$x^{2n} + x^{2n-1} + x^ {2n-2} +ldots+ x + 1$ is irreducible for any $nin mathbb N$ in $F_2[x]$. True or...
1
$begingroup$
Will the polynomials of the following set $A$ be irreducible in $F_2[x]$?
$A = [x^{2n} + x^{2n-1} + x^ {2n-2} + ldots+ x + 1 : nin mathbb N]$
Can anyone please give me hints how to proceed?
Every term is there. I meant every polynomial of degree $2n$ has $2n+1$ terms.
abstract-algebra polynomials ring-theory irreducible-polynomials
asked Dec 14 '18 at 11:58
cmicmi
1,141312
$endgroup$
add a comment |
1
$begingroup$
Will the polynomials of the following set $A$ be irreducible in $F_2[x]$?
$A = [x^{2n} + x^{2n-1} + x^ {2n-2} + ldots+ x + 1 : nin mathbb N]$
Can anyone please give me hints how to proceed?
Every term is there. I meant every polynomial of degree $2n$ has $2n+1$ terms.
abstract-algebra polynomials ring-theory irreducible-polynomials
asked Dec 14 '18 at 11:58
cmicmi
1,141312
$endgroup$
$begingroup$
Are you intending to omit any terms among the $x^i$ and if yes then which?
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:06
$begingroup$
No no every term is there. I mean every polynomial of degree $2n$ has $2n+1$ terms.@SorinTirc
$endgroup$
– cmi
Dec 14 '18 at 12:17
1
$begingroup$
I am asking this because you omitted $x^{2n-2}$ so maybe you should correct that and then my answer below should be an answer to your question
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:21
add a comment |
1
1
1
1
$begingroup$
Will the polynomials of the following set $A$ be irreducible in $F_2[x]$?
$A = [x^{2n} + x^{2n-1} + x^ {2n-2} + ldots+ x + 1 : nin mathbb N]$
Can anyone please give me hints how to proceed?
Every term is there. I meant every polynomial of degree $2n$ has $2n+1$ terms.
abstract-algebra polynomials ring-theory irreducible-polynomials
asked Dec 14 '18 at 11:58
cmicmi
1,141312
$endgroup$
Will the polynomials of the following set $A$ be irreducible in $F_2[x]$?
$A = [x^{2n} + x^{2n-1} + x^ {2n-2} + ldots+ x + 1 : nin mathbb N]$
Can anyone please give me hints how to proceed?
Every term is there. I meant every polynomial of degree $2n$ has $2n+1$ terms.
abstract-algebra polynomials ring-theory irreducible-polynomials
abstract-algebra polynomials ring-theory irreducible-polynomials
asked Dec 14 '18 at 11:58
cmicmi
1,141312
asked Dec 14 '18 at 11:58
cmicmi
1,141312
edited Dec 14 '18 at 15:09
cmi
asked Dec 14 '18 at 11:58
cmicmi
1,141312
asked Dec 14 '18 at 11:58
cmicmi
1,141312
asked Dec 14 '18 at 11:58
cmicmi
1,141312
1,141312
$begingroup$
Are you intending to omit any terms among the $x^i$ and if yes then which?
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:06
$begingroup$
No no every term is there. I mean every polynomial of degree $2n$ has $2n+1$ terms.@SorinTirc
$endgroup$
– cmi
Dec 14 '18 at 12:17
1
$begingroup$
I am asking this because you omitted $x^{2n-2}$ so maybe you should correct that and then my answer below should be an answer to your question
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:21
add a comment |
$begingroup$
Are you intending to omit any terms among the $x^i$ and if yes then which?
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:06
$begingroup$
No no every term is there. I mean every polynomial of degree $2n$ has $2n+1$ terms.@SorinTirc
$endgroup$
– cmi
Dec 14 '18 at 12:17
1
$begingroup$
I am asking this because you omitted $x^{2n-2}$ so maybe you should correct that and then my answer below should be an answer to your question
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:21
$begingroup$
Are you intending to omit any terms among the $x^i$ and if yes then which?
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:06
$begingroup$
Are you intending to omit any terms among the $x^i$ and if yes then which?
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:06
$begingroup$
No no every term is there. I mean every polynomial of degree $2n$ has $2n+1$ terms.@SorinTirc
$endgroup$
– cmi
Dec 14 '18 at 12:17
$begingroup$
No no every term is there. I mean every polynomial of degree $2n$ has $2n+1$ terms.@SorinTirc
$endgroup$
– cmi
Dec 14 '18 at 12:17
1
1
$begingroup$
I am asking this because you omitted $x^{2n-2}$ so maybe you should correct that and then my answer below should be an answer to your question
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:21
$begingroup$
I am asking this because you omitted $x^{2n-2}$ so maybe you should correct that and then my answer below should be an answer to your question
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:21
add a comment |
1 Answer
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It's false in general, for example for $n=4$ : $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 = (x^2+x+1)(x^6+x^3+1)$.
In general $frac{x^n-1}{x-1}$ is irreducible iff n is prime(the cyclotomic poly) over $ mathbb{Q}[X]$
answered Dec 14 '18 at 12:02
Sorin TircSorin Tirc
1,875213
$endgroup$
$begingroup$
I think it's intended to omit the terms $x^6, x^4, x^2$.
$endgroup$
– Slade
Dec 14 '18 at 12:02
$begingroup$
oh, didn't see that, but then I am not quite sure exactly which terms are omitted tbh
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:04
$begingroup$
All the even terms are omitted except the first one.
$endgroup$
– Slade
Dec 14 '18 at 12:04
$begingroup$
and the last one $x^0$ then too maybe :) ? In whice case the n=2 poly is reducible :)
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:05
$begingroup$
Can we say if a polynomial is irreducible in $mathbb Q[x]$ so will be in $F_2[x]$?@SorinTirc
$endgroup$
– cmi
Dec 14 '18 at 15:08
|
show 3 more comments
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2
$begingroup$
It's false in general, for example for $n=4$ : $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 = (x^2+x+1)(x^6+x^3+1)$.
In general $frac{x^n-1}{x-1}$ is irreducible iff n is prime(the cyclotomic poly) over $ mathbb{Q}[X]$
answered Dec 14 '18 at 12:02
Sorin TircSorin Tirc
1,875213
$endgroup$
$begingroup$
I think it's intended to omit the terms $x^6, x^4, x^2$.
$endgroup$
– Slade
Dec 14 '18 at 12:02
$begingroup$
oh, didn't see that, but then I am not quite sure exactly which terms are omitted tbh
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:04
$begingroup$
All the even terms are omitted except the first one.
$endgroup$
– Slade
Dec 14 '18 at 12:04
$begingroup$
and the last one $x^0$ then too maybe :) ? In whice case the n=2 poly is reducible :)
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:05
$begingroup$
Can we say if a polynomial is irreducible in $mathbb Q[x]$ so will be in $F_2[x]$?@SorinTirc
$endgroup$
– cmi
Dec 14 '18 at 15:08
|
show 3 more comments
2
$begingroup$
It's false in general, for example for $n=4$ : $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 = (x^2+x+1)(x^6+x^3+1)$.
In general $frac{x^n-1}{x-1}$ is irreducible iff n is prime(the cyclotomic poly) over $ mathbb{Q}[X]$
answered Dec 14 '18 at 12:02
Sorin TircSorin Tirc
1,875213
$endgroup$
$begingroup$
I think it's intended to omit the terms $x^6, x^4, x^2$.
$endgroup$
– Slade
Dec 14 '18 at 12:02
$begingroup$
oh, didn't see that, but then I am not quite sure exactly which terms are omitted tbh
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:04
$begingroup$
All the even terms are omitted except the first one.
$endgroup$
– Slade
Dec 14 '18 at 12:04
$begingroup$
and the last one $x^0$ then too maybe :) ? In whice case the n=2 poly is reducible :)
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:05
$begingroup$
Can we say if a polynomial is irreducible in $mathbb Q[x]$ so will be in $F_2[x]$?@SorinTirc
$endgroup$
– cmi
Dec 14 '18 at 15:08
|
show 3 more comments
2
2
2
$begingroup$
It's false in general, for example for $n=4$ : $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 = (x^2+x+1)(x^6+x^3+1)$.
In general $frac{x^n-1}{x-1}$ is irreducible iff n is prime(the cyclotomic poly) over $ mathbb{Q}[X]$
answered Dec 14 '18 at 12:02
Sorin TircSorin Tirc
1,875213
$endgroup$
It's false in general, for example for $n=4$ : $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 = (x^2+x+1)(x^6+x^3+1)$.
In general $frac{x^n-1}{x-1}$ is irreducible iff n is prime(the cyclotomic poly) over $ mathbb{Q}[X]$
answered Dec 14 '18 at 12:02
Sorin TircSorin Tirc
1,875213
edited Dec 14 '18 at 12:02
answered Dec 14 '18 at 12:02
Sorin TircSorin Tirc
1,875213
answered Dec 14 '18 at 12:02
Sorin TircSorin Tirc
1,875213
answered Dec 14 '18 at 12:02
Sorin TircSorin Tirc
1,875213
1,875213
$begingroup$
I think it's intended to omit the terms $x^6, x^4, x^2$.
$endgroup$
– Slade
Dec 14 '18 at 12:02
$begingroup$
oh, didn't see that, but then I am not quite sure exactly which terms are omitted tbh
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:04
$begingroup$
All the even terms are omitted except the first one.
$endgroup$
– Slade
Dec 14 '18 at 12:04
$begingroup$
and the last one $x^0$ then too maybe :) ? In whice case the n=2 poly is reducible :)
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:05
$begingroup$
Can we say if a polynomial is irreducible in $mathbb Q[x]$ so will be in $F_2[x]$?@SorinTirc
$endgroup$
– cmi
Dec 14 '18 at 15:08
|
show 3 more comments
$begingroup$
I think it's intended to omit the terms $x^6, x^4, x^2$.
$endgroup$
– Slade
Dec 14 '18 at 12:02
$begingroup$
oh, didn't see that, but then I am not quite sure exactly which terms are omitted tbh
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:04
$begingroup$
All the even terms are omitted except the first one.
$endgroup$
– Slade
Dec 14 '18 at 12:04
$begingroup$
and the last one $x^0$ then too maybe :) ? In whice case the n=2 poly is reducible :)
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:05
$begingroup$
Can we say if a polynomial is irreducible in $mathbb Q[x]$ so will be in $F_2[x]$?@SorinTirc
$endgroup$
– cmi
Dec 14 '18 at 15:08
$begingroup$
I think it's intended to omit the terms $x^6, x^4, x^2$.
$endgroup$
– Slade
Dec 14 '18 at 12:02
$begingroup$
I think it's intended to omit the terms $x^6, x^4, x^2$.
$endgroup$
– Slade
Dec 14 '18 at 12:02
$begingroup$
oh, didn't see that, but then I am not quite sure exactly which terms are omitted tbh
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:04
$begingroup$
oh, didn't see that, but then I am not quite sure exactly which terms are omitted tbh
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:04
$begingroup$
All the even terms are omitted except the first one.
$endgroup$
– Slade
Dec 14 '18 at 12:04
$begingroup$
All the even terms are omitted except the first one.
$endgroup$
– Slade
Dec 14 '18 at 12:04
$begingroup$
and the last one $x^0$ then too maybe :) ? In whice case the n=2 poly is reducible :)
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:05
$begingroup$
and the last one $x^0$ then too maybe :) ? In whice case the n=2 poly is reducible :)
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:05
$begingroup$
Can we say if a polynomial is irreducible in $mathbb Q[x]$ so will be in $F_2[x]$?@SorinTirc
$endgroup$
– cmi
Dec 14 '18 at 15:08
$begingroup$
Can we say if a polynomial is irreducible in $mathbb Q[x]$ so will be in $F_2[x]$?@SorinTirc
$endgroup$
– cmi
Dec 14 '18 at 15:08
|
show 3 more comments
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$begingroup$
Are you intending to omit any terms among the $x^i$ and if yes then which?
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:06
$begingroup$
No no every term is there. I mean every polynomial of degree $2n$ has $2n+1$ terms.@SorinTirc
$endgroup$
– cmi
Dec 14 '18 at 12:17
1
$begingroup$
I am asking this because you omitted $x^{2n-2}$ so maybe you should correct that and then my answer below should be an answer to your question
$endgroup$
– Sorin Tirc
Dec 14 '18 at 12:21