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

25.1k12665




25.1k12665












  • $begingroup$
    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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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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