Polynomials are irreducible in $mathbb{Z}[x]$ if they are irreducible in $mathbb{Z}/pmathbb{Z}$, why p needs...












2












$begingroup$


The proof says we define the modulo function as the natural homomorphism,
$$ barphi: mathbb Z to mathbb Z / pmathbb Z$$
Then if $h in mathbb Z[x]$ is reducible, the identity
$$ barphi(h) = barphi(fg) = barphi(f) barphi(g) $$
proves $h (text{mod} p)$ is also reducible in $mathbb Z_p$



However, this proof tells nothing about $p$ being prime. And consider
$$ barphi: mathbb Z to mathbb Z / 4mathbb Z $$
is also a homomorphism, defined by
$$ z mapsto z + 4mathbb Z$$






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    $2x^2+4$ is irreducible over $mathbb{Z}$, but over $mathbb{Z}_4$ it splits as $(2x)x$.
    $endgroup$
    – Randall
    Dec 10 '18 at 13:47










  • $begingroup$
    You're right, there's no need for $p$ to be prime.
    $endgroup$
    – Stephen
    Dec 10 '18 at 13:56










  • $begingroup$
    @Randall That does not mean the statement is wrong, since from the statement, any polynomial reducible in $mathbb Z_4$ says nothing about its irreducibility in $mathbb Z$
    $endgroup$
    – Astrick Harren
    Dec 10 '18 at 13:57










  • $begingroup$
    @AstrickHarren good point
    $endgroup$
    – Randall
    Dec 10 '18 at 14:05
















2












$begingroup$


The proof says we define the modulo function as the natural homomorphism,
$$ barphi: mathbb Z to mathbb Z / pmathbb Z$$
Then if $h in mathbb Z[x]$ is reducible, the identity
$$ barphi(h) = barphi(fg) = barphi(f) barphi(g) $$
proves $h (text{mod} p)$ is also reducible in $mathbb Z_p$



However, this proof tells nothing about $p$ being prime. And consider
$$ barphi: mathbb Z to mathbb Z / 4mathbb Z $$
is also a homomorphism, defined by
$$ z mapsto z + 4mathbb Z$$






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    $2x^2+4$ is irreducible over $mathbb{Z}$, but over $mathbb{Z}_4$ it splits as $(2x)x$.
    $endgroup$
    – Randall
    Dec 10 '18 at 13:47










  • $begingroup$
    You're right, there's no need for $p$ to be prime.
    $endgroup$
    – Stephen
    Dec 10 '18 at 13:56










  • $begingroup$
    @Randall That does not mean the statement is wrong, since from the statement, any polynomial reducible in $mathbb Z_4$ says nothing about its irreducibility in $mathbb Z$
    $endgroup$
    – Astrick Harren
    Dec 10 '18 at 13:57










  • $begingroup$
    @AstrickHarren good point
    $endgroup$
    – Randall
    Dec 10 '18 at 14:05














2












2








2


0



$begingroup$


The proof says we define the modulo function as the natural homomorphism,
$$ barphi: mathbb Z to mathbb Z / pmathbb Z$$
Then if $h in mathbb Z[x]$ is reducible, the identity
$$ barphi(h) = barphi(fg) = barphi(f) barphi(g) $$
proves $h (text{mod} p)$ is also reducible in $mathbb Z_p$



However, this proof tells nothing about $p$ being prime. And consider
$$ barphi: mathbb Z to mathbb Z / 4mathbb Z $$
is also a homomorphism, defined by
$$ z mapsto z + 4mathbb Z$$






abstract-algebra







share|cite|improve this question











$endgroup$




The proof says we define the modulo function as the natural homomorphism,
$$ barphi: mathbb Z to mathbb Z / pmathbb Z$$
Then if $h in mathbb Z[x]$ is reducible, the identity
$$ barphi(h) = barphi(fg) = barphi(f) barphi(g) $$
proves $h (text{mod} p)$ is also reducible in $mathbb Z_p$



However, this proof tells nothing about $p$ being prime. And consider
$$ barphi: mathbb Z to mathbb Z / 4mathbb Z $$
is also a homomorphism, defined by
$$ z mapsto z + 4mathbb Z$$






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 13:54







Astrick Harren

















asked Dec 10 '18 at 13:42









Astrick HarrenAstrick Harren

614




614












  • $begingroup$
    $2x^2+4$ is irreducible over $mathbb{Z}$, but over $mathbb{Z}_4$ it splits as $(2x)x$.
    $endgroup$
    – Randall
    Dec 10 '18 at 13:47










  • $begingroup$
    You're right, there's no need for $p$ to be prime.
    $endgroup$
    – Stephen
    Dec 10 '18 at 13:56










  • $begingroup$
    @Randall That does not mean the statement is wrong, since from the statement, any polynomial reducible in $mathbb Z_4$ says nothing about its irreducibility in $mathbb Z$
    $endgroup$
    – Astrick Harren
    Dec 10 '18 at 13:57










  • $begingroup$
    @AstrickHarren good point
    $endgroup$
    – Randall
    Dec 10 '18 at 14:05


















  • $begingroup$
    $2x^2+4$ is irreducible over $mathbb{Z}$, but over $mathbb{Z}_4$ it splits as $(2x)x$.
    $endgroup$
    – Randall
    Dec 10 '18 at 13:47










  • $begingroup$
    You're right, there's no need for $p$ to be prime.
    $endgroup$
    – Stephen
    Dec 10 '18 at 13:56










  • $begingroup$
    @Randall That does not mean the statement is wrong, since from the statement, any polynomial reducible in $mathbb Z_4$ says nothing about its irreducibility in $mathbb Z$
    $endgroup$
    – Astrick Harren
    Dec 10 '18 at 13:57










  • $begingroup$
    @AstrickHarren good point
    $endgroup$
    – Randall
    Dec 10 '18 at 14:05
















$begingroup$
$2x^2+4$ is irreducible over $mathbb{Z}$, but over $mathbb{Z}_4$ it splits as $(2x)x$.
$endgroup$
– Randall
Dec 10 '18 at 13:47




$begingroup$
$2x^2+4$ is irreducible over $mathbb{Z}$, but over $mathbb{Z}_4$ it splits as $(2x)x$.
$endgroup$
– Randall
Dec 10 '18 at 13:47












$begingroup$
You're right, there's no need for $p$ to be prime.
$endgroup$
– Stephen
Dec 10 '18 at 13:56




$begingroup$
You're right, there's no need for $p$ to be prime.
$endgroup$
– Stephen
Dec 10 '18 at 13:56












$begingroup$
@Randall That does not mean the statement is wrong, since from the statement, any polynomial reducible in $mathbb Z_4$ says nothing about its irreducibility in $mathbb Z$
$endgroup$
– Astrick Harren
Dec 10 '18 at 13:57




$begingroup$
@Randall That does not mean the statement is wrong, since from the statement, any polynomial reducible in $mathbb Z_4$ says nothing about its irreducibility in $mathbb Z$
$endgroup$
– Astrick Harren
Dec 10 '18 at 13:57












$begingroup$
@AstrickHarren good point
$endgroup$
– Randall
Dec 10 '18 at 14:05




$begingroup$
@AstrickHarren good point
$endgroup$
– Randall
Dec 10 '18 at 14:05










1 Answer
1






active

oldest

votes


















3












$begingroup$

The question of reducibility in $R[X]$ is more complicated when $R$ is not a field, so we usually restrict our attention to fields.



For example, $5x+1=(2x+1)(3x+1)$ is reducible in $mathbb{Z}/6mathbb{Z}$. When $p$ is not prime, reducibility is too common to be interesting.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does factorization in monic polynomials change something ? $mathbb{Z}/(n)[x]$ factorizes in product of $mathbb{Z}/(p_i^{k_i})[x]$. If $h$ is irreducible $bmod p^k$ is it irreducible modulo $q$ for many primes $q$ ?
    $endgroup$
    – reuns
    Dec 10 '18 at 14:02












  • $begingroup$
    @reuns Even irreducibility mod $mathbb{Z}$, which is much stronger, is not sufficient to be irreducible modulo any prime. But I think that no polynomial is irreducible modulo a compositie integer...
    $endgroup$
    – Slade
    Dec 10 '18 at 14:29










  • $begingroup$
    That's why we should restrict to monic factorization (ask that $n$ is coprime with the leading coefficient and look at $f equiv a_d gh bmod n$). If $h$ is irreducible $bmod p$ then it is irreducible modulo $q$ for many primes $q$ by Chebotarev density theorem
    $endgroup$
    – reuns
    Dec 10 '18 at 14:51












  • $begingroup$
    $x^2+1$ is irreducible modulo $4$. I think $f = (x-sqrt{2}-sqrt{3})(x+sqrt{2}-sqrt{3})(x-sqrt{2}+sqrt{3})(x+sqrt{2}+sqrt{3})$ is irreducible in $mathbb{Z}[x]$, reducible modulo every prime, and reducible in $mathbb{Q}_p$ for $p ge 5$ (as $mathbb{Q}_p(sqrt{2},sqrt{3})/mathbb{Q}_p$ is unramified). As $(2 | 3) = -1$, $mathbb{Q}_3(sqrt{2},sqrt{3})/mathbb{Q}_3$ is a tower of a non-trivial unramified and ramified extension, thus $f$ is irreducible in $mathbb{Q}_3$ whence irreducible in $mathbb{Z}/(3^k)$ for some $k$
    $endgroup$
    – reuns
    Dec 10 '18 at 15:17













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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

The question of reducibility in $R[X]$ is more complicated when $R$ is not a field, so we usually restrict our attention to fields.



For example, $5x+1=(2x+1)(3x+1)$ is reducible in $mathbb{Z}/6mathbb{Z}$. When $p$ is not prime, reducibility is too common to be interesting.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does factorization in monic polynomials change something ? $mathbb{Z}/(n)[x]$ factorizes in product of $mathbb{Z}/(p_i^{k_i})[x]$. If $h$ is irreducible $bmod p^k$ is it irreducible modulo $q$ for many primes $q$ ?
    $endgroup$
    – reuns
    Dec 10 '18 at 14:02












  • $begingroup$
    @reuns Even irreducibility mod $mathbb{Z}$, which is much stronger, is not sufficient to be irreducible modulo any prime. But I think that no polynomial is irreducible modulo a compositie integer...
    $endgroup$
    – Slade
    Dec 10 '18 at 14:29










  • $begingroup$
    That's why we should restrict to monic factorization (ask that $n$ is coprime with the leading coefficient and look at $f equiv a_d gh bmod n$). If $h$ is irreducible $bmod p$ then it is irreducible modulo $q$ for many primes $q$ by Chebotarev density theorem
    $endgroup$
    – reuns
    Dec 10 '18 at 14:51












  • $begingroup$
    $x^2+1$ is irreducible modulo $4$. I think $f = (x-sqrt{2}-sqrt{3})(x+sqrt{2}-sqrt{3})(x-sqrt{2}+sqrt{3})(x+sqrt{2}+sqrt{3})$ is irreducible in $mathbb{Z}[x]$, reducible modulo every prime, and reducible in $mathbb{Q}_p$ for $p ge 5$ (as $mathbb{Q}_p(sqrt{2},sqrt{3})/mathbb{Q}_p$ is unramified). As $(2 | 3) = -1$, $mathbb{Q}_3(sqrt{2},sqrt{3})/mathbb{Q}_3$ is a tower of a non-trivial unramified and ramified extension, thus $f$ is irreducible in $mathbb{Q}_3$ whence irreducible in $mathbb{Z}/(3^k)$ for some $k$
    $endgroup$
    – reuns
    Dec 10 '18 at 15:17


















3












$begingroup$

The question of reducibility in $R[X]$ is more complicated when $R$ is not a field, so we usually restrict our attention to fields.



For example, $5x+1=(2x+1)(3x+1)$ is reducible in $mathbb{Z}/6mathbb{Z}$. When $p$ is not prime, reducibility is too common to be interesting.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does factorization in monic polynomials change something ? $mathbb{Z}/(n)[x]$ factorizes in product of $mathbb{Z}/(p_i^{k_i})[x]$. If $h$ is irreducible $bmod p^k$ is it irreducible modulo $q$ for many primes $q$ ?
    $endgroup$
    – reuns
    Dec 10 '18 at 14:02












  • $begingroup$
    @reuns Even irreducibility mod $mathbb{Z}$, which is much stronger, is not sufficient to be irreducible modulo any prime. But I think that no polynomial is irreducible modulo a compositie integer...
    $endgroup$
    – Slade
    Dec 10 '18 at 14:29










  • $begingroup$
    That's why we should restrict to monic factorization (ask that $n$ is coprime with the leading coefficient and look at $f equiv a_d gh bmod n$). If $h$ is irreducible $bmod p$ then it is irreducible modulo $q$ for many primes $q$ by Chebotarev density theorem
    $endgroup$
    – reuns
    Dec 10 '18 at 14:51












  • $begingroup$
    $x^2+1$ is irreducible modulo $4$. I think $f = (x-sqrt{2}-sqrt{3})(x+sqrt{2}-sqrt{3})(x-sqrt{2}+sqrt{3})(x+sqrt{2}+sqrt{3})$ is irreducible in $mathbb{Z}[x]$, reducible modulo every prime, and reducible in $mathbb{Q}_p$ for $p ge 5$ (as $mathbb{Q}_p(sqrt{2},sqrt{3})/mathbb{Q}_p$ is unramified). As $(2 | 3) = -1$, $mathbb{Q}_3(sqrt{2},sqrt{3})/mathbb{Q}_3$ is a tower of a non-trivial unramified and ramified extension, thus $f$ is irreducible in $mathbb{Q}_3$ whence irreducible in $mathbb{Z}/(3^k)$ for some $k$
    $endgroup$
    – reuns
    Dec 10 '18 at 15:17
















3












3








3





$begingroup$

The question of reducibility in $R[X]$ is more complicated when $R$ is not a field, so we usually restrict our attention to fields.



For example, $5x+1=(2x+1)(3x+1)$ is reducible in $mathbb{Z}/6mathbb{Z}$. When $p$ is not prime, reducibility is too common to be interesting.






share|cite|improve this answer









$endgroup$



The question of reducibility in $R[X]$ is more complicated when $R$ is not a field, so we usually restrict our attention to fields.



For example, $5x+1=(2x+1)(3x+1)$ is reducible in $mathbb{Z}/6mathbb{Z}$. When $p$ is not prime, reducibility is too common to be interesting.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 13:57









SladeSlade

25k12665




25k12665












  • $begingroup$
    Does factorization in monic polynomials change something ? $mathbb{Z}/(n)[x]$ factorizes in product of $mathbb{Z}/(p_i^{k_i})[x]$. If $h$ is irreducible $bmod p^k$ is it irreducible modulo $q$ for many primes $q$ ?
    $endgroup$
    – reuns
    Dec 10 '18 at 14:02












  • $begingroup$
    @reuns Even irreducibility mod $mathbb{Z}$, which is much stronger, is not sufficient to be irreducible modulo any prime. But I think that no polynomial is irreducible modulo a compositie integer...
    $endgroup$
    – Slade
    Dec 10 '18 at 14:29










  • $begingroup$
    That's why we should restrict to monic factorization (ask that $n$ is coprime with the leading coefficient and look at $f equiv a_d gh bmod n$). If $h$ is irreducible $bmod p$ then it is irreducible modulo $q$ for many primes $q$ by Chebotarev density theorem
    $endgroup$
    – reuns
    Dec 10 '18 at 14:51












  • $begingroup$
    $x^2+1$ is irreducible modulo $4$. I think $f = (x-sqrt{2}-sqrt{3})(x+sqrt{2}-sqrt{3})(x-sqrt{2}+sqrt{3})(x+sqrt{2}+sqrt{3})$ is irreducible in $mathbb{Z}[x]$, reducible modulo every prime, and reducible in $mathbb{Q}_p$ for $p ge 5$ (as $mathbb{Q}_p(sqrt{2},sqrt{3})/mathbb{Q}_p$ is unramified). As $(2 | 3) = -1$, $mathbb{Q}_3(sqrt{2},sqrt{3})/mathbb{Q}_3$ is a tower of a non-trivial unramified and ramified extension, thus $f$ is irreducible in $mathbb{Q}_3$ whence irreducible in $mathbb{Z}/(3^k)$ for some $k$
    $endgroup$
    – reuns
    Dec 10 '18 at 15:17




















  • $begingroup$
    Does factorization in monic polynomials change something ? $mathbb{Z}/(n)[x]$ factorizes in product of $mathbb{Z}/(p_i^{k_i})[x]$. If $h$ is irreducible $bmod p^k$ is it irreducible modulo $q$ for many primes $q$ ?
    $endgroup$
    – reuns
    Dec 10 '18 at 14:02












  • $begingroup$
    @reuns Even irreducibility mod $mathbb{Z}$, which is much stronger, is not sufficient to be irreducible modulo any prime. But I think that no polynomial is irreducible modulo a compositie integer...
    $endgroup$
    – Slade
    Dec 10 '18 at 14:29










  • $begingroup$
    That's why we should restrict to monic factorization (ask that $n$ is coprime with the leading coefficient and look at $f equiv a_d gh bmod n$). If $h$ is irreducible $bmod p$ then it is irreducible modulo $q$ for many primes $q$ by Chebotarev density theorem
    $endgroup$
    – reuns
    Dec 10 '18 at 14:51












  • $begingroup$
    $x^2+1$ is irreducible modulo $4$. I think $f = (x-sqrt{2}-sqrt{3})(x+sqrt{2}-sqrt{3})(x-sqrt{2}+sqrt{3})(x+sqrt{2}+sqrt{3})$ is irreducible in $mathbb{Z}[x]$, reducible modulo every prime, and reducible in $mathbb{Q}_p$ for $p ge 5$ (as $mathbb{Q}_p(sqrt{2},sqrt{3})/mathbb{Q}_p$ is unramified). As $(2 | 3) = -1$, $mathbb{Q}_3(sqrt{2},sqrt{3})/mathbb{Q}_3$ is a tower of a non-trivial unramified and ramified extension, thus $f$ is irreducible in $mathbb{Q}_3$ whence irreducible in $mathbb{Z}/(3^k)$ for some $k$
    $endgroup$
    – reuns
    Dec 10 '18 at 15:17


















$begingroup$
Does factorization in monic polynomials change something ? $mathbb{Z}/(n)[x]$ factorizes in product of $mathbb{Z}/(p_i^{k_i})[x]$. If $h$ is irreducible $bmod p^k$ is it irreducible modulo $q$ for many primes $q$ ?
$endgroup$
– reuns
Dec 10 '18 at 14:02






$begingroup$
Does factorization in monic polynomials change something ? $mathbb{Z}/(n)[x]$ factorizes in product of $mathbb{Z}/(p_i^{k_i})[x]$. If $h$ is irreducible $bmod p^k$ is it irreducible modulo $q$ for many primes $q$ ?
$endgroup$
– reuns
Dec 10 '18 at 14:02














$begingroup$
@reuns Even irreducibility mod $mathbb{Z}$, which is much stronger, is not sufficient to be irreducible modulo any prime. But I think that no polynomial is irreducible modulo a compositie integer...
$endgroup$
– Slade
Dec 10 '18 at 14:29




$begingroup$
@reuns Even irreducibility mod $mathbb{Z}$, which is much stronger, is not sufficient to be irreducible modulo any prime. But I think that no polynomial is irreducible modulo a compositie integer...
$endgroup$
– Slade
Dec 10 '18 at 14:29












$begingroup$
That's why we should restrict to monic factorization (ask that $n$ is coprime with the leading coefficient and look at $f equiv a_d gh bmod n$). If $h$ is irreducible $bmod p$ then it is irreducible modulo $q$ for many primes $q$ by Chebotarev density theorem
$endgroup$
– reuns
Dec 10 '18 at 14:51






$begingroup$
That's why we should restrict to monic factorization (ask that $n$ is coprime with the leading coefficient and look at $f equiv a_d gh bmod n$). If $h$ is irreducible $bmod p$ then it is irreducible modulo $q$ for many primes $q$ by Chebotarev density theorem
$endgroup$
– reuns
Dec 10 '18 at 14:51














$begingroup$
$x^2+1$ is irreducible modulo $4$. I think $f = (x-sqrt{2}-sqrt{3})(x+sqrt{2}-sqrt{3})(x-sqrt{2}+sqrt{3})(x+sqrt{2}+sqrt{3})$ is irreducible in $mathbb{Z}[x]$, reducible modulo every prime, and reducible in $mathbb{Q}_p$ for $p ge 5$ (as $mathbb{Q}_p(sqrt{2},sqrt{3})/mathbb{Q}_p$ is unramified). As $(2 | 3) = -1$, $mathbb{Q}_3(sqrt{2},sqrt{3})/mathbb{Q}_3$ is a tower of a non-trivial unramified and ramified extension, thus $f$ is irreducible in $mathbb{Q}_3$ whence irreducible in $mathbb{Z}/(3^k)$ for some $k$
$endgroup$
– reuns
Dec 10 '18 at 15:17






$begingroup$
$x^2+1$ is irreducible modulo $4$. I think $f = (x-sqrt{2}-sqrt{3})(x+sqrt{2}-sqrt{3})(x-sqrt{2}+sqrt{3})(x+sqrt{2}+sqrt{3})$ is irreducible in $mathbb{Z}[x]$, reducible modulo every prime, and reducible in $mathbb{Q}_p$ for $p ge 5$ (as $mathbb{Q}_p(sqrt{2},sqrt{3})/mathbb{Q}_p$ is unramified). As $(2 | 3) = -1$, $mathbb{Q}_3(sqrt{2},sqrt{3})/mathbb{Q}_3$ is a tower of a non-trivial unramified and ramified extension, thus $f$ is irreducible in $mathbb{Q}_3$ whence irreducible in $mathbb{Z}/(3^k)$ for some $k$
$endgroup$
– reuns
Dec 10 '18 at 15:17




















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Sphinx de Gizeh

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Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
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Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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