrt5)$ the fifth cyclotomic polynomial factors. See my comments here
    – Jyrki Lahtonen
    Dec 29 '18 at 9:44










  • @ancientmathematician I still don't get this... Why cannot I state $zeta_n mapsto zeta_n^r, (n, r) = 1$ as an automorphism? Observe the group of nth roots of unity is isomorphic to $Z_n$, therefore every automorphism of $Z_n$ must define a $mathbb{Q}$-automorphism of $mathbb{Q}(zeta_n)$?
    – Astrick Harren
    Dec 29 '18 at 10:31








  • 4




    Group automorphisms are not necessarily field automorphisms. Look at what @JyrkiLahtonen has pointed you to.
    – ancientmathematician
    Dec 29 '18 at 10:34






  • 1




    Your proof does not seem to use the fact that the base field $F$ is $Bbb Q$.
    – Lubin
    Jan 2 at 1:32














4












4








4







Some lemma ahead:




Lemma Let $E/F$ be the field extension. Then all F-automorphisms defines a permutation of roots of any polynomials in $F$






Proof For any root $alpha$ of any polynomial $p(x) in F[x]$, it is straightforward that
$$ p(sigma(alpha)) = sigma(p(alpha)) = 0$$
for any $sigma in text{Gal}(E/F)$.
Therefore $sigma(alpha)$ is indeed a root of $p(x)$.




The proof of irreducibility of $Psi_n(x)$ over $mathbb{Q}[x]$:
$$Psi_n(x) := prod_{(i,n)=1}(x - zeta^i_n)$$
It is proven $Psi_n(x) in mathbb{Z}[x]$.
Consider the group of nth roots of unity, cyclic and isomorphic to $mathbb{Z}_n$.



First note that any $mathbb{Q}$-automorphism must permute the group of nth roots of unity (actually defines an automorphism of that group). It is easy to check any automorphism on $mathbb{Z}_n$ can and can only map a unit to another. Likewise any $mathbb{Q}$-automorphism must map a primitive to another primitive, and we can identify any $mathbb{Q}$-automorphism: $zeta_n mapsto zeta_n^r, (n, r) = 1$ solely by r.



By the lemma, either none of the elements of the primitive nth roots of unity is a root of a polynomial, or all of them. In other words, a minimum polynomial for any $zeta^i_n$ must have degree more than $Psi_n$, therefore must be $Psi_n$. By the definition of minimum polynomial, $Psi_n$ is irreducible.



Is this proof correct?






abstract-algebra galois-theory







share|cite|improve this question















Some lemma ahead:




Lemma Let $E/F$ be the field extension. Then all F-automorphisms defines a permutation of roots of any polynomials in $F$






Proof For any root $alpha$ of any polynomial $p(x) in F[x]$, it is straightforward that
$$ p(sigma(alpha)) = sigma(p(alpha)) = 0$$
for any $sigma in text{Gal}(E/F)$.
Therefore $sigma(alpha)$ is indeed a root of $p(x)$.




The proof of irreducibility of $Psi_n(x)$ over $mathbb{Q}[x]$:
$$Psi_n(x) := prod_{(i,n)=1}(x - zeta^i_n)$$
It is proven $Psi_n(x) in mathbb{Z}[x]$.
Consider the group of nth roots of unity, cyclic and isomorphic to $mathbb{Z}_n$.



First note that any $mathbb{Q}$-automorphism must permute the group of nth roots of unity (actually defines an automorphism of that group). It is easy to check any automorphism on $mathbb{Z}_n$ can and can only map a unit to another. Likewise any $mathbb{Q}$-automorphism must map a primitive to another primitive, and we can identify any $mathbb{Q}$-automorphism: $zeta_n mapsto zeta_n^r, (n, r) = 1$ solely by r.



By the lemma, either none of the elements of the primitive nth roots of unity is a root of a polynomial, or all of them. In other words, a minimum polynomial for any $zeta^i_n$ must have degree more than $Psi_n$, therefore must be $Psi_n$. By the definition of minimum polynomial, $Psi_n$ is irreducible.



Is this proof correct?






abstract-algebra galois-theory




abstract-algebra galois-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 10:36







Astrick Harren

















asked Dec 29 '18 at 8:50









Astrick HarrenAstrick Harren

614




614








  • 2




    Until you know the cyclotomic polynomial is urreducible you don't know that every $zeta_n mapsto zeta_n^r, (n, r) = 1$ is an automorphism.
    – ancientmathematician
    Dec 29 '18 at 9:24






  • 2




    What @ancientmathematician says. A census of $Bbb{Q}$-automorphisms is equivalent to irreducibility. Otherwise you can't know that all choices of $r$ are actually automorphisms. For a different way of looking at the weak link in your argument I suggest the following thought experiment. What would change if you look at it over a base field other than $Bbb{Q}$? Remember that over $Bbb{Q}(i)$ the eighth cyclotomic polynomial factors $x^4+1=(x^2+i)(x^2-i)$. And over $Bbb{Q}(sqrt5)$ the fifth cyclotomic polynomial factors. See my comments here
    – Jyrki Lahtonen
    Dec 29 '18 at 9:44










  • @ancientmathematician I still don't get this... Why cannot I state $zeta_n mapsto zeta_n^r, (n, r) = 1$ as an automorphism? Observe the group of nth roots of unity is isomorphic to $Z_n$, therefore every automorphism of $Z_n$ must define a $mathbb{Q}$-automorphism of $mathbb{Q}(zeta_n)$?
    – Astrick Harren
    Dec 29 '18 at 10:31








  • 4




    Group automorphisms are not necessarily field automorphisms. Look at what @JyrkiLahtonen has pointed you to.
    – ancientmathematician
    Dec 29 '18 at 10:34






  • 1




    Your proof does not seem to use the fact that the base field $F$ is $Bbb Q$.
    – Lubin
    Jan 2 at 1:32














  • 2




    Until you know the cyclotomic polynomial is urreducible you don't know that every $zeta_n mapsto zeta_n^r, (n, r) = 1$ is an automorphism.
    – ancientmathematician
    Dec 29 '18 at 9:24






  • 2




    What @ancientmathematician says. A census of $Bbb{Q}$-automorphisms is equivalent to irreducibility. Otherwise you can't know that all choices of $r$ are actually automorphisms. For a different way of looking at the weak link in your argument I suggest the following thought experiment. What would change if you look at it over a base field other than $Bbb{Q}$? Remember that over $Bbb{Q}(i)$ the eighth cyclotomic polynomial factors $x^4+1=(x^2+i)(x^2-i)$. And over $Bbb{Q}(sqrt5)$ the fifth cyclotomic polynomial factors. See my comments here
    – Jyrki Lahtonen
    Dec 29 '18 at 9:44










  • @ancientmathematician I still don't get this... Why cannot I state $zeta_n mapsto zeta_n^r, (n, r) = 1$ as an automorphism? Observe the group of nth roots of unity is isomorphic to $Z_n$, therefore every automorphism of $Z_n$ must define a $mathbb{Q}$-automorphism of $mathbb{Q}(zeta_n)$?
    – Astrick Harren
    Dec 29 '18 at 10:31








  • 4




    Group automorphisms are not necessarily field automorphisms. Look at what @JyrkiLahtonen has pointed you to.
    – ancientmathematician
    Dec 29 '18 at 10:34






  • 1




    Your proof does not seem to use the fact that the base field $F$ is $Bbb Q$.
    – Lubin
    Jan 2 at 1:32








2




2




Until you know the cyclotomic polynomial is urreducible you don't know that every $zeta_n mapsto zeta_n^r, (n, r) = 1$ is an automorphism.
– ancientmathematician
Dec 29 '18 at 9:24




Until you know the cyclotomic polynomial is urreducible you don't know that every $zeta_n mapsto zeta_n^r, (n, r) = 1$ is an automorphism.
– ancientmathematician
Dec 29 '18 at 9:24




2




2




What @ancientmathematician says. A census of $Bbb{Q}$-automorphisms is equivalent to irreducibility. Otherwise you can't know that all choices of $r$ are actually automorphisms. For a different way of looking at the weak link in your argument I suggest the following thought experiment. What would change if you look at it over a base field other than $Bbb{Q}$? Remember that over $Bbb{Q}(i)$ the eighth cyclotomic polynomial factors $x^4+1=(x^2+i)(x^2-i)$. And over $Bbb{Q}(sqrt5)$ the fifth cyclotomic polynomial factors. See my comments here
– Jyrki Lahtonen
Dec 29 '18 at 9:44




What @ancientmathematician says. A census of $Bbb{Q}$-automorphisms is equivalent to irreducibility. Otherwise you can't know that all choices of $r$ are actually automorphisms. For a different way of looking at the weak link in your argument I suggest the following thought experiment. What would change if you look at it over a base field other than $Bbb{Q}$? Remember that over $Bbb{Q}(i)$ the eighth cyclotomic polynomial factors $x^4+1=(x^2+i)(x^2-i)$. And over $Bbb{Q}(sqrt5)$ the fifth cyclotomic polynomial factors. See my comments here
– Jyrki Lahtonen
Dec 29 '18 at 9:44












@ancientmathematician I still don't get this... Why cannot I state $zeta_n mapsto zeta_n^r, (n, r) = 1$ as an automorphism? Observe the group of nth roots of unity is isomorphic to $Z_n$, therefore every automorphism of $Z_n$ must define a $mathbb{Q}$-automorphism of $mathbb{Q}(zeta_n)$?
– Astrick Harren
Dec 29 '18 at 10:31






@ancientmathematician I still don't get this... Why cannot I state $zeta_n mapsto zeta_n^r, (n, r) = 1$ as an automorphism? Observe the group of nth roots of unity is isomorphic to $Z_n$, therefore every automorphism of $Z_n$ must define a $mathbb{Q}$-automorphism of $mathbb{Q}(zeta_n)$?
– Astrick Harren
Dec 29 '18 at 10:31






4




4




Group automorphisms are not necessarily field automorphisms. Look at what @JyrkiLahtonen has pointed you to.
– ancientmathematician
Dec 29 '18 at 10:34




Group automorphisms are not necessarily field automorphisms. Look at what @JyrkiLahtonen has pointed you to.
– ancientmathematician
Dec 29 '18 at 10:34




1




1




Your proof does not seem to use the fact that the base field $F$ is $Bbb Q$.
– Lubin
Jan 2 at 1:32




Your proof does not seem to use the fact that the base field $F$ is $Bbb Q$.
– Lubin
Jan 2 at 1:32










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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055660%2fstrange-proof-on-the-irreducibility-of-psix-the-cyclotomic-polynomial%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
















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055660%2fstrange-proof-on-the-irreducibility-of-psix-the-cyclotomic-polynomial%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Questions related to Moebius Transform of Characteristic Function of the Primes

Image
2 $begingroup$ Consider the function defined in (1) below related to the fundamental prime counting function $pi(x)$ . Note that A143519(n) is not multiplicative. (1) $quad f(x)=sumlimits_{n=1}^{x}A143519(n)$ https://oeis.org/A143519 The following plot illustrates $f(x)$ defined in formula (1) above. Figure (1) : Illustration of $f(x)$ defined in formula (1) The integer zeros of $f(x)$ for $xle 10,000$ are listed in (2) below. (2) $quad$ {1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109} The zero crossings of $f(x)$ for $xle 10,000$ where $f(x)$ doesn't settle at zero are listed in (3) below. (3) $quad$ {14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682} Question (1) : Does $f(x)$ have a finite number of integer zeros, and if so ...
Read more

List of scandals in India

Image
List of scandals in India From Wikipedia, the free encyclopedia Jump to navigation Jump to search This is a dynamic list and may never be able to satisfy particular standards for completeness. You can help by expanding it with reliably sourced entries. This article has multiple issues. Please help improve it or discuss these issues on the talk page . (Learn how and when to remove these template messages) This article has an unclear citation style . The references used may be made clearer with a different or consistent style of citation and footnoting. ( January 2014 ) (Learn how and when to remove this template message) This list is in reverse order. As a consequence it appears to be slanted towards recent events. Please try to keep recent events in historical perspective and add more content related to non-recent events. ( May 2013 ) (Learn how and when to remove this template message) ...
Read more

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

Image
0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
Read more