Irreducible polynomial over $mathbb{Q}[x]$












0












$begingroup$


Question:



Check if:
$$
f(x) = x^4 + 4x^3 + 6x^2 + 2x + 1
$$
is reducible or irreducible over $mathbb{Q}[x]$




My Answer [Edited]:



Suppose that $f(x)$ is irreducible over $Z_p[x]$ for a prime $p$, therefore it's going to be irreducible over $mathbb{Q}[x]$. Let $p=2$. Therefore:
$$
f_2(x) = f(x) mod{2} = x^4 + 1 = a_2(x)cdot b_2(x) in mathbb{Z}_2[x]
$$
Since $x^4+1$ have roots in $mathbb{Z}_2[x]$, there are two cases to check... First let's suppose that $a_2(x),b_2(x)$ have degree equal to two, therefore:
begin{align*}
x^4+1&=(a_2x^2+a_1x+a_0)cdot(b_2x^2+b_1x+b_0)=\
&=(a_2b_2)x^4+(a_2b_1+a_1b_2)x^3+(a_2b_0+a_1b_1+a_0b_2)x^2+(a_1b_0+a_0b_1)x^+(a_0b_0)
end{align*}
Since $a_i,b_i in mathbb{Z}_2 rightarrow a_i,b_iin {0,1}$, it follows that:
$$
a_0b_0=1 longrightarrow a_0=b_0=1\
a_1+b_1=0 longrightarrow a_1=b_1=0\
a_2b_2=1 longrightarrow a_2=b_2=1
$$
Hence:
begin{align*}
a_2(x)&=x^2+1\
b_2(x)&=x^2+1\
a_2(x)cdot b_2(x) &= (x^2+1)^2 = x^4 + 2x + 1 overbrace{longrightarrow}^{in mathbb{Z}_2} x^4 + 1
end{align*}



Now let's check WLOG the case for $a_2(x)$ having degree equal to $3$ and $b_2(x)$ having degree equal to $1$:



begin{align*}
x^4+1 &= a_2(x)cdot b_2(x) = (a_3x^3+a_2x^2+a_1x+a_0)cdot(b_1x+b_0)\
&= (b_1a_3)x^4 + (b_1a_2+b_0a_3)x^3 + (b_1a_1+b_0a_2)x^2 + (b_1a_0+b_0a_1)x + (b_0a_0)
end{align*}



By the same reasoning as before:
begin{align*}
b_0a_0=1 longrightarrow b_0=a_0=1\
b_1a_3=1 longrightarrow b_1=a_3=1\
a_2+b_0 = 0 longrightarrow a_2=b_0=0\
a_1+a_2 = 0 longrightarrow a_1=a_2=0\
end{align*}



Conclusion, it is reducible over $mathbb{Z}_2[x]$ and therefore I can't conclude anything.








polynomials ring-theory field-theory irreducible-polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    Seems to me you have shown that $f(x)$ is reducible over $Bbb Z_2[x]$, not whether or not it's reducible over $Bbb Q[x]$.
    $endgroup$
    – Rolf Hoyer
    Dec 6 '18 at 3:34










  • $begingroup$
    @RolfHoyer Yeah, I was checking what I've done... I just know that if a monic polynomial is reducible over $mathbb{Q}[x]$, then it is going to be reducible over $mathbb{Z}_2[x]$... But if that polynomial is irreducible, I can't assume the same... How you would approach that question?
    $endgroup$
    – Bruno Reis
    Dec 6 '18 at 3:37










  • $begingroup$
    Why doesn't $x^4+1$ have roots mod $2$?
    $endgroup$
    – Michael Burr
    Dec 6 '18 at 3:38










  • $begingroup$
    @MichaelBurr in mod 2, $x in {0,1}$ and from that we have $f(0) = 1$ and $f(1)=2=0$... Yeah, you are correct... We have roots...
    $endgroup$
    – Bruno Reis
    Dec 6 '18 at 3:39






  • 1




    $begingroup$
    @AnikBhowmick Shifting is an invertable operation, so irreduciblity must be preserved. In this case, factoring commutes with shifting.
    $endgroup$
    – Michael Burr
    Dec 6 '18 at 3:45


















0












$begingroup$


Question:



Check if:
$$
f(x) = x^4 + 4x^3 + 6x^2 + 2x + 1
$$
is reducible or irreducible over $mathbb{Q}[x]$




My Answer [Edited]:



Suppose that $f(x)$ is irreducible over $Z_p[x]$ for a prime $p$, therefore it's going to be irreducible over $mathbb{Q}[x]$. Let $p=2$. Therefore:
$$
f_2(x) = f(x) mod{2} = x^4 + 1 = a_2(x)cdot b_2(x) in mathbb{Z}_2[x]
$$
Since $x^4+1$ have roots in $mathbb{Z}_2[x]$, there are two cases to check... First let's suppose that $a_2(x),b_2(x)$ have degree equal to two, therefore:
begin{align*}
x^4+1&=(a_2x^2+a_1x+a_0)cdot(b_2x^2+b_1x+b_0)=\
&=(a_2b_2)x^4+(a_2b_1+a_1b_2)x^3+(a_2b_0+a_1b_1+a_0b_2)x^2+(a_1b_0+a_0b_1)x^+(a_0b_0)
end{align*}
Since $a_i,b_i in mathbb{Z}_2 rightarrow a_i,b_iin {0,1}$, it follows that:
$$
a_0b_0=1 longrightarrow a_0=b_0=1\
a_1+b_1=0 longrightarrow a_1=b_1=0\
a_2b_2=1 longrightarrow a_2=b_2=1
$$
Hence:
begin{align*}
a_2(x)&=x^2+1\
b_2(x)&=x^2+1\
a_2(x)cdot b_2(x) &= (x^2+1)^2 = x^4 + 2x + 1 overbrace{longrightarrow}^{in mathbb{Z}_2} x^4 + 1
end{align*}



Now let's check WLOG the case for $a_2(x)$ having degree equal to $3$ and $b_2(x)$ having degree equal to $1$:



begin{align*}
x^4+1 &= a_2(x)cdot b_2(x) = (a_3x^3+a_2x^2+a_1x+a_0)cdot(b_1x+b_0)\
&= (b_1a_3)x^4 + (b_1a_2+b_0a_3)x^3 + (b_1a_1+b_0a_2)x^2 + (b_1a_0+b_0a_1)x + (b_0a_0)
end{align*}



By the same reasoning as before:
begin{align*}
b_0a_0=1 longrightarrow b_0=a_0=1\
b_1a_3=1 longrightarrow b_1=a_3=1\
a_2+b_0 = 0 longrightarrow a_2=b_0=0\
a_1+a_2 = 0 longrightarrow a_1=a_2=0\
end{align*}



Conclusion, it is reducible over $mathbb{Z}_2[x]$ and therefore I can't conclude anything.








polynomials ring-theory field-theory irreducible-polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    Seems to me you have shown that $f(x)$ is reducible over $Bbb Z_2[x]$, not whether or not it's reducible over $Bbb Q[x]$.
    $endgroup$
    – Rolf Hoyer
    Dec 6 '18 at 3:34










  • $begingroup$
    @RolfHoyer Yeah, I was checking what I've done... I just know that if a monic polynomial is reducible over $mathbb{Q}[x]$, then it is going to be reducible over $mathbb{Z}_2[x]$... But if that polynomial is irreducible, I can't assume the same... How you would approach that question?
    $endgroup$
    – Bruno Reis
    Dec 6 '18 at 3:37










  • $begingroup$
    Why doesn't $x^4+1$ have roots mod $2$?
    $endgroup$
    – Michael Burr
    Dec 6 '18 at 3:38










  • $begingroup$
    @MichaelBurr in mod 2, $x in {0,1}$ and from that we have $f(0) = 1$ and $f(1)=2=0$... Yeah, you are correct... We have roots...
    $endgroup$
    – Bruno Reis
    Dec 6 '18 at 3:39






  • 1




    $begingroup$
    @AnikBhowmick Shifting is an invertable operation, so irreduciblity must be preserved. In this case, factoring commutes with shifting.
    $endgroup$
    – Michael Burr
    Dec 6 '18 at 3:45
















0












0








0





$begingroup$


Question:



Check if:
$$
f(x) = x^4 + 4x^3 + 6x^2 + 2x + 1
$$
is reducible or irreducible over $mathbb{Q}[x]$




My Answer [Edited]:



Suppose that $f(x)$ is irreducible over $Z_p[x]$ for a prime $p$, therefore it's going to be irreducible over $mathbb{Q}[x]$. Let $p=2$. Therefore:
$$
f_2(x) = f(x) mod{2} = x^4 + 1 = a_2(x)cdot b_2(x) in mathbb{Z}_2[x]
$$
Since $x^4+1$ have roots in $mathbb{Z}_2[x]$, there are two cases to check... First let's suppose that $a_2(x),b_2(x)$ have degree equal to two, therefore:
begin{align*}
x^4+1&=(a_2x^2+a_1x+a_0)cdot(b_2x^2+b_1x+b_0)=\
&=(a_2b_2)x^4+(a_2b_1+a_1b_2)x^3+(a_2b_0+a_1b_1+a_0b_2)x^2+(a_1b_0+a_0b_1)x^+(a_0b_0)
end{align*}
Since $a_i,b_i in mathbb{Z}_2 rightarrow a_i,b_iin {0,1}$, it follows that:
$$
a_0b_0=1 longrightarrow a_0=b_0=1\
a_1+b_1=0 longrightarrow a_1=b_1=0\
a_2b_2=1 longrightarrow a_2=b_2=1
$$
Hence:
begin{align*}
a_2(x)&=x^2+1\
b_2(x)&=x^2+1\
a_2(x)cdot b_2(x) &= (x^2+1)^2 = x^4 + 2x + 1 overbrace{longrightarrow}^{in mathbb{Z}_2} x^4 + 1
end{align*}



Now let's check WLOG the case for $a_2(x)$ having degree equal to $3$ and $b_2(x)$ having degree equal to $1$:



begin{align*}
x^4+1 &= a_2(x)cdot b_2(x) = (a_3x^3+a_2x^2+a_1x+a_0)cdot(b_1x+b_0)\
&= (b_1a_3)x^4 + (b_1a_2+b_0a_3)x^3 + (b_1a_1+b_0a_2)x^2 + (b_1a_0+b_0a_1)x + (b_0a_0)
end{align*}



By the same reasoning as before:
begin{align*}
b_0a_0=1 longrightarrow b_0=a_0=1\
b_1a_3=1 longrightarrow b_1=a_3=1\
a_2+b_0 = 0 longrightarrow a_2=b_0=0\
a_1+a_2 = 0 longrightarrow a_1=a_2=0\
end{align*}



Conclusion, it is reducible over $mathbb{Z}_2[x]$ and therefore I can't conclude anything.








polynomials ring-theory field-theory irreducible-polynomials







share|cite|improve this question











$endgroup$




Question:



Check if:
$$
f(x) = x^4 + 4x^3 + 6x^2 + 2x + 1
$$
is reducible or irreducible over $mathbb{Q}[x]$




My Answer [Edited]:



Suppose that $f(x)$ is irreducible over $Z_p[x]$ for a prime $p$, therefore it's going to be irreducible over $mathbb{Q}[x]$. Let $p=2$. Therefore:
$$
f_2(x) = f(x) mod{2} = x^4 + 1 = a_2(x)cdot b_2(x) in mathbb{Z}_2[x]
$$
Since $x^4+1$ have roots in $mathbb{Z}_2[x]$, there are two cases to check... First let's suppose that $a_2(x),b_2(x)$ have degree equal to two, therefore:
begin{align*}
x^4+1&=(a_2x^2+a_1x+a_0)cdot(b_2x^2+b_1x+b_0)=\
&=(a_2b_2)x^4+(a_2b_1+a_1b_2)x^3+(a_2b_0+a_1b_1+a_0b_2)x^2+(a_1b_0+a_0b_1)x^+(a_0b_0)
end{align*}
Since $a_i,b_i in mathbb{Z}_2 rightarrow a_i,b_iin {0,1}$, it follows that:
$$
a_0b_0=1 longrightarrow a_0=b_0=1\
a_1+b_1=0 longrightarrow a_1=b_1=0\
a_2b_2=1 longrightarrow a_2=b_2=1
$$
Hence:
begin{align*}
a_2(x)&=x^2+1\
b_2(x)&=x^2+1\
a_2(x)cdot b_2(x) &= (x^2+1)^2 = x^4 + 2x + 1 overbrace{longrightarrow}^{in mathbb{Z}_2} x^4 + 1
end{align*}



Now let's check WLOG the case for $a_2(x)$ having degree equal to $3$ and $b_2(x)$ having degree equal to $1$:



begin{align*}
x^4+1 &= a_2(x)cdot b_2(x) = (a_3x^3+a_2x^2+a_1x+a_0)cdot(b_1x+b_0)\
&= (b_1a_3)x^4 + (b_1a_2+b_0a_3)x^3 + (b_1a_1+b_0a_2)x^2 + (b_1a_0+b_0a_1)x + (b_0a_0)
end{align*}



By the same reasoning as before:
begin{align*}
b_0a_0=1 longrightarrow b_0=a_0=1\
b_1a_3=1 longrightarrow b_1=a_3=1\
a_2+b_0 = 0 longrightarrow a_2=b_0=0\
a_1+a_2 = 0 longrightarrow a_1=a_2=0\
end{align*}



Conclusion, it is reducible over $mathbb{Z}_2[x]$ and therefore I can't conclude anything.








polynomials ring-theory field-theory irreducible-polynomials




polynomials ring-theory field-theory irreducible-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 3:57







Bruno Reis

















asked Dec 6 '18 at 3:30









Bruno ReisBruno Reis

982418




982418












  • $begingroup$
    Seems to me you have shown that $f(x)$ is reducible over $Bbb Z_2[x]$, not whether or not it's reducible over $Bbb Q[x]$.
    $endgroup$
    – Rolf Hoyer
    Dec 6 '18 at 3:34










  • $begingroup$
    @RolfHoyer Yeah, I was checking what I've done... I just know that if a monic polynomial is reducible over $mathbb{Q}[x]$, then it is going to be reducible over $mathbb{Z}_2[x]$... But if that polynomial is irreducible, I can't assume the same... How you would approach that question?
    $endgroup$
    – Bruno Reis
    Dec 6 '18 at 3:37










  • $begingroup$
    Why doesn't $x^4+1$ have roots mod $2$?
    $endgroup$
    – Michael Burr
    Dec 6 '18 at 3:38










  • $begingroup$
    @MichaelBurr in mod 2, $x in {0,1}$ and from that we have $f(0) = 1$ and $f(1)=2=0$... Yeah, you are correct... We have roots...
    $endgroup$
    – Bruno Reis
    Dec 6 '18 at 3:39






  • 1




    $begingroup$
    @AnikBhowmick Shifting is an invertable operation, so irreduciblity must be preserved. In this case, factoring commutes with shifting.
    $endgroup$
    – Michael Burr
    Dec 6 '18 at 3:45




















  • $begingroup$
    Seems to me you have shown that $f(x)$ is reducible over $Bbb Z_2[x]$, not whether or not it's reducible over $Bbb Q[x]$.
    $endgroup$
    – Rolf Hoyer
    Dec 6 '18 at 3:34










  • $begingroup$
    @RolfHoyer Yeah, I was checking what I've done... I just know that if a monic polynomial is reducible over $mathbb{Q}[x]$, then it is going to be reducible over $mathbb{Z}_2[x]$... But if that polynomial is irreducible, I can't assume the same... How you would approach that question?
    $endgroup$
    – Bruno Reis
    Dec 6 '18 at 3:37










  • $begingroup$
    Why doesn't $x^4+1$ have roots mod $2$?
    $endgroup$
    – Michael Burr
    Dec 6 '18 at 3:38










  • $begingroup$
    @MichaelBurr in mod 2, $x in {0,1}$ and from that we have $f(0) = 1$ and $f(1)=2=0$... Yeah, you are correct... We have roots...
    $endgroup$
    – Bruno Reis
    Dec 6 '18 at 3:39






  • 1




    $begingroup$
    @AnikBhowmick Shifting is an invertable operation, so irreduciblity must be preserved. In this case, factoring commutes with shifting.
    $endgroup$
    – Michael Burr
    Dec 6 '18 at 3:45


















$begingroup$
Seems to me you have shown that $f(x)$ is reducible over $Bbb Z_2[x]$, not whether or not it's reducible over $Bbb Q[x]$.
$endgroup$
– Rolf Hoyer
Dec 6 '18 at 3:34




$begingroup$
Seems to me you have shown that $f(x)$ is reducible over $Bbb Z_2[x]$, not whether or not it's reducible over $Bbb Q[x]$.
$endgroup$
– Rolf Hoyer
Dec 6 '18 at 3:34












$begingroup$
@RolfHoyer Yeah, I was checking what I've done... I just know that if a monic polynomial is reducible over $mathbb{Q}[x]$, then it is going to be reducible over $mathbb{Z}_2[x]$... But if that polynomial is irreducible, I can't assume the same... How you would approach that question?
$endgroup$
– Bruno Reis
Dec 6 '18 at 3:37




$begingroup$
@RolfHoyer Yeah, I was checking what I've done... I just know that if a monic polynomial is reducible over $mathbb{Q}[x]$, then it is going to be reducible over $mathbb{Z}_2[x]$... But if that polynomial is irreducible, I can't assume the same... How you would approach that question?
$endgroup$
– Bruno Reis
Dec 6 '18 at 3:37












$begingroup$
Why doesn't $x^4+1$ have roots mod $2$?
$endgroup$
– Michael Burr
Dec 6 '18 at 3:38




$begingroup$
Why doesn't $x^4+1$ have roots mod $2$?
$endgroup$
– Michael Burr
Dec 6 '18 at 3:38












$begingroup$
@MichaelBurr in mod 2, $x in {0,1}$ and from that we have $f(0) = 1$ and $f(1)=2=0$... Yeah, you are correct... We have roots...
$endgroup$
– Bruno Reis
Dec 6 '18 at 3:39




$begingroup$
@MichaelBurr in mod 2, $x in {0,1}$ and from that we have $f(0) = 1$ and $f(1)=2=0$... Yeah, you are correct... We have roots...
$endgroup$
– Bruno Reis
Dec 6 '18 at 3:39




1




1




$begingroup$
@AnikBhowmick Shifting is an invertable operation, so irreduciblity must be preserved. In this case, factoring commutes with shifting.
$endgroup$
– Michael Burr
Dec 6 '18 at 3:45






$begingroup$
@AnikBhowmick Shifting is an invertable operation, so irreduciblity must be preserved. In this case, factoring commutes with shifting.
$endgroup$
– Michael Burr
Dec 6 '18 at 3:45












2 Answers
2






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oldest

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2












$begingroup$

The expression for $f(x)$ looks pretty much like a binomial expansion: to be precise,
$$f(x)=(x+1)^4-2x .$$
Hence
$$f(x-1)=x^4-2x+2 ;$$
this is irreducible by Eisenstein, therefore $f(x)$ is also irreducible.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Your polynomial is very close to $(x+1)^4$. Making the substitution, $x rightarrow y-1$, we get $ y^4 - 2y + 2$, which is irreducible by Eisenstein's criterion.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
      2






      active

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      active

      oldest

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      $begingroup$

      The expression for $f(x)$ looks pretty much like a binomial expansion: to be precise,
      $$f(x)=(x+1)^4-2x .$$
      Hence
      $$f(x-1)=x^4-2x+2 ;$$
      this is irreducible by Eisenstein, therefore $f(x)$ is also irreducible.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        The expression for $f(x)$ looks pretty much like a binomial expansion: to be precise,
        $$f(x)=(x+1)^4-2x .$$
        Hence
        $$f(x-1)=x^4-2x+2 ;$$
        this is irreducible by Eisenstein, therefore $f(x)$ is also irreducible.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          The expression for $f(x)$ looks pretty much like a binomial expansion: to be precise,
          $$f(x)=(x+1)^4-2x .$$
          Hence
          $$f(x-1)=x^4-2x+2 ;$$
          this is irreducible by Eisenstein, therefore $f(x)$ is also irreducible.






          share|cite|improve this answer









          $endgroup$



          The expression for $f(x)$ looks pretty much like a binomial expansion: to be precise,
          $$f(x)=(x+1)^4-2x .$$
          Hence
          $$f(x-1)=x^4-2x+2 ;$$
          this is irreducible by Eisenstein, therefore $f(x)$ is also irreducible.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 3:45









          DavidDavid

          68.1k664126




          68.1k664126























              2












              $begingroup$

              Your polynomial is very close to $(x+1)^4$. Making the substitution, $x rightarrow y-1$, we get $ y^4 - 2y + 2$, which is irreducible by Eisenstein's criterion.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Your polynomial is very close to $(x+1)^4$. Making the substitution, $x rightarrow y-1$, we get $ y^4 - 2y + 2$, which is irreducible by Eisenstein's criterion.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Your polynomial is very close to $(x+1)^4$. Making the substitution, $x rightarrow y-1$, we get $ y^4 - 2y + 2$, which is irreducible by Eisenstein's criterion.






                  share|cite|improve this answer









                  $endgroup$



                  Your polynomial is very close to $(x+1)^4$. Making the substitution, $x rightarrow y-1$, we get $ y^4 - 2y + 2$, which is irreducible by Eisenstein's criterion.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 3:48









                  Eric TowersEric Towers

                  32.5k22370




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