On the irreducibility of a polynomial and Gauss lemma












0












$begingroup$


Let $P (X) = X^5 − 6X + 3$.



Prove that it is irreducible over $mathbb{Q}$.



In the solution I have for this exercice, I have litterally:



P is irreducible over $mathbb{F}_5$, therefore, by Gauss Lemma, it is irreducible over $mathbb{Q}$.



I am lacking the necessary knowledge about irreducibility by reducing modulo p. All I know is that there is a bijection between roots of P in $overline{mathbb{Q}}$ and the roots of $overline{P}$ in $overline{mathbb{F}_p}$.



I also know this generalization of Gauss Lemma: in a factorial ring A, with its fraction field K, a primitive $P$ is irreducible in $A[X] Leftrightarrow $ $P$ is irreducible in K[X].



How does this apply to P above?



Thank you for any directions or help.






abstract-algebra







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Because $P$ is irreducible over $mathbb{Z}_5 Rightarrow P$ is irreducible over $mathbb{Z}$. And then $P$ is irreducible over $mathbb{Q}$ by Gauss's Lemma.
    $endgroup$
    – mouthetics
    Dec 23 '18 at 3:50










  • $begingroup$
    Thank you! is this a general fact? could please give some link to the litterature or a proof that P is irreducible over $mathbb{F}_pRightarrow$ P is irreducible over $mathbb{Z}$?
    $endgroup$
    – PerelMan
    Dec 23 '18 at 4:17








  • 2




    $begingroup$
    @PerelMan A factorisation over $Bbb Z$ induces one over $Bbb F_p$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 23 '18 at 5:19






  • 2




    $begingroup$
    PerelMan, I try and describe the use of modular reduction in irreducibility proofs here. Also observe that Eisenstein's criterion with $p=3$ works here as well (and is the go-to technique for many).
    $endgroup$
    – Jyrki Lahtonen
    Dec 23 '18 at 7:57
















0












$begingroup$


Let $P (X) = X^5 − 6X + 3$.



Prove that it is irreducible over $mathbb{Q}$.



In the solution I have for this exercice, I have litterally:



P is irreducible over $mathbb{F}_5$, therefore, by Gauss Lemma, it is irreducible over $mathbb{Q}$.



I am lacking the necessary knowledge about irreducibility by reducing modulo p. All I know is that there is a bijection between roots of P in $overline{mathbb{Q}}$ and the roots of $overline{P}$ in $overline{mathbb{F}_p}$.



I also know this generalization of Gauss Lemma: in a factorial ring A, with its fraction field K, a primitive $P$ is irreducible in $A[X] Leftrightarrow $ $P$ is irreducible in K[X].



How does this apply to P above?



Thank you for any directions or help.






abstract-algebra







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Because $P$ is irreducible over $mathbb{Z}_5 Rightarrow P$ is irreducible over $mathbb{Z}$. And then $P$ is irreducible over $mathbb{Q}$ by Gauss's Lemma.
    $endgroup$
    – mouthetics
    Dec 23 '18 at 3:50










  • $begingroup$
    Thank you! is this a general fact? could please give some link to the litterature or a proof that P is irreducible over $mathbb{F}_pRightarrow$ P is irreducible over $mathbb{Z}$?
    $endgroup$
    – PerelMan
    Dec 23 '18 at 4:17








  • 2




    $begingroup$
    @PerelMan A factorisation over $Bbb Z$ induces one over $Bbb F_p$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 23 '18 at 5:19






  • 2




    $begingroup$
    PerelMan, I try and describe the use of modular reduction in irreducibility proofs here. Also observe that Eisenstein's criterion with $p=3$ works here as well (and is the go-to technique for many).
    $endgroup$
    – Jyrki Lahtonen
    Dec 23 '18 at 7:57














0












0








0





$begingroup$


Let $P (X) = X^5 − 6X + 3$.



Prove that it is irreducible over $mathbb{Q}$.



In the solution I have for this exercice, I have litterally:



P is irreducible over $mathbb{F}_5$, therefore, by Gauss Lemma, it is irreducible over $mathbb{Q}$.



I am lacking the necessary knowledge about irreducibility by reducing modulo p. All I know is that there is a bijection between roots of P in $overline{mathbb{Q}}$ and the roots of $overline{P}$ in $overline{mathbb{F}_p}$.



I also know this generalization of Gauss Lemma: in a factorial ring A, with its fraction field K, a primitive $P$ is irreducible in $A[X] Leftrightarrow $ $P$ is irreducible in K[X].



How does this apply to P above?



Thank you for any directions or help.






abstract-algebra







share|cite|improve this question









$endgroup$




Let $P (X) = X^5 − 6X + 3$.



Prove that it is irreducible over $mathbb{Q}$.



In the solution I have for this exercice, I have litterally:



P is irreducible over $mathbb{F}_5$, therefore, by Gauss Lemma, it is irreducible over $mathbb{Q}$.



I am lacking the necessary knowledge about irreducibility by reducing modulo p. All I know is that there is a bijection between roots of P in $overline{mathbb{Q}}$ and the roots of $overline{P}$ in $overline{mathbb{F}_p}$.



I also know this generalization of Gauss Lemma: in a factorial ring A, with its fraction field K, a primitive $P$ is irreducible in $A[X] Leftrightarrow $ $P$ is irreducible in K[X].



How does this apply to P above?



Thank you for any directions or help.






abstract-algebra




abstract-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 23 '18 at 3:40









PerelManPerelMan

538311




538311








  • 3




    $begingroup$
    Because $P$ is irreducible over $mathbb{Z}_5 Rightarrow P$ is irreducible over $mathbb{Z}$. And then $P$ is irreducible over $mathbb{Q}$ by Gauss's Lemma.
    $endgroup$
    – mouthetics
    Dec 23 '18 at 3:50










  • $begingroup$
    Thank you! is this a general fact? could please give some link to the litterature or a proof that P is irreducible over $mathbb{F}_pRightarrow$ P is irreducible over $mathbb{Z}$?
    $endgroup$
    – PerelMan
    Dec 23 '18 at 4:17








  • 2




    $begingroup$
    @PerelMan A factorisation over $Bbb Z$ induces one over $Bbb F_p$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 23 '18 at 5:19






  • 2




    $begingroup$
    PerelMan, I try and describe the use of modular reduction in irreducibility proofs here. Also observe that Eisenstein's criterion with $p=3$ works here as well (and is the go-to technique for many).
    $endgroup$
    – Jyrki Lahtonen
    Dec 23 '18 at 7:57














  • 3




    $begingroup$
    Because $P$ is irreducible over $mathbb{Z}_5 Rightarrow P$ is irreducible over $mathbb{Z}$. And then $P$ is irreducible over $mathbb{Q}$ by Gauss's Lemma.
    $endgroup$
    – mouthetics
    Dec 23 '18 at 3:50










  • $begingroup$
    Thank you! is this a general fact? could please give some link to the litterature or a proof that P is irreducible over $mathbb{F}_pRightarrow$ P is irreducible over $mathbb{Z}$?
    $endgroup$
    – PerelMan
    Dec 23 '18 at 4:17








  • 2




    $begingroup$
    @PerelMan A factorisation over $Bbb Z$ induces one over $Bbb F_p$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 23 '18 at 5:19






  • 2




    $begingroup$
    PerelMan, I try and describe the use of modular reduction in irreducibility proofs here. Also observe that Eisenstein's criterion with $p=3$ works here as well (and is the go-to technique for many).
    $endgroup$
    – Jyrki Lahtonen
    Dec 23 '18 at 7:57








3




3




$begingroup$
Because $P$ is irreducible over $mathbb{Z}_5 Rightarrow P$ is irreducible over $mathbb{Z}$. And then $P$ is irreducible over $mathbb{Q}$ by Gauss's Lemma.
$endgroup$
– mouthetics
Dec 23 '18 at 3:50




$begingroup$
Because $P$ is irreducible over $mathbb{Z}_5 Rightarrow P$ is irreducible over $mathbb{Z}$. And then $P$ is irreducible over $mathbb{Q}$ by Gauss's Lemma.
$endgroup$
– mouthetics
Dec 23 '18 at 3:50












$begingroup$
Thank you! is this a general fact? could please give some link to the litterature or a proof that P is irreducible over $mathbb{F}_pRightarrow$ P is irreducible over $mathbb{Z}$?
$endgroup$
– PerelMan
Dec 23 '18 at 4:17






$begingroup$
Thank you! is this a general fact? could please give some link to the litterature or a proof that P is irreducible over $mathbb{F}_pRightarrow$ P is irreducible over $mathbb{Z}$?
$endgroup$
– PerelMan
Dec 23 '18 at 4:17






2




2




$begingroup$
@PerelMan A factorisation over $Bbb Z$ induces one over $Bbb F_p$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:19




$begingroup$
@PerelMan A factorisation over $Bbb Z$ induces one over $Bbb F_p$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:19




2




2




$begingroup$
PerelMan, I try and describe the use of modular reduction in irreducibility proofs here. Also observe that Eisenstein's criterion with $p=3$ works here as well (and is the go-to technique for many).
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 7:57




$begingroup$
PerelMan, I try and describe the use of modular reduction in irreducibility proofs here. Also observe that Eisenstein's criterion with $p=3$ works here as well (and is the go-to technique for many).
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 7:57










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There is a general result: Let $R$ be a ring and $I$ be a proper ideal. Fix a nonconstant monic polynomial $p(x) in R[x]$. If $overline{p(x)} in (R/I)[x]$ is irreducible, then $p(x)$ is irreducible in $R[x]$.



You can prove this by considering the contrapositive.






share|cite|improve this answer









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    $begingroup$

    There is a general result: Let $R$ be a ring and $I$ be a proper ideal. Fix a nonconstant monic polynomial $p(x) in R[x]$. If $overline{p(x)} in (R/I)[x]$ is irreducible, then $p(x)$ is irreducible in $R[x]$.



    You can prove this by considering the contrapositive.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      There is a general result: Let $R$ be a ring and $I$ be a proper ideal. Fix a nonconstant monic polynomial $p(x) in R[x]$. If $overline{p(x)} in (R/I)[x]$ is irreducible, then $p(x)$ is irreducible in $R[x]$.



      You can prove this by considering the contrapositive.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        There is a general result: Let $R$ be a ring and $I$ be a proper ideal. Fix a nonconstant monic polynomial $p(x) in R[x]$. If $overline{p(x)} in (R/I)[x]$ is irreducible, then $p(x)$ is irreducible in $R[x]$.



        You can prove this by considering the contrapositive.






        share|cite|improve this answer









        $endgroup$



        There is a general result: Let $R$ be a ring and $I$ be a proper ideal. Fix a nonconstant monic polynomial $p(x) in R[x]$. If $overline{p(x)} in (R/I)[x]$ is irreducible, then $p(x)$ is irreducible in $R[x]$.



        You can prove this by considering the contrapositive.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 23 '18 at 5:18









        Aniruddh AgarwalAniruddh Agarwal

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