Mixing
The polynomial $x^6+x^3+1$ is irreducible over $mathbb{Q}[x]$
$begingroup$
To solve Dummit-Foote's exercise I want to check that whether the polynomial $p(x)= x^6+x^3+1in mathbb{Z}[x]$ is reducible over $mathbb{Q}$ or not.
I've seen that this polynomial has no zero in $mathbb{Q}$. But the degree of this polynomial is other than 2 and 3. Therefore, I can not say directly that this polynomial is irreducible over $mathbb{Q}$.
How do I show that the polynomial $x^6+x^3+1in mathbb{Z}[x]$ is irreducible over $mathbb{Q}$? Can anyone give me some hints?
Thank You in advance.
abstract-algebra ring-theory field-theory irreducible-polynomials
asked Jan 31 at 16:21
BijanDattaBijanDatta
309113
$endgroup$
add a comment |
$begingroup$
To solve Dummit-Foote's exercise I want to check that whether the polynomial $p(x)= x^6+x^3+1in mathbb{Z}[x]$ is reducible over $mathbb{Q}$ or not.
I've seen that this polynomial has no zero in $mathbb{Q}$. But the degree of this polynomial is other than 2 and 3. Therefore, I can not say directly that this polynomial is irreducible over $mathbb{Q}$.
How do I show that the polynomial $x^6+x^3+1in mathbb{Z}[x]$ is irreducible over $mathbb{Q}$? Can anyone give me some hints?
Thank You in advance.
abstract-algebra ring-theory field-theory irreducible-polynomials
asked Jan 31 at 16:21
BijanDattaBijanDatta
309113
$endgroup$
4
$begingroup$
Please add context to your question. Here are some tips on how to ask a good question.
$endgroup$
– André 3000
Jan 31 at 16:29
1
$begingroup$
Hey dude, that's three Problem-Statement-Questions asking for "show irreducible" in two hours. Before you carpet bomb us with any more, please go back and add context to the questions, or at least learn from the earlier responses before you pepper us with more. Thanks
$endgroup$
– rschwieb
Jan 31 at 18:06
$begingroup$
I'm really very sorry for my unexpected manner of questioning. I think I've edited my question properly now. Thanks every one for commenting here.
$endgroup$
– BijanDatta
Feb 1 at 8:17
add a comment |
$begingroup$
To solve Dummit-Foote's exercise I want to check that whether the polynomial $p(x)= x^6+x^3+1in mathbb{Z}[x]$ is reducible over $mathbb{Q}$ or not.
I've seen that this polynomial has no zero in $mathbb{Q}$. But the degree of this polynomial is other than 2 and 3. Therefore, I can not say directly that this polynomial is irreducible over $mathbb{Q}$.
How do I show that the polynomial $x^6+x^3+1in mathbb{Z}[x]$ is irreducible over $mathbb{Q}$? Can anyone give me some hints?
Thank You in advance.
abstract-algebra ring-theory field-theory irreducible-polynomials
asked Jan 31 at 16:21
BijanDattaBijanDatta
309113
$endgroup$
To solve Dummit-Foote's exercise I want to check that whether the polynomial $p(x)= x^6+x^3+1in mathbb{Z}[x]$ is reducible over $mathbb{Q}$ or not.
I've seen that this polynomial has no zero in $mathbb{Q}$. But the degree of this polynomial is other than 2 and 3. Therefore, I can not say directly that this polynomial is irreducible over $mathbb{Q}$.
How do I show that the polynomial $x^6+x^3+1in mathbb{Z}[x]$ is irreducible over $mathbb{Q}$? Can anyone give me some hints?
Thank You in advance.
abstract-algebra ring-theory field-theory irreducible-polynomials
abstract-algebra ring-theory field-theory irreducible-polynomials
asked Jan 31 at 16:21
BijanDattaBijanDatta
309113
asked Jan 31 at 16:21
BijanDattaBijanDatta
309113
edited Feb 1 at 8:11
BijanDatta
asked Jan 31 at 16:21
BijanDattaBijanDatta
309113
asked Jan 31 at 16:21
BijanDattaBijanDatta
309113
asked Jan 31 at 16:21
BijanDattaBijanDatta
309113
309113
4
$begingroup$
Please add context to your question. Here are some tips on how to ask a good question.
$endgroup$
– André 3000
Jan 31 at 16:29
1
$begingroup$
Hey dude, that's three Problem-Statement-Questions asking for "show irreducible" in two hours. Before you carpet bomb us with any more, please go back and add context to the questions, or at least learn from the earlier responses before you pepper us with more. Thanks
$endgroup$
– rschwieb
Jan 31 at 18:06
$begingroup$
I'm really very sorry for my unexpected manner of questioning. I think I've edited my question properly now. Thanks every one for commenting here.
$endgroup$
– BijanDatta
Feb 1 at 8:17
add a comment |
4
$begingroup$
Please add context to your question. Here are some tips on how to ask a good question.
$endgroup$
– André 3000
Jan 31 at 16:29
1
$begingroup$
Hey dude, that's three Problem-Statement-Questions asking for "show irreducible" in two hours. Before you carpet bomb us with any more, please go back and add context to the questions, or at least learn from the earlier responses before you pepper us with more. Thanks
$endgroup$
– rschwieb
Jan 31 at 18:06
$begingroup$
I'm really very sorry for my unexpected manner of questioning. I think I've edited my question properly now. Thanks every one for commenting here.
$endgroup$
– BijanDatta
Feb 1 at 8:17
4
4
$begingroup$
Please add context to your question. Here are some tips on how to ask a good question.
$endgroup$
– André 3000
Jan 31 at 16:29
$begingroup$
Please add context to your question. Here are some tips on how to ask a good question.
$endgroup$
– André 3000
Jan 31 at 16:29
1
1
$begingroup$
Hey dude, that's three Problem-Statement-Questions asking for "show irreducible" in two hours. Before you carpet bomb us with any more, please go back and add context to the questions, or at least learn from the earlier responses before you pepper us with more. Thanks
$endgroup$
– rschwieb
Jan 31 at 18:06
$begingroup$
Hey dude, that's three Problem-Statement-Questions asking for "show irreducible" in two hours. Before you carpet bomb us with any more, please go back and add context to the questions, or at least learn from the earlier responses before you pepper us with more. Thanks
$endgroup$
– rschwieb
Jan 31 at 18:06
$begingroup$
I'm really very sorry for my unexpected manner of questioning. I think I've edited my question properly now. Thanks every one for commenting here.
$endgroup$
– BijanDatta
Feb 1 at 8:17
$begingroup$
I'm really very sorry for my unexpected manner of questioning. I think I've edited my question properly now. Thanks every one for commenting here.
$endgroup$
– BijanDatta
Feb 1 at 8:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I would do the following:
Note that $p(x)$ is irreducible if and only if $p(x+1)$ is irreducible.
Compute $p(x+1)$. We have
begin{align}
p(x+1)&=(x+1)^{6}+(x+1)^{3}+1\
&=sum_{k=0}^{6}binom{6}{k}x^{k}+sum_{k=0}^{3}binom{3}{k}x^{k}+1\
&=(x^{6}+6x^{5}+15x^{4}+20x^{3}+15x^{2}+6x+1)+(x^{3}+3x^{2}+3x+1)+1\
&=x^{6}+6x^{5}+15x^{4}+21x^{3}+18x^{2}+9x+3
end{align}
Can you see that this polynomial is irreducible over $mathbb{Q}$?
answered Feb 1 at 12:58
studiosusstudiosus
2,294815
$endgroup$
$begingroup$
Eisenstein criterion when p=3, right?
$endgroup$
– pendermath
Feb 1 at 13:43
$begingroup$
@pendermath Exactly
$endgroup$
– studiosus
Feb 1 at 13:44
$begingroup$
However, how did you come up with the substitution p(x+1)? There are many of these to prove (eg, 1, -1, 2, -2, etc.), and success is not even guaranteed.
$endgroup$
– pendermath
Feb 1 at 13:45
add a comment |
$begingroup$
studiosus answer has a much better argument, but in my experience my students always had trouble finding good substitutitions. So here is a harder, less elegant, but more staightforward way.
Note that it is an integer polynomial, so it is enough to decide irreduciblity in $mathbb{Z}[X]$ by one of the many results which claim the name 'Gauss Lemma'.
If $p$ were reducible over $mathbb{Z}$, then it would also be reducible after 'going mod 2', i.e. in $mathbb{F}_2[X]$. However in $mathbb{F}_2[X]$ it is easy to go through the different cases. Clearly $overline{p}$ has no roots, i.e. no factors of degree 1. The only irreducible polynomial of degree $2$ is $X^2+X+1$ and one can simply check that $p=(X^2+X+1)(X^4+X^3)+1$. Also there are just two irreducible polynomials of degree 3, namely $X^3+X+1$ and $X^3+X^2+1$. All possible products of those two come out to be different from $p$.
So $p$ does not have factors of degree 1, 2, or 3. So it must be irreducible in $mathbb{F}_2[X]$, hence in $mathbb{Z}[X]$, hence in $mathbb{Q}[X]$.
answered Feb 1 at 13:38
Rene RecktenwaldRene Recktenwald
1808
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
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active
oldest
votes
$begingroup$
I would do the following:
Note that $p(x)$ is irreducible if and only if $p(x+1)$ is irreducible.
Compute $p(x+1)$. We have
begin{align}
p(x+1)&=(x+1)^{6}+(x+1)^{3}+1\
&=sum_{k=0}^{6}binom{6}{k}x^{k}+sum_{k=0}^{3}binom{3}{k}x^{k}+1\
&=(x^{6}+6x^{5}+15x^{4}+20x^{3}+15x^{2}+6x+1)+(x^{3}+3x^{2}+3x+1)+1\
&=x^{6}+6x^{5}+15x^{4}+21x^{3}+18x^{2}+9x+3
end{align}
Can you see that this polynomial is irreducible over $mathbb{Q}$?
answered Feb 1 at 12:58
studiosusstudiosus
2,294815
$endgroup$
$begingroup$
Eisenstein criterion when p=3, right?
$endgroup$
– pendermath
Feb 1 at 13:43
$begingroup$
@pendermath Exactly
$endgroup$
– studiosus
Feb 1 at 13:44
$begingroup$
However, how did you come up with the substitution p(x+1)? There are many of these to prove (eg, 1, -1, 2, -2, etc.), and success is not even guaranteed.
$endgroup$
– pendermath
Feb 1 at 13:45
add a comment |
$begingroup$
I would do the following:
Note that $p(x)$ is irreducible if and only if $p(x+1)$ is irreducible.
Compute $p(x+1)$. We have
begin{align}
p(x+1)&=(x+1)^{6}+(x+1)^{3}+1\
&=sum_{k=0}^{6}binom{6}{k}x^{k}+sum_{k=0}^{3}binom{3}{k}x^{k}+1\
&=(x^{6}+6x^{5}+15x^{4}+20x^{3}+15x^{2}+6x+1)+(x^{3}+3x^{2}+3x+1)+1\
&=x^{6}+6x^{5}+15x^{4}+21x^{3}+18x^{2}+9x+3
end{align}
Can you see that this polynomial is irreducible over $mathbb{Q}$?
answered Feb 1 at 12:58
studiosusstudiosus
2,294815
$endgroup$
$begingroup$
Eisenstein criterion when p=3, right?
$endgroup$
– pendermath
Feb 1 at 13:43
$begingroup$
@pendermath Exactly
$endgroup$
– studiosus
Feb 1 at 13:44
$begingroup$
However, how did you come up with the substitution p(x+1)? There are many of these to prove (eg, 1, -1, 2, -2, etc.), and success is not even guaranteed.
$endgroup$
– pendermath
Feb 1 at 13:45
add a comment |
$begingroup$
I would do the following:
Note that $p(x)$ is irreducible if and only if $p(x+1)$ is irreducible.
Compute $p(x+1)$. We have
begin{align}
p(x+1)&=(x+1)^{6}+(x+1)^{3}+1\
&=sum_{k=0}^{6}binom{6}{k}x^{k}+sum_{k=0}^{3}binom{3}{k}x^{k}+1\
&=(x^{6}+6x^{5}+15x^{4}+20x^{3}+15x^{2}+6x+1)+(x^{3}+3x^{2}+3x+1)+1\
&=x^{6}+6x^{5}+15x^{4}+21x^{3}+18x^{2}+9x+3
end{align}
Can you see that this polynomial is irreducible over $mathbb{Q}$?
answered Feb 1 at 12:58
studiosusstudiosus
2,294815
$endgroup$
I would do the following:
Note that $p(x)$ is irreducible if and only if $p(x+1)$ is irreducible.
Compute $p(x+1)$. We have
begin{align}
p(x+1)&=(x+1)^{6}+(x+1)^{3}+1\
&=sum_{k=0}^{6}binom{6}{k}x^{k}+sum_{k=0}^{3}binom{3}{k}x^{k}+1\
&=(x^{6}+6x^{5}+15x^{4}+20x^{3}+15x^{2}+6x+1)+(x^{3}+3x^{2}+3x+1)+1\
&=x^{6}+6x^{5}+15x^{4}+21x^{3}+18x^{2}+9x+3
end{align}
Can you see that this polynomial is irreducible over $mathbb{Q}$?
answered Feb 1 at 12:58
studiosusstudiosus
2,294815
answered Feb 1 at 12:58
studiosusstudiosus
2,294815
answered Feb 1 at 12:58
studiosusstudiosus
2,294815
answered Feb 1 at 12:58
studiosusstudiosus
2,294815
2,294815
$begingroup$
Eisenstein criterion when p=3, right?
$endgroup$
– pendermath
Feb 1 at 13:43
$begingroup$
@pendermath Exactly
$endgroup$
– studiosus
Feb 1 at 13:44
$begingroup$
However, how did you come up with the substitution p(x+1)? There are many of these to prove (eg, 1, -1, 2, -2, etc.), and success is not even guaranteed.
$endgroup$
– pendermath
Feb 1 at 13:45
add a comment |
$begingroup$
Eisenstein criterion when p=3, right?
$endgroup$
– pendermath
Feb 1 at 13:43
$begingroup$
@pendermath Exactly
$endgroup$
– studiosus
Feb 1 at 13:44
$begingroup$
However, how did you come up with the substitution p(x+1)? There are many of these to prove (eg, 1, -1, 2, -2, etc.), and success is not even guaranteed.
$endgroup$
– pendermath
Feb 1 at 13:45
$begingroup$
Eisenstein criterion when p=3, right?
$endgroup$
– pendermath
Feb 1 at 13:43
$begingroup$
Eisenstein criterion when p=3, right?
$endgroup$
– pendermath
Feb 1 at 13:43
$begingroup$
@pendermath Exactly
$endgroup$
– studiosus
Feb 1 at 13:44
$begingroup$
@pendermath Exactly
$endgroup$
– studiosus
Feb 1 at 13:44
$begingroup$
However, how did you come up with the substitution p(x+1)? There are many of these to prove (eg, 1, -1, 2, -2, etc.), and success is not even guaranteed.
$endgroup$
– pendermath
Feb 1 at 13:45
$begingroup$
However, how did you come up with the substitution p(x+1)? There are many of these to prove (eg, 1, -1, 2, -2, etc.), and success is not even guaranteed.
$endgroup$
– pendermath
Feb 1 at 13:45
add a comment |
$begingroup$
studiosus answer has a much better argument, but in my experience my students always had trouble finding good substitutitions. So here is a harder, less elegant, but more staightforward way.
Note that it is an integer polynomial, so it is enough to decide irreduciblity in $mathbb{Z}[X]$ by one of the many results which claim the name 'Gauss Lemma'.
If $p$ were reducible over $mathbb{Z}$, then it would also be reducible after 'going mod 2', i.e. in $mathbb{F}_2[X]$. However in $mathbb{F}_2[X]$ it is easy to go through the different cases. Clearly $overline{p}$ has no roots, i.e. no factors of degree 1. The only irreducible polynomial of degree $2$ is $X^2+X+1$ and one can simply check that $p=(X^2+X+1)(X^4+X^3)+1$. Also there are just two irreducible polynomials of degree 3, namely $X^3+X+1$ and $X^3+X^2+1$. All possible products of those two come out to be different from $p$.
So $p$ does not have factors of degree 1, 2, or 3. So it must be irreducible in $mathbb{F}_2[X]$, hence in $mathbb{Z}[X]$, hence in $mathbb{Q}[X]$.
answered Feb 1 at 13:38
Rene RecktenwaldRene Recktenwald
1808
$endgroup$
add a comment |
$begingroup$
studiosus answer has a much better argument, but in my experience my students always had trouble finding good substitutitions. So here is a harder, less elegant, but more staightforward way.
Note that it is an integer polynomial, so it is enough to decide irreduciblity in $mathbb{Z}[X]$ by one of the many results which claim the name 'Gauss Lemma'.
If $p$ were reducible over $mathbb{Z}$, then it would also be reducible after 'going mod 2', i.e. in $mathbb{F}_2[X]$. However in $mathbb{F}_2[X]$ it is easy to go through the different cases. Clearly $overline{p}$ has no roots, i.e. no factors of degree 1. The only irreducible polynomial of degree $2$ is $X^2+X+1$ and one can simply check that $p=(X^2+X+1)(X^4+X^3)+1$. Also there are just two irreducible polynomials of degree 3, namely $X^3+X+1$ and $X^3+X^2+1$. All possible products of those two come out to be different from $p$.
So $p$ does not have factors of degree 1, 2, or 3. So it must be irreducible in $mathbb{F}_2[X]$, hence in $mathbb{Z}[X]$, hence in $mathbb{Q}[X]$.
answered Feb 1 at 13:38
Rene RecktenwaldRene Recktenwald
1808
$endgroup$
add a comment |
$begingroup$
studiosus answer has a much better argument, but in my experience my students always had trouble finding good substitutitions. So here is a harder, less elegant, but more staightforward way.
Note that it is an integer polynomial, so it is enough to decide irreduciblity in $mathbb{Z}[X]$ by one of the many results which claim the name 'Gauss Lemma'.
If $p$ were reducible over $mathbb{Z}$, then it would also be reducible after 'going mod 2', i.e. in $mathbb{F}_2[X]$. However in $mathbb{F}_2[X]$ it is easy to go through the different cases. Clearly $overline{p}$ has no roots, i.e. no factors of degree 1. The only irreducible polynomial of degree $2$ is $X^2+X+1$ and one can simply check that $p=(X^2+X+1)(X^4+X^3)+1$. Also there are just two irreducible polynomials of degree 3, namely $X^3+X+1$ and $X^3+X^2+1$. All possible products of those two come out to be different from $p$.
So $p$ does not have factors of degree 1, 2, or 3. So it must be irreducible in $mathbb{F}_2[X]$, hence in $mathbb{Z}[X]$, hence in $mathbb{Q}[X]$.
answered Feb 1 at 13:38
Rene RecktenwaldRene Recktenwald
1808
$endgroup$
studiosus answer has a much better argument, but in my experience my students always had trouble finding good substitutitions. So here is a harder, less elegant, but more staightforward way.
Note that it is an integer polynomial, so it is enough to decide irreduciblity in $mathbb{Z}[X]$ by one of the many results which claim the name 'Gauss Lemma'.
If $p$ were reducible over $mathbb{Z}$, then it would also be reducible after 'going mod 2', i.e. in $mathbb{F}_2[X]$. However in $mathbb{F}_2[X]$ it is easy to go through the different cases. Clearly $overline{p}$ has no roots, i.e. no factors of degree 1. The only irreducible polynomial of degree $2$ is $X^2+X+1$ and one can simply check that $p=(X^2+X+1)(X^4+X^3)+1$. Also there are just two irreducible polynomials of degree 3, namely $X^3+X+1$ and $X^3+X^2+1$. All possible products of those two come out to be different from $p$.
So $p$ does not have factors of degree 1, 2, or 3. So it must be irreducible in $mathbb{F}_2[X]$, hence in $mathbb{Z}[X]$, hence in $mathbb{Q}[X]$.
answered Feb 1 at 13:38
Rene RecktenwaldRene Recktenwald
1808
answered Feb 1 at 13:38
Rene RecktenwaldRene Recktenwald
1808
answered Feb 1 at 13:38
Rene RecktenwaldRene Recktenwald
1808
answered Feb 1 at 13:38
Rene RecktenwaldRene Recktenwald
1808
1808
add a comment |
add a comment |
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4
$begingroup$
Please add context to your question. Here are some tips on how to ask a good question.
$endgroup$
– André 3000
Jan 31 at 16:29
1
$begingroup$
Hey dude, that's three Problem-Statement-Questions asking for "show irreducible" in two hours. Before you carpet bomb us with any more, please go back and add context to the questions, or at least learn from the earlier responses before you pepper us with more. Thanks
$endgroup$
– rschwieb
Jan 31 at 18:06
$begingroup$
I'm really very sorry for my unexpected manner of questioning. I think I've edited my question properly now. Thanks every one for commenting here.
$endgroup$
– BijanDatta
Feb 1 at 8:17