Prove that if $p notequiv 1 hspace{0.2 cm} (5)$ then $f(x) = x^{5} - 2$ has a unique solution in...
$begingroup$
To prove the statetament, i thought to define a linear application $$ phi : mathbb{F}_{p}^{*} longmapsto mathbb{F}_{p}^{*}$$
Define by : $f(x) = x^{5}$, studying the kernel of $phi$ I noticed that from $x^{5} equiv 1 hspace{0.2 cm}(5)$ $phi$ was injective (Because from the hp. we have $5 nmid p - 1 $, which translates in $p notequiv 1 hspace{0.2 cm} (5)$).
It follows that this $phi$ is an automorphism of $mathbb{F}_{p}^{*}$, in particular it is surjective,
From here i know that for every $p notequiv 1 hspace{0.2 cm} (5)$ it exists $x in mathbb{F}_{p}^{*}$ such that $x^{5} = 2$.
Now i know my $f(x) = (x - alpha)p(x)$ splits into one linear factor, and one of $deg p = 4$.
I'd like to conclude saying that $p$ doesn't split and so that the solution is unique, but don't see how,
Any help or tip would be appreciated,
Thanks.
group-theory field-theory finite-fields splitting-field automorphism-group
asked Jan 5 at 20:29
jacopoburellijacopoburelli
1707
$endgroup$
add a comment |
$begingroup$
To prove the statetament, i thought to define a linear application $$ phi : mathbb{F}_{p}^{*} longmapsto mathbb{F}_{p}^{*}$$
Define by : $f(x) = x^{5}$, studying the kernel of $phi$ I noticed that from $x^{5} equiv 1 hspace{0.2 cm}(5)$ $phi$ was injective (Because from the hp. we have $5 nmid p - 1 $, which translates in $p notequiv 1 hspace{0.2 cm} (5)$).
It follows that this $phi$ is an automorphism of $mathbb{F}_{p}^{*}$, in particular it is surjective,
From here i know that for every $p notequiv 1 hspace{0.2 cm} (5)$ it exists $x in mathbb{F}_{p}^{*}$ such that $x^{5} = 2$.
Now i know my $f(x) = (x - alpha)p(x)$ splits into one linear factor, and one of $deg p = 4$.
I'd like to conclude saying that $p$ doesn't split and so that the solution is unique, but don't see how,
Any help or tip would be appreciated,
Thanks.
group-theory field-theory finite-fields splitting-field automorphism-group
asked Jan 5 at 20:29
jacopoburellijacopoburelli
1707
$endgroup$
1
$begingroup$
Just use that $phi$ is injective.
$endgroup$
– Mark Bennet
Jan 5 at 20:34
$begingroup$
Since $x in mathbb{F}_p^* longmapsto x^5 in mathbb{F}_p^*$ is bijective, $f$ is a bijection from $mathbb{F}_p$ to itself. So it has a unique root. Compare $f$ with its derivative to prove that the root is simple iff $p neq 5$.
$endgroup$
– Mindlack
Jan 5 at 20:39
$begingroup$
So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
$endgroup$
– greedoid
Jan 5 at 20:44
add a comment |
$begingroup$
To prove the statetament, i thought to define a linear application $$ phi : mathbb{F}_{p}^{*} longmapsto mathbb{F}_{p}^{*}$$
Define by : $f(x) = x^{5}$, studying the kernel of $phi$ I noticed that from $x^{5} equiv 1 hspace{0.2 cm}(5)$ $phi$ was injective (Because from the hp. we have $5 nmid p - 1 $, which translates in $p notequiv 1 hspace{0.2 cm} (5)$).
It follows that this $phi$ is an automorphism of $mathbb{F}_{p}^{*}$, in particular it is surjective,
From here i know that for every $p notequiv 1 hspace{0.2 cm} (5)$ it exists $x in mathbb{F}_{p}^{*}$ such that $x^{5} = 2$.
Now i know my $f(x) = (x - alpha)p(x)$ splits into one linear factor, and one of $deg p = 4$.
I'd like to conclude saying that $p$ doesn't split and so that the solution is unique, but don't see how,
Any help or tip would be appreciated,
Thanks.
group-theory field-theory finite-fields splitting-field automorphism-group
asked Jan 5 at 20:29
jacopoburellijacopoburelli
1707
$endgroup$
To prove the statetament, i thought to define a linear application $$ phi : mathbb{F}_{p}^{*} longmapsto mathbb{F}_{p}^{*}$$
Define by : $f(x) = x^{5}$, studying the kernel of $phi$ I noticed that from $x^{5} equiv 1 hspace{0.2 cm}(5)$ $phi$ was injective (Because from the hp. we have $5 nmid p - 1 $, which translates in $p notequiv 1 hspace{0.2 cm} (5)$).
It follows that this $phi$ is an automorphism of $mathbb{F}_{p}^{*}$, in particular it is surjective,
From here i know that for every $p notequiv 1 hspace{0.2 cm} (5)$ it exists $x in mathbb{F}_{p}^{*}$ such that $x^{5} = 2$.
Now i know my $f(x) = (x - alpha)p(x)$ splits into one linear factor, and one of $deg p = 4$.
I'd like to conclude saying that $p$ doesn't split and so that the solution is unique, but don't see how,
Any help or tip would be appreciated,
Thanks.
group-theory field-theory finite-fields splitting-field automorphism-group
group-theory field-theory finite-fields splitting-field automorphism-group
asked Jan 5 at 20:29
jacopoburellijacopoburelli
1707
asked Jan 5 at 20:29
jacopoburellijacopoburelli
1707
asked Jan 5 at 20:29
jacopoburellijacopoburelli
1707
asked Jan 5 at 20:29
jacopoburellijacopoburelli
1707
asked Jan 5 at 20:29
jacopoburellijacopoburelli
1707
1707
1
$begingroup$
Just use that $phi$ is injective.
$endgroup$
– Mark Bennet
Jan 5 at 20:34
$begingroup$
Since $x in mathbb{F}_p^* longmapsto x^5 in mathbb{F}_p^*$ is bijective, $f$ is a bijection from $mathbb{F}_p$ to itself. So it has a unique root. Compare $f$ with its derivative to prove that the root is simple iff $p neq 5$.
$endgroup$
– Mindlack
Jan 5 at 20:39
$begingroup$
So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
$endgroup$
– greedoid
Jan 5 at 20:44
add a comment |
1
$begingroup$
Just use that $phi$ is injective.
$endgroup$
– Mark Bennet
Jan 5 at 20:34
$begingroup$
Since $x in mathbb{F}_p^* longmapsto x^5 in mathbb{F}_p^*$ is bijective, $f$ is a bijection from $mathbb{F}_p$ to itself. So it has a unique root. Compare $f$ with its derivative to prove that the root is simple iff $p neq 5$.
$endgroup$
– Mindlack
Jan 5 at 20:39
$begingroup$
So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
$endgroup$
– greedoid
Jan 5 at 20:44
1
1
$begingroup$
Just use that $phi$ is injective.
$endgroup$
– Mark Bennet
Jan 5 at 20:34
$begingroup$
Just use that $phi$ is injective.
$endgroup$
– Mark Bennet
Jan 5 at 20:34
$begingroup$
Since $x in mathbb{F}_p^* longmapsto x^5 in mathbb{F}_p^*$ is bijective, $f$ is a bijection from $mathbb{F}_p$ to itself. So it has a unique root. Compare $f$ with its derivative to prove that the root is simple iff $p neq 5$.
$endgroup$
– Mindlack
Jan 5 at 20:39
$begingroup$
Since $x in mathbb{F}_p^* longmapsto x^5 in mathbb{F}_p^*$ is bijective, $f$ is a bijection from $mathbb{F}_p$ to itself. So it has a unique root. Compare $f$ with its derivative to prove that the root is simple iff $p neq 5$.
$endgroup$
– Mindlack
Jan 5 at 20:39
$begingroup$
So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
$endgroup$
– greedoid
Jan 5 at 20:44
$begingroup$
So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
$endgroup$
– greedoid
Jan 5 at 20:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint If $alpha$ is a multiple root, then $f'(alpha)=0$.
Alternate solution Use the fact that there exists a primitive root $b$ modulo $p$.
Write $x=b^k, 2=b^alpha$ and solve for $k$.
answered Jan 5 at 20:32
N. S.N. S.
104k7114209
$endgroup$
add a comment |
$begingroup$
Let $p= 5k+r$ where $rin{0,2,3,4}$. Let us prove that $${0^5-2,1^5-2,...,(p-1)^5-2}=_{pmod p} {0,1,2,...,p-1}$$
Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
Since by Fermat theorem we have $a^{5k+r-1}equiv _p1$ we deduce:$$a^{5k}equiv _pb^{5k}implies a^{r-1}equiv_pb^{r-1} $$
Case 1: $r=0$ (so $p=5$) then $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.
Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.
Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.
Case 4: $r=4$ then $a^3equiv_p b^3$, then $a^6equiv_5 b^6 equiv a^5b implies pmid a^5(a-b) implies pmid a implies pmid b implies a=b$ a contradiction.
answered Jan 5 at 21:00
greedoidgreedoid
46.1k1160117
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Hint If $alpha$ is a multiple root, then $f'(alpha)=0$.
Alternate solution Use the fact that there exists a primitive root $b$ modulo $p$.
Write $x=b^k, 2=b^alpha$ and solve for $k$.
answered Jan 5 at 20:32
N. S.N. S.
104k7114209
$endgroup$
add a comment |
$begingroup$
Hint If $alpha$ is a multiple root, then $f'(alpha)=0$.
Alternate solution Use the fact that there exists a primitive root $b$ modulo $p$.
Write $x=b^k, 2=b^alpha$ and solve for $k$.
answered Jan 5 at 20:32
N. S.N. S.
104k7114209
$endgroup$
add a comment |
$begingroup$
Hint If $alpha$ is a multiple root, then $f'(alpha)=0$.
Alternate solution Use the fact that there exists a primitive root $b$ modulo $p$.
Write $x=b^k, 2=b^alpha$ and solve for $k$.
answered Jan 5 at 20:32
N. S.N. S.
104k7114209
$endgroup$
Hint If $alpha$ is a multiple root, then $f'(alpha)=0$.
Alternate solution Use the fact that there exists a primitive root $b$ modulo $p$.
Write $x=b^k, 2=b^alpha$ and solve for $k$.
answered Jan 5 at 20:32
N. S.N. S.
104k7114209
answered Jan 5 at 20:32
N. S.N. S.
104k7114209
answered Jan 5 at 20:32
N. S.N. S.
104k7114209
answered Jan 5 at 20:32
N. S.N. S.
104k7114209
104k7114209
add a comment |
add a comment |
$begingroup$
Let $p= 5k+r$ where $rin{0,2,3,4}$. Let us prove that $${0^5-2,1^5-2,...,(p-1)^5-2}=_{pmod p} {0,1,2,...,p-1}$$
Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
Since by Fermat theorem we have $a^{5k+r-1}equiv _p1$ we deduce:$$a^{5k}equiv _pb^{5k}implies a^{r-1}equiv_pb^{r-1} $$
Case 1: $r=0$ (so $p=5$) then $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.
Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.
Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.
Case 4: $r=4$ then $a^3equiv_p b^3$, then $a^6equiv_5 b^6 equiv a^5b implies pmid a^5(a-b) implies pmid a implies pmid b implies a=b$ a contradiction.
answered Jan 5 at 21:00
greedoidgreedoid
46.1k1160117
$endgroup$
add a comment |
$begingroup$
Let $p= 5k+r$ where $rin{0,2,3,4}$. Let us prove that $${0^5-2,1^5-2,...,(p-1)^5-2}=_{pmod p} {0,1,2,...,p-1}$$
Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
Since by Fermat theorem we have $a^{5k+r-1}equiv _p1$ we deduce:$$a^{5k}equiv _pb^{5k}implies a^{r-1}equiv_pb^{r-1} $$
Case 1: $r=0$ (so $p=5$) then $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.
Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.
Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.
Case 4: $r=4$ then $a^3equiv_p b^3$, then $a^6equiv_5 b^6 equiv a^5b implies pmid a^5(a-b) implies pmid a implies pmid b implies a=b$ a contradiction.
answered Jan 5 at 21:00
greedoidgreedoid
46.1k1160117
$endgroup$
add a comment |
$begingroup$
Let $p= 5k+r$ where $rin{0,2,3,4}$. Let us prove that $${0^5-2,1^5-2,...,(p-1)^5-2}=_{pmod p} {0,1,2,...,p-1}$$
Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
Since by Fermat theorem we have $a^{5k+r-1}equiv _p1$ we deduce:$$a^{5k}equiv _pb^{5k}implies a^{r-1}equiv_pb^{r-1} $$
Case 1: $r=0$ (so $p=5$) then $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.
Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.
Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.
Case 4: $r=4$ then $a^3equiv_p b^3$, then $a^6equiv_5 b^6 equiv a^5b implies pmid a^5(a-b) implies pmid a implies pmid b implies a=b$ a contradiction.
answered Jan 5 at 21:00
greedoidgreedoid
46.1k1160117
$endgroup$
Let $p= 5k+r$ where $rin{0,2,3,4}$. Let us prove that $${0^5-2,1^5-2,...,(p-1)^5-2}=_{pmod p} {0,1,2,...,p-1}$$
Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
Since by Fermat theorem we have $a^{5k+r-1}equiv _p1$ we deduce:$$a^{5k}equiv _pb^{5k}implies a^{r-1}equiv_pb^{r-1} $$
Case 1: $r=0$ (so $p=5$) then $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.
Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.
Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.
Case 4: $r=4$ then $a^3equiv_p b^3$, then $a^6equiv_5 b^6 equiv a^5b implies pmid a^5(a-b) implies pmid a implies pmid b implies a=b$ a contradiction.
answered Jan 5 at 21:00
greedoidgreedoid
46.1k1160117
answered Jan 5 at 21:00
greedoidgreedoid
46.1k1160117
answered Jan 5 at 21:00
greedoidgreedoid
46.1k1160117
answered Jan 5 at 21:00
greedoidgreedoid
46.1k1160117
46.1k1160117
add a comment |
add a comment |
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1
$begingroup$
Just use that $phi$ is injective.
$endgroup$
– Mark Bennet
Jan 5 at 20:34
$begingroup$
Since $x in mathbb{F}_p^* longmapsto x^5 in mathbb{F}_p^*$ is bijective, $f$ is a bijection from $mathbb{F}_p$ to itself. So it has a unique root. Compare $f$ with its derivative to prove that the root is simple iff $p neq 5$.
$endgroup$
– Mindlack
Jan 5 at 20:39
$begingroup$
So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
$endgroup$
– greedoid
Jan 5 at 20:44