Prove that a polynomial of the form $x^p−x−a in F[x]$ is either irreducible or splits in $F$.












0












$begingroup$


Let $F$ be a field with char$F = p$. Prove that a polynomial of the form $x^p−x−a in F[x]$ is either irreducible or splits in $F$.



I've seen a few different ways of proving this with $x^p-a$. I know that $a in F$ being a root of $F(x)$ means $a^p=a$, so $x^p−a=(x−a)^p$. But I don't think this property works for $x^p-x-a$.






abstract-algebra irreducible-polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    Close to being a duplicate of this.
    $endgroup$
    – Jyrki Lahtonen
    May 3 '18 at 18:52
















0












$begingroup$


Let $F$ be a field with char$F = p$. Prove that a polynomial of the form $x^p−x−a in F[x]$ is either irreducible or splits in $F$.



I've seen a few different ways of proving this with $x^p-a$. I know that $a in F$ being a root of $F(x)$ means $a^p=a$, so $x^p−a=(x−a)^p$. But I don't think this property works for $x^p-x-a$.






abstract-algebra irreducible-polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    Close to being a duplicate of this.
    $endgroup$
    – Jyrki Lahtonen
    May 3 '18 at 18:52














0












0








0


1



$begingroup$


Let $F$ be a field with char$F = p$. Prove that a polynomial of the form $x^p−x−a in F[x]$ is either irreducible or splits in $F$.



I've seen a few different ways of proving this with $x^p-a$. I know that $a in F$ being a root of $F(x)$ means $a^p=a$, so $x^p−a=(x−a)^p$. But I don't think this property works for $x^p-x-a$.






abstract-algebra irreducible-polynomials







share|cite|improve this question









$endgroup$




Let $F$ be a field with char$F = p$. Prove that a polynomial of the form $x^p−x−a in F[x]$ is either irreducible or splits in $F$.



I've seen a few different ways of proving this with $x^p-a$. I know that $a in F$ being a root of $F(x)$ means $a^p=a$, so $x^p−a=(x−a)^p$. But I don't think this property works for $x^p-x-a$.






abstract-algebra irreducible-polynomials




abstract-algebra irreducible-polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked May 2 '18 at 16:01









TimTim

305




305












  • $begingroup$
    Close to being a duplicate of this.
    $endgroup$
    – Jyrki Lahtonen
    May 3 '18 at 18:52


















  • $begingroup$
    Close to being a duplicate of this.
    $endgroup$
    – Jyrki Lahtonen
    May 3 '18 at 18:52
















$begingroup$
Close to being a duplicate of this.
$endgroup$
– Jyrki Lahtonen
May 3 '18 at 18:52




$begingroup$
Close to being a duplicate of this.
$endgroup$
– Jyrki Lahtonen
May 3 '18 at 18:52










1 Answer
1






active

oldest

votes


















3












$begingroup$

This is an Artin-Schreier polynomial. If $alpha$ is a root,
the other roots are $alpha+k$ for $kinBbb F_p$.
If $alphanotin F$ then for some nonzero $kinBbb F_p$,
$alpha+k$ is a conjugate of $alpha$ over $F$. Then $alpha+2k$
is a conjugate of $alpha+k$ and so also a conjugate of $alpha$, etc.
Then all $alpha+mk$ are conjugates of $alpha$, and these are all
the roots. Therefore $x^p-x-a$ is irreducible in this case.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
    $endgroup$
    – Tim
    May 2 '18 at 17:02










  • $begingroup$
    Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
    $endgroup$
    – Lubin
    May 2 '18 at 18:15











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









3












$begingroup$

This is an Artin-Schreier polynomial. If $alpha$ is a root,
the other roots are $alpha+k$ for $kinBbb F_p$.
If $alphanotin F$ then for some nonzero $kinBbb F_p$,
$alpha+k$ is a conjugate of $alpha$ over $F$. Then $alpha+2k$
is a conjugate of $alpha+k$ and so also a conjugate of $alpha$, etc.
Then all $alpha+mk$ are conjugates of $alpha$, and these are all
the roots. Therefore $x^p-x-a$ is irreducible in this case.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
    $endgroup$
    – Tim
    May 2 '18 at 17:02










  • $begingroup$
    Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
    $endgroup$
    – Lubin
    May 2 '18 at 18:15
















3












$begingroup$

This is an Artin-Schreier polynomial. If $alpha$ is a root,
the other roots are $alpha+k$ for $kinBbb F_p$.
If $alphanotin F$ then for some nonzero $kinBbb F_p$,
$alpha+k$ is a conjugate of $alpha$ over $F$. Then $alpha+2k$
is a conjugate of $alpha+k$ and so also a conjugate of $alpha$, etc.
Then all $alpha+mk$ are conjugates of $alpha$, and these are all
the roots. Therefore $x^p-x-a$ is irreducible in this case.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
    $endgroup$
    – Tim
    May 2 '18 at 17:02










  • $begingroup$
    Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
    $endgroup$
    – Lubin
    May 2 '18 at 18:15














3












3








3





$begingroup$

This is an Artin-Schreier polynomial. If $alpha$ is a root,
the other roots are $alpha+k$ for $kinBbb F_p$.
If $alphanotin F$ then for some nonzero $kinBbb F_p$,
$alpha+k$ is a conjugate of $alpha$ over $F$. Then $alpha+2k$
is a conjugate of $alpha+k$ and so also a conjugate of $alpha$, etc.
Then all $alpha+mk$ are conjugates of $alpha$, and these are all
the roots. Therefore $x^p-x-a$ is irreducible in this case.






share|cite|improve this answer









$endgroup$



This is an Artin-Schreier polynomial. If $alpha$ is a root,
the other roots are $alpha+k$ for $kinBbb F_p$.
If $alphanotin F$ then for some nonzero $kinBbb F_p$,
$alpha+k$ is a conjugate of $alpha$ over $F$. Then $alpha+2k$
is a conjugate of $alpha+k$ and so also a conjugate of $alpha$, etc.
Then all $alpha+mk$ are conjugates of $alpha$, and these are all
the roots. Therefore $x^p-x-a$ is irreducible in this case.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered May 2 '18 at 16:08









Lord Shark the UnknownLord Shark the Unknown

104k1160132




104k1160132












  • $begingroup$
    So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
    $endgroup$
    – Tim
    May 2 '18 at 17:02










  • $begingroup$
    Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
    $endgroup$
    – Lubin
    May 2 '18 at 18:15


















  • $begingroup$
    So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
    $endgroup$
    – Tim
    May 2 '18 at 17:02










  • $begingroup$
    Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
    $endgroup$
    – Lubin
    May 2 '18 at 18:15
















$begingroup$
So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
$endgroup$
– Tim
May 2 '18 at 17:02




$begingroup$
So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
$endgroup$
– Tim
May 2 '18 at 17:02












$begingroup$
Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
$endgroup$
– Lubin
May 2 '18 at 18:15




$begingroup$
Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
$endgroup$
– Lubin
May 2 '18 at 18:15


















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