$F$-Automorphism of $F(w)$ induces automorphism on ${1,w,w^2,…,w^{n-1} }$ only if $w$ primitive n-th root?
1
$begingroup$
I just stumble across a theorem in Galois theory, that says that if F is a field of characteristic p and $w$ is a primitive n-th root of unity with $p nmid n$ than if we take a $F$-automorphism $sigma$ of $F(w)$, then $sigma$ induces an automorphism $sigma_0$ of ${1,w,w^2,...,w^{n-1}}$ by restricting $sigma$ to this set. But I was wondering : What could go wrong if $w$ was just an algebraic element over $F$ and not a primitive root of unity?
galois-theory
asked Dec 1 '18 at 15:03
roi_saumonroi_saumon
45628
$endgroup$
add a comment |
1
$begingroup$
I just stumble across a theorem in Galois theory, that says that if F is a field of characteristic p and $w$ is a primitive n-th root of unity with $p nmid n$ than if we take a $F$-automorphism $sigma$ of $F(w)$, then $sigma$ induces an automorphism $sigma_0$ of ${1,w,w^2,...,w^{n-1}}$ by restricting $sigma$ to this set. But I was wondering : What could go wrong if $w$ was just an algebraic element over $F$ and not a primitive root of unity?
galois-theory
asked Dec 1 '18 at 15:03
roi_saumonroi_saumon
45628
$endgroup$
$begingroup$
The point is that ${1,w,w^2,ldots, w^{n-1}}$ is a group under multiplication. This clearly is not the case when $w$ is just algebraic.
$endgroup$
– freakish
Dec 1 '18 at 17:21
$begingroup$
@freakish: Surely the powers of any element of $F(w)$ form a group under multiplication irrespective of whether $w$ is a root of unity or not. I think this is more about the group of units of order dividing $n$ being unique, and hence fixed by $sigma$ (as a set).
$endgroup$
– Jyrki Lahtonen
Dec 1 '18 at 20:12
1
$begingroup$
So what could wrong is that unless $w$ has finite order it may happen that $sigma(w)$ is not a power of $w$ at all. For example, let $F=Bbb{F}_2(x)$ be a field of rational functions, $x$ transcental over the prime field. Assume that $w$ satisfies the equation $w^2+w=x$. Then $w$ is algebraic over $F$ and the only non-trivial automorphism of $F(w)/F$ is determined by $sigma(w)=w+1$. We cannot have $w+1=w^k$ for any $kinBbb{Z}$ for then $w$ would be algebraic over the prime field.
$endgroup$
– Jyrki Lahtonen
Dec 1 '18 at 20:19
$begingroup$
Thanks, but why the only non-trivial F-automorphism is $sigma(w)=w+1$? Shouldn't it send $w$ to a root of the minimal polynomial $y^2+y-x$?
$endgroup$
– roi_saumon
Dec 1 '18 at 21:10
1
$begingroup$
The roots of $y^2+y-x$ are $w$ and $w+1$. Characteristic two is essential.
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 5:30
add a comment |
1
1
1
$begingroup$
I just stumble across a theorem in Galois theory, that says that if F is a field of characteristic p and $w$ is a primitive n-th root of unity with $p nmid n$ than if we take a $F$-automorphism $sigma$ of $F(w)$, then $sigma$ induces an automorphism $sigma_0$ of ${1,w,w^2,...,w^{n-1}}$ by restricting $sigma$ to this set. But I was wondering : What could go wrong if $w$ was just an algebraic element over $F$ and not a primitive root of unity?
galois-theory
asked Dec 1 '18 at 15:03
roi_saumonroi_saumon
45628
$endgroup$
I just stumble across a theorem in Galois theory, that says that if F is a field of characteristic p and $w$ is a primitive n-th root of unity with $p nmid n$ than if we take a $F$-automorphism $sigma$ of $F(w)$, then $sigma$ induces an automorphism $sigma_0$ of ${1,w,w^2,...,w^{n-1}}$ by restricting $sigma$ to this set. But I was wondering : What could go wrong if $w$ was just an algebraic element over $F$ and not a primitive root of unity?
galois-theory
galois-theory
asked Dec 1 '18 at 15:03
roi_saumonroi_saumon
45628
asked Dec 1 '18 at 15:03
roi_saumonroi_saumon
45628
asked Dec 1 '18 at 15:03
roi_saumonroi_saumon
45628
asked Dec 1 '18 at 15:03
roi_saumonroi_saumon
45628
asked Dec 1 '18 at 15:03
roi_saumonroi_saumon
45628
45628
$begingroup$
The point is that ${1,w,w^2,ldots, w^{n-1}}$ is a group under multiplication. This clearly is not the case when $w$ is just algebraic.
$endgroup$
– freakish
Dec 1 '18 at 17:21
$begingroup$
@freakish: Surely the powers of any element of $F(w)$ form a group under multiplication irrespective of whether $w$ is a root of unity or not. I think this is more about the group of units of order dividing $n$ being unique, and hence fixed by $sigma$ (as a set).
$endgroup$
– Jyrki Lahtonen
Dec 1 '18 at 20:12
1
$begingroup$
So what could wrong is that unless $w$ has finite order it may happen that $sigma(w)$ is not a power of $w$ at all. For example, let $F=Bbb{F}_2(x)$ be a field of rational functions, $x$ transcental over the prime field. Assume that $w$ satisfies the equation $w^2+w=x$. Then $w$ is algebraic over $F$ and the only non-trivial automorphism of $F(w)/F$ is determined by $sigma(w)=w+1$. We cannot have $w+1=w^k$ for any $kinBbb{Z}$ for then $w$ would be algebraic over the prime field.
$endgroup$
– Jyrki Lahtonen
Dec 1 '18 at 20:19
$begingroup$
Thanks, but why the only non-trivial F-automorphism is $sigma(w)=w+1$? Shouldn't it send $w$ to a root of the minimal polynomial $y^2+y-x$?
$endgroup$
– roi_saumon
Dec 1 '18 at 21:10
1
$begingroup$
The roots of $y^2+y-x$ are $w$ and $w+1$. Characteristic two is essential.
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 5:30
add a comment |
$begingroup$
The point is that ${1,w,w^2,ldots, w^{n-1}}$ is a group under multiplication. This clearly is not the case when $w$ is just algebraic.
$endgroup$
– freakish
Dec 1 '18 at 17:21
$begingroup$
@freakish: Surely the powers of any element of $F(w)$ form a group under multiplication irrespective of whether $w$ is a root of unity or not. I think this is more about the group of units of order dividing $n$ being unique, and hence fixed by $sigma$ (as a set).
$endgroup$
– Jyrki Lahtonen
Dec 1 '18 at 20:12
1
$begingroup$
So what could wrong is that unless $w$ has finite order it may happen that $sigma(w)$ is not a power of $w$ at all. For example, let $F=Bbb{F}_2(x)$ be a field of rational functions, $x$ transcental over the prime field. Assume that $w$ satisfies the equation $w^2+w=x$. Then $w$ is algebraic over $F$ and the only non-trivial automorphism of $F(w)/F$ is determined by $sigma(w)=w+1$. We cannot have $w+1=w^k$ for any $kinBbb{Z}$ for then $w$ would be algebraic over the prime field.
$endgroup$
– Jyrki Lahtonen
Dec 1 '18 at 20:19
$begingroup$
Thanks, but why the only non-trivial F-automorphism is $sigma(w)=w+1$? Shouldn't it send $w$ to a root of the minimal polynomial $y^2+y-x$?
$endgroup$
– roi_saumon
Dec 1 '18 at 21:10
1
$begingroup$
The roots of $y^2+y-x$ are $w$ and $w+1$. Characteristic two is essential.
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 5:30
$begingroup$
The point is that ${1,w,w^2,ldots, w^{n-1}}$ is a group under multiplication. This clearly is not the case when $w$ is just algebraic.
$endgroup$
– freakish
Dec 1 '18 at 17:21
$begingroup$
The point is that ${1,w,w^2,ldots, w^{n-1}}$ is a group under multiplication. This clearly is not the case when $w$ is just algebraic.
$endgroup$
– freakish
Dec 1 '18 at 17:21
$begingroup$
@freakish: Surely the powers of any element of $F(w)$ form a group under multiplication irrespective of whether $w$ is a root of unity or not. I think this is more about the group of units of order dividing $n$ being unique, and hence fixed by $sigma$ (as a set).
$endgroup$
– Jyrki Lahtonen
Dec 1 '18 at 20:12
$begingroup$
@freakish: Surely the powers of any element of $F(w)$ form a group under multiplication irrespective of whether $w$ is a root of unity or not. I think this is more about the group of units of order dividing $n$ being unique, and hence fixed by $sigma$ (as a set).
$endgroup$
– Jyrki Lahtonen
Dec 1 '18 at 20:12
1
1
$begingroup$
So what could wrong is that unless $w$ has finite order it may happen that $sigma(w)$ is not a power of $w$ at all. For example, let $F=Bbb{F}_2(x)$ be a field of rational functions, $x$ transcental over the prime field. Assume that $w$ satisfies the equation $w^2+w=x$. Then $w$ is algebraic over $F$ and the only non-trivial automorphism of $F(w)/F$ is determined by $sigma(w)=w+1$. We cannot have $w+1=w^k$ for any $kinBbb{Z}$ for then $w$ would be algebraic over the prime field.
$endgroup$
– Jyrki Lahtonen
Dec 1 '18 at 20:19
$begingroup$
So what could wrong is that unless $w$ has finite order it may happen that $sigma(w)$ is not a power of $w$ at all. For example, let $F=Bbb{F}_2(x)$ be a field of rational functions, $x$ transcental over the prime field. Assume that $w$ satisfies the equation $w^2+w=x$. Then $w$ is algebraic over $F$ and the only non-trivial automorphism of $F(w)/F$ is determined by $sigma(w)=w+1$. We cannot have $w+1=w^k$ for any $kinBbb{Z}$ for then $w$ would be algebraic over the prime field.
$endgroup$
– Jyrki Lahtonen
Dec 1 '18 at 20:19
$begingroup$
Thanks, but why the only non-trivial F-automorphism is $sigma(w)=w+1$? Shouldn't it send $w$ to a root of the minimal polynomial $y^2+y-x$?
$endgroup$
– roi_saumon
Dec 1 '18 at 21:10
$begingroup$
Thanks, but why the only non-trivial F-automorphism is $sigma(w)=w+1$? Shouldn't it send $w$ to a root of the minimal polynomial $y^2+y-x$?
$endgroup$
– roi_saumon
Dec 1 '18 at 21:10
1
1
$begingroup$
The roots of $y^2+y-x$ are $w$ and $w+1$. Characteristic two is essential.
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 5:30
$begingroup$
The roots of $y^2+y-x$ are $w$ and $w+1$. Characteristic two is essential.
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 5:30
add a comment |
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$begingroup$
The point is that ${1,w,w^2,ldots, w^{n-1}}$ is a group under multiplication. This clearly is not the case when $w$ is just algebraic.
$endgroup$
– freakish
Dec 1 '18 at 17:21
$begingroup$
@freakish: Surely the powers of any element of $F(w)$ form a group under multiplication irrespective of whether $w$ is a root of unity or not. I think this is more about the group of units of order dividing $n$ being unique, and hence fixed by $sigma$ (as a set).
$endgroup$
– Jyrki Lahtonen
Dec 1 '18 at 20:12
1
$begingroup$
So what could wrong is that unless $w$ has finite order it may happen that $sigma(w)$ is not a power of $w$ at all. For example, let $F=Bbb{F}_2(x)$ be a field of rational functions, $x$ transcental over the prime field. Assume that $w$ satisfies the equation $w^2+w=x$. Then $w$ is algebraic over $F$ and the only non-trivial automorphism of $F(w)/F$ is determined by $sigma(w)=w+1$. We cannot have $w+1=w^k$ for any $kinBbb{Z}$ for then $w$ would be algebraic over the prime field.
$endgroup$
– Jyrki Lahtonen
Dec 1 '18 at 20:19
$begingroup$
Thanks, but why the only non-trivial F-automorphism is $sigma(w)=w+1$? Shouldn't it send $w$ to a root of the minimal polynomial $y^2+y-x$?
$endgroup$
– roi_saumon
Dec 1 '18 at 21:10
1
$begingroup$
The roots of $y^2+y-x$ are $w$ and $w+1$. Characteristic two is essential.
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 5:30