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Polynomial Version for Fermat Last Theorem with $mathbbZ_p$
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How to prove the polynomial version for Fermat Last Theorem with $Z_p$, that is prove there exists no non-constant coprime solution $f(x), g(x), h(x) in Z_p[x]$ satisfies the following equation $$f(x)^n + g(x)^n = h(x)^n$$,
I have already proven the polynomial version for Fermat Last Theorem in $mathbbC[x]$. It can be used if necessary. Any hint will be helpful. Thanks in advance.
abstract-algebra
asked Jul 16 at 8:17
user577443
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up vote
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How to prove the polynomial version for Fermat Last Theorem with $Z_p$, that is prove there exists no non-constant coprime solution $f(x), g(x), h(x) in Z_p[x]$ satisfies the following equation $$f(x)^n + g(x)^n = h(x)^n$$,
I have already proven the polynomial version for Fermat Last Theorem in $mathbbC[x]$. It can be used if necessary. Any hint will be helpful. Thanks in advance.
abstract-algebra
asked Jul 16 at 8:17
user577443
112
With $mathbb Z_p$ do you mean integers modulo $p$ or the $p$-adic integers? In the former case, it's false: $x^p+(x+1)^p=(2x+1)^p$. In the latter case, the same proof as for $mathbb C[x]$ should work, perhaps with some modifications, but I don't know the details of your proof so can't elaborate.
– Wojowu
Jul 16 at 9:03
I mean $mathbbZ_p$ is the integer mod p, and the power is n,which is a number that is coprime with p, (In other words, it is not p or a multiple of p)
– user577443
Jul 16 at 12:43
Check out Mason's theorem, of which this is an immediate corollary. en.wikipedia.org/wiki/Mason%E2%80%93Stothers_theorem
– Badam Baplan
Jul 16 at 14:07
Is that only works for $mathbbC$?
– user577443
Jul 16 at 15:52
The same thing works for arbitrary characteristics by way of a quick reduction. I detailed it below, hope it helps.
– Badam Baplan
Jul 16 at 18:22
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
How to prove the polynomial version for Fermat Last Theorem with $Z_p$, that is prove there exists no non-constant coprime solution $f(x), g(x), h(x) in Z_p[x]$ satisfies the following equation $$f(x)^n + g(x)^n = h(x)^n$$,
I have already proven the polynomial version for Fermat Last Theorem in $mathbbC[x]$. It can be used if necessary. Any hint will be helpful. Thanks in advance.
abstract-algebra
asked Jul 16 at 8:17
user577443
112
How to prove the polynomial version for Fermat Last Theorem with $Z_p$, that is prove there exists no non-constant coprime solution $f(x), g(x), h(x) in Z_p[x]$ satisfies the following equation $$f(x)^n + g(x)^n = h(x)^n$$,
I have already proven the polynomial version for Fermat Last Theorem in $mathbbC[x]$. It can be used if necessary. Any hint will be helpful. Thanks in advance.
abstract-algebra
asked Jul 16 at 8:17
user577443
112
edited Jul 16 at 8:22
asked Jul 16 at 8:17
user577443
112
asked Jul 16 at 8:17
user577443
112
asked Jul 16 at 8:17
user577443
112
112
With $mathbb Z_p$ do you mean integers modulo $p$ or the $p$-adic integers? In the former case, it's false: $x^p+(x+1)^p=(2x+1)^p$. In the latter case, the same proof as for $mathbb C[x]$ should work, perhaps with some modifications, but I don't know the details of your proof so can't elaborate.
– Wojowu
Jul 16 at 9:03
I mean $mathbbZ_p$ is the integer mod p, and the power is n,which is a number that is coprime with p, (In other words, it is not p or a multiple of p)
– user577443
Jul 16 at 12:43
Check out Mason's theorem, of which this is an immediate corollary. en.wikipedia.org/wiki/Mason%E2%80%93Stothers_theorem
– Badam Baplan
Jul 16 at 14:07
Is that only works for $mathbbC$?
– user577443
Jul 16 at 15:52
The same thing works for arbitrary characteristics by way of a quick reduction. I detailed it below, hope it helps.
– Badam Baplan
Jul 16 at 18:22
add a comment |Â
With $mathbb Z_p$ do you mean integers modulo $p$ or the $p$-adic integers? In the former case, it's false: $x^p+(x+1)^p=(2x+1)^p$. In the latter case, the same proof as for $mathbb C[x]$ should work, perhaps with some modifications, but I don't know the details of your proof so can't elaborate.
– Wojowu
Jul 16 at 9:03
I mean $mathbbZ_p$ is the integer mod p, and the power is n,which is a number that is coprime with p, (In other words, it is not p or a multiple of p)
– user577443
Jul 16 at 12:43
Check out Mason's theorem, of which this is an immediate corollary. en.wikipedia.org/wiki/Mason%E2%80%93Stothers_theorem
– Badam Baplan
Jul 16 at 14:07
Is that only works for $mathbbC$?
– user577443
Jul 16 at 15:52
The same thing works for arbitrary characteristics by way of a quick reduction. I detailed it below, hope it helps.
– Badam Baplan
Jul 16 at 18:22
With $mathbb Z_p$ do you mean integers modulo $p$ or the $p$-adic integers? In the former case, it's false: $x^p+(x+1)^p=(2x+1)^p$. In the latter case, the same proof as for $mathbb C[x]$ should work, perhaps with some modifications, but I don't know the details of your proof so can't elaborate.
– Wojowu
Jul 16 at 9:03
With $mathbb Z_p$ do you mean integers modulo $p$ or the $p$-adic integers? In the former case, it's false: $x^p+(x+1)^p=(2x+1)^p$. In the latter case, the same proof as for $mathbb C[x]$ should work, perhaps with some modifications, but I don't know the details of your proof so can't elaborate.
– Wojowu
Jul 16 at 9:03
I mean $mathbbZ_p$ is the integer mod p, and the power is n,which is a number that is coprime with p, (In other words, it is not p or a multiple of p)
– user577443
Jul 16 at 12:43
I mean $mathbbZ_p$ is the integer mod p, and the power is n,which is a number that is coprime with p, (In other words, it is not p or a multiple of p)
– user577443
Jul 16 at 12:43
Check out Mason's theorem, of which this is an immediate corollary. en.wikipedia.org/wiki/Mason%E2%80%93Stothers_theorem
– Badam Baplan
Jul 16 at 14:07
Check out Mason's theorem, of which this is an immediate corollary. en.wikipedia.org/wiki/Mason%E2%80%93Stothers_theorem
– Badam Baplan
Jul 16 at 14:07
Is that only works for $mathbbC$?
– user577443
Jul 16 at 15:52
Is that only works for $mathbbC$?
– user577443
Jul 16 at 15:52
The same thing works for arbitrary characteristics by way of a quick reduction. I detailed it below, hope it helps.
– Badam Baplan
Jul 16 at 18:22
The same thing works for arbitrary characteristics by way of a quick reduction. I detailed it below, hope it helps.
– Badam Baplan
Jul 16 at 18:22
add a comment |Â
1 Answer
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Let's derive it as a corollary to the Mason-Stothers theorem, which states
Let $F$ a field and $f, g, h in F[x]$ coprime with $f + g = h$ and at least one of $f', g', h'$ nonzero. Then
$$maxdeg(f),deg(g),deg(h) leq degbig(textrad (fgh)big)-1 $$
where $textrad(f) equiv prodlimits_textirred p mid f p$
In fields of characteristic $0$, like $mathbbC$, having nonvanishing derivative is equivalent to being nonconstant. But in fields of characteristic $p > 0$, this is not the case. However, things aren't very much more complicated, for our purposes: if $F$ has characteristic $p > 0$, then the kernel of the derivative $F[x] rightarrow F[x]$ is precisely $F[x^p]$. We can use this fact to reduce the proof for characteristic $p>0$ to the proof for characteristic $0$.
Now suppose we are in a field $F$ of characteristic $p > 0$. If we have $f^n + g^n = h^n$, $gcd(n,p) = 1$, then $(f^n)' = nf'f^n-1 =0$ iff $f' = 0$ iff $f in F[x^p]$, and similarly for $(g^n)'$ and $(h^n)'$.
We thus observe that
There exists $m in N$ such that, setting $barf = f(x^1/p^m), barg = g(x^1/p^m), barh = h(x^1/p^m)$, we have that $barf, barg, barh in F[x]$ are coprime, $barf^n + barg^n = barh^n$, and one of $barf, barg, barh$ has nonvanishing derivative.
Indeed if all of $f^n, g^n, h^n$ have vanishing derivatives, then $f,g,h in F[x^p]$, and setting $barf = f(x^1/p), barg = g(x^1/p), barh = h(x^1/p)$ (which are certainly in $F[x]$) we get a relation $barf^n + barg^n = barh^n$. If all of $barf, barg, barh$ have vanishing derivatives, we can repeat the process, on so and.... you can confirm that it terminates after finitely many steps with the desired polynomials.
The above reduction gives us a relation
$$barf^n + barg^n = barh^n$$
which satisfies all the conditions of Mason-Stothers as stated above. The argument is now the same as in the case of $F$ having characteristic $0$. The bounds $$degbig(textrad(barf^n barg^n barh^nbig) - 1 < deg(barf) + deg(barg) + deg(barh)$$ and
$$max deg( barf^n), deg( barg^n), deg( barh^n ) geq fracn3big(deg( barf) + deg( barg) + deg( barh ) big) $$
are immediate. When $n > 2$ this blatantly contradicts the conclusion of Mason-Stothers, yielding the "Fermat's last theorem" for polynomials over fields of arbitrary characteristic.
answered Jul 16 at 18:20
Badam Baplan
3,331721
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let's derive it as a corollary to the Mason-Stothers theorem, which states
Let $F$ a field and $f, g, h in F[x]$ coprime with $f + g = h$ and at least one of $f', g', h'$ nonzero. Then
$$maxdeg(f),deg(g),deg(h) leq degbig(textrad (fgh)big)-1 $$
where $textrad(f) equiv prodlimits_textirred p mid f p$
In fields of characteristic $0$, like $mathbbC$, having nonvanishing derivative is equivalent to being nonconstant. But in fields of characteristic $p > 0$, this is not the case. However, things aren't very much more complicated, for our purposes: if $F$ has characteristic $p > 0$, then the kernel of the derivative $F[x] rightarrow F[x]$ is precisely $F[x^p]$. We can use this fact to reduce the proof for characteristic $p>0$ to the proof for characteristic $0$.
Now suppose we are in a field $F$ of characteristic $p > 0$. If we have $f^n + g^n = h^n$, $gcd(n,p) = 1$, then $(f^n)' = nf'f^n-1 =0$ iff $f' = 0$ iff $f in F[x^p]$, and similarly for $(g^n)'$ and $(h^n)'$.
We thus observe that
There exists $m in N$ such that, setting $barf = f(x^1/p^m), barg = g(x^1/p^m), barh = h(x^1/p^m)$, we have that $barf, barg, barh in F[x]$ are coprime, $barf^n + barg^n = barh^n$, and one of $barf, barg, barh$ has nonvanishing derivative.
Indeed if all of $f^n, g^n, h^n$ have vanishing derivatives, then $f,g,h in F[x^p]$, and setting $barf = f(x^1/p), barg = g(x^1/p), barh = h(x^1/p)$ (which are certainly in $F[x]$) we get a relation $barf^n + barg^n = barh^n$. If all of $barf, barg, barh$ have vanishing derivatives, we can repeat the process, on so and.... you can confirm that it terminates after finitely many steps with the desired polynomials.
The above reduction gives us a relation
$$barf^n + barg^n = barh^n$$
which satisfies all the conditions of Mason-Stothers as stated above. The argument is now the same as in the case of $F$ having characteristic $0$. The bounds $$degbig(textrad(barf^n barg^n barh^nbig) - 1 < deg(barf) + deg(barg) + deg(barh)$$ and
$$max deg( barf^n), deg( barg^n), deg( barh^n ) geq fracn3big(deg( barf) + deg( barg) + deg( barh ) big) $$
are immediate. When $n > 2$ this blatantly contradicts the conclusion of Mason-Stothers, yielding the "Fermat's last theorem" for polynomials over fields of arbitrary characteristic.
answered Jul 16 at 18:20
Badam Baplan
3,331721
add a comment |Â
up vote
1
down vote
Let's derive it as a corollary to the Mason-Stothers theorem, which states
Let $F$ a field and $f, g, h in F[x]$ coprime with $f + g = h$ and at least one of $f', g', h'$ nonzero. Then
$$maxdeg(f),deg(g),deg(h) leq degbig(textrad (fgh)big)-1 $$
where $textrad(f) equiv prodlimits_textirred p mid f p$
In fields of characteristic $0$, like $mathbbC$, having nonvanishing derivative is equivalent to being nonconstant. But in fields of characteristic $p > 0$, this is not the case. However, things aren't very much more complicated, for our purposes: if $F$ has characteristic $p > 0$, then the kernel of the derivative $F[x] rightarrow F[x]$ is precisely $F[x^p]$. We can use this fact to reduce the proof for characteristic $p>0$ to the proof for characteristic $0$.
Now suppose we are in a field $F$ of characteristic $p > 0$. If we have $f^n + g^n = h^n$, $gcd(n,p) = 1$, then $(f^n)' = nf'f^n-1 =0$ iff $f' = 0$ iff $f in F[x^p]$, and similarly for $(g^n)'$ and $(h^n)'$.
We thus observe that
There exists $m in N$ such that, setting $barf = f(x^1/p^m), barg = g(x^1/p^m), barh = h(x^1/p^m)$, we have that $barf, barg, barh in F[x]$ are coprime, $barf^n + barg^n = barh^n$, and one of $barf, barg, barh$ has nonvanishing derivative.
Indeed if all of $f^n, g^n, h^n$ have vanishing derivatives, then $f,g,h in F[x^p]$, and setting $barf = f(x^1/p), barg = g(x^1/p), barh = h(x^1/p)$ (which are certainly in $F[x]$) we get a relation $barf^n + barg^n = barh^n$. If all of $barf, barg, barh$ have vanishing derivatives, we can repeat the process, on so and.... you can confirm that it terminates after finitely many steps with the desired polynomials.
The above reduction gives us a relation
$$barf^n + barg^n = barh^n$$
which satisfies all the conditions of Mason-Stothers as stated above. The argument is now the same as in the case of $F$ having characteristic $0$. The bounds $$degbig(textrad(barf^n barg^n barh^nbig) - 1 < deg(barf) + deg(barg) + deg(barh)$$ and
$$max deg( barf^n), deg( barg^n), deg( barh^n ) geq fracn3big(deg( barf) + deg( barg) + deg( barh ) big) $$
are immediate. When $n > 2$ this blatantly contradicts the conclusion of Mason-Stothers, yielding the "Fermat's last theorem" for polynomials over fields of arbitrary characteristic.
answered Jul 16 at 18:20
Badam Baplan
3,331721
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let's derive it as a corollary to the Mason-Stothers theorem, which states
Let $F$ a field and $f, g, h in F[x]$ coprime with $f + g = h$ and at least one of $f', g', h'$ nonzero. Then
$$maxdeg(f),deg(g),deg(h) leq degbig(textrad (fgh)big)-1 $$
where $textrad(f) equiv prodlimits_textirred p mid f p$
In fields of characteristic $0$, like $mathbbC$, having nonvanishing derivative is equivalent to being nonconstant. But in fields of characteristic $p > 0$, this is not the case. However, things aren't very much more complicated, for our purposes: if $F$ has characteristic $p > 0$, then the kernel of the derivative $F[x] rightarrow F[x]$ is precisely $F[x^p]$. We can use this fact to reduce the proof for characteristic $p>0$ to the proof for characteristic $0$.
Now suppose we are in a field $F$ of characteristic $p > 0$. If we have $f^n + g^n = h^n$, $gcd(n,p) = 1$, then $(f^n)' = nf'f^n-1 =0$ iff $f' = 0$ iff $f in F[x^p]$, and similarly for $(g^n)'$ and $(h^n)'$.
We thus observe that
There exists $m in N$ such that, setting $barf = f(x^1/p^m), barg = g(x^1/p^m), barh = h(x^1/p^m)$, we have that $barf, barg, barh in F[x]$ are coprime, $barf^n + barg^n = barh^n$, and one of $barf, barg, barh$ has nonvanishing derivative.
Indeed if all of $f^n, g^n, h^n$ have vanishing derivatives, then $f,g,h in F[x^p]$, and setting $barf = f(x^1/p), barg = g(x^1/p), barh = h(x^1/p)$ (which are certainly in $F[x]$) we get a relation $barf^n + barg^n = barh^n$. If all of $barf, barg, barh$ have vanishing derivatives, we can repeat the process, on so and.... you can confirm that it terminates after finitely many steps with the desired polynomials.
The above reduction gives us a relation
$$barf^n + barg^n = barh^n$$
which satisfies all the conditions of Mason-Stothers as stated above. The argument is now the same as in the case of $F$ having characteristic $0$. The bounds $$degbig(textrad(barf^n barg^n barh^nbig) - 1 < deg(barf) + deg(barg) + deg(barh)$$ and
$$max deg( barf^n), deg( barg^n), deg( barh^n ) geq fracn3big(deg( barf) + deg( barg) + deg( barh ) big) $$
are immediate. When $n > 2$ this blatantly contradicts the conclusion of Mason-Stothers, yielding the "Fermat's last theorem" for polynomials over fields of arbitrary characteristic.
answered Jul 16 at 18:20
Badam Baplan
3,331721
Let's derive it as a corollary to the Mason-Stothers theorem, which states
Let $F$ a field and $f, g, h in F[x]$ coprime with $f + g = h$ and at least one of $f', g', h'$ nonzero. Then
$$maxdeg(f),deg(g),deg(h) leq degbig(textrad (fgh)big)-1 $$
where $textrad(f) equiv prodlimits_textirred p mid f p$
In fields of characteristic $0$, like $mathbbC$, having nonvanishing derivative is equivalent to being nonconstant. But in fields of characteristic $p > 0$, this is not the case. However, things aren't very much more complicated, for our purposes: if $F$ has characteristic $p > 0$, then the kernel of the derivative $F[x] rightarrow F[x]$ is precisely $F[x^p]$. We can use this fact to reduce the proof for characteristic $p>0$ to the proof for characteristic $0$.
Now suppose we are in a field $F$ of characteristic $p > 0$. If we have $f^n + g^n = h^n$, $gcd(n,p) = 1$, then $(f^n)' = nf'f^n-1 =0$ iff $f' = 0$ iff $f in F[x^p]$, and similarly for $(g^n)'$ and $(h^n)'$.
We thus observe that
There exists $m in N$ such that, setting $barf = f(x^1/p^m), barg = g(x^1/p^m), barh = h(x^1/p^m)$, we have that $barf, barg, barh in F[x]$ are coprime, $barf^n + barg^n = barh^n$, and one of $barf, barg, barh$ has nonvanishing derivative.
Indeed if all of $f^n, g^n, h^n$ have vanishing derivatives, then $f,g,h in F[x^p]$, and setting $barf = f(x^1/p), barg = g(x^1/p), barh = h(x^1/p)$ (which are certainly in $F[x]$) we get a relation $barf^n + barg^n = barh^n$. If all of $barf, barg, barh$ have vanishing derivatives, we can repeat the process, on so and.... you can confirm that it terminates after finitely many steps with the desired polynomials.
The above reduction gives us a relation
$$barf^n + barg^n = barh^n$$
which satisfies all the conditions of Mason-Stothers as stated above. The argument is now the same as in the case of $F$ having characteristic $0$. The bounds $$degbig(textrad(barf^n barg^n barh^nbig) - 1 < deg(barf) + deg(barg) + deg(barh)$$ and
$$max deg( barf^n), deg( barg^n), deg( barh^n ) geq fracn3big(deg( barf) + deg( barg) + deg( barh ) big) $$
are immediate. When $n > 2$ this blatantly contradicts the conclusion of Mason-Stothers, yielding the "Fermat's last theorem" for polynomials over fields of arbitrary characteristic.
answered Jul 16 at 18:20
Badam Baplan
3,331721
answered Jul 16 at 18:20
Badam Baplan
3,331721
answered Jul 16 at 18:20
Badam Baplan
3,331721
answered Jul 16 at 18:20
Badam Baplan
3,331721
3,331721
add a comment |Â
add a comment |Â
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With $mathbb Z_p$ do you mean integers modulo $p$ or the $p$-adic integers? In the former case, it's false: $x^p+(x+1)^p=(2x+1)^p$. In the latter case, the same proof as for $mathbb C[x]$ should work, perhaps with some modifications, but I don't know the details of your proof so can't elaborate.
– Wojowu
Jul 16 at 9:03
I mean $mathbbZ_p$ is the integer mod p, and the power is n,which is a number that is coprime with p, (In other words, it is not p or a multiple of p)
– user577443
Jul 16 at 12:43
Check out Mason's theorem, of which this is an immediate corollary. en.wikipedia.org/wiki/Mason%E2%80%93Stothers_theorem
– Badam Baplan
Jul 16 at 14:07
Is that only works for $mathbbC$?
– user577443
Jul 16 at 15:52
The same thing works for arbitrary characteristics by way of a quick reduction. I detailed it below, hope it helps.
– Badam Baplan
Jul 16 at 18:22