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How to understand the Artin-Schreier correspondence?
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Let $K$ be a field of characteristic $p > 0$. Then it is due to Artin and Schreier that the assignment
$$c in K mapsto text{Splitting field } L_c text{ of } X^p-X+c$$
induces a bijection between the non-trivial elements in $K/{a^p-a mid a in K}$ and the $K$-isomorphism classes of Galois extensions of degree $p$ over $K$.
In particular, this should imply that if $c, c' in K$ are such that $L_c$ and $L_{c'}$ are $K$-isomorphic, then there is $k in K$ such that $k^p-k = c-c'$.
However, what about the following example (which was suggested by user8268 in this question): Let $p > 2$ and $c in K setminus {a^p-a mid a in K}$ and let $alpha in L_c$ be a root of $x^p-x+c$. Then the roots of $x^p-x+2c$ are given by $2alpha + u$, where $u$ ranges through $mathbb{F}_p subseteq K$, hence $L_c = L_{2c}$. But $2c - c = c notin {a^p-a mid a in K}$.
How is this compatible with the Artin-Schreier correspondence? I am grateful for any help!
EDIT 1: Note that the Artin-Schreier correspondence is usually proved by constructing an inverse map, as it was done e.g. in this answer.
field-theory galois-theory splitting-field galois-extensions
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Let $K$ be a field of characteristic $p > 0$. Then it is due to Artin and Schreier that the assignment
$$c in K mapsto text{Splitting field } L_c text{ of } X^p-X+c$$
induces a bijection between the non-trivial elements in $K/{a^p-a mid a in K}$ and the $K$-isomorphism classes of Galois extensions of degree $p$ over $K$.
In particular, this should imply that if $c, c' in K$ are such that $L_c$ and $L_{c'}$ are $K$-isomorphic, then there is $k in K$ such that $k^p-k = c-c'$.
However, what about the following example (which was suggested by user8268 in this question): Let $p > 2$ and $c in K setminus {a^p-a mid a in K}$ and let $alpha in L_c$ be a root of $x^p-x+c$. Then the roots of $x^p-x+2c$ are given by $2alpha + u$, where $u$ ranges through $mathbb{F}_p subseteq K$, hence $L_c = L_{2c}$. But $2c - c = c notin {a^p-a mid a in K}$.
How is this compatible with the Artin-Schreier correspondence? I am grateful for any help!
EDIT 1: Note that the Artin-Schreier correspondence is usually proved by constructing an inverse map, as it was done e.g. in this answer.
field-theory galois-theory splitting-field galois-extensions
asked 2 days ago
Algebrus
407210
This question has an open bounty worth +50
reputation from Algebrus ending in 6 days.
This question has not received enough attention.
2
The bijection is rather between the Galois extensions of $K$ of order $p$ and $bigl(Ksetminus{a^p-a|ain K}bigr)/Bbb F_p^*$. (If $lambdainBbb F_p^*$ then $L_{lambda c}=L_c$.)
– user8268
2 days ago
@user8268: Yes, this seems to be the case, but why is the AS-correspondence always stated as I did it above?
– Algebrus
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $K$ be a field of characteristic $p > 0$. Then it is due to Artin and Schreier that the assignment
$$c in K mapsto text{Splitting field } L_c text{ of } X^p-X+c$$
induces a bijection between the non-trivial elements in $K/{a^p-a mid a in K}$ and the $K$-isomorphism classes of Galois extensions of degree $p$ over $K$.
In particular, this should imply that if $c, c' in K$ are such that $L_c$ and $L_{c'}$ are $K$-isomorphic, then there is $k in K$ such that $k^p-k = c-c'$.
However, what about the following example (which was suggested by user8268 in this question): Let $p > 2$ and $c in K setminus {a^p-a mid a in K}$ and let $alpha in L_c$ be a root of $x^p-x+c$. Then the roots of $x^p-x+2c$ are given by $2alpha + u$, where $u$ ranges through $mathbb{F}_p subseteq K$, hence $L_c = L_{2c}$. But $2c - c = c notin {a^p-a mid a in K}$.
How is this compatible with the Artin-Schreier correspondence? I am grateful for any help!
EDIT 1: Note that the Artin-Schreier correspondence is usually proved by constructing an inverse map, as it was done e.g. in this answer.
field-theory galois-theory splitting-field galois-extensions
asked 2 days ago
Algebrus
407210
Let $K$ be a field of characteristic $p > 0$. Then it is due to Artin and Schreier that the assignment
$$c in K mapsto text{Splitting field } L_c text{ of } X^p-X+c$$
induces a bijection between the non-trivial elements in $K/{a^p-a mid a in K}$ and the $K$-isomorphism classes of Galois extensions of degree $p$ over $K$.
In particular, this should imply that if $c, c' in K$ are such that $L_c$ and $L_{c'}$ are $K$-isomorphic, then there is $k in K$ such that $k^p-k = c-c'$.
However, what about the following example (which was suggested by user8268 in this question): Let $p > 2$ and $c in K setminus {a^p-a mid a in K}$ and let $alpha in L_c$ be a root of $x^p-x+c$. Then the roots of $x^p-x+2c$ are given by $2alpha + u$, where $u$ ranges through $mathbb{F}_p subseteq K$, hence $L_c = L_{2c}$. But $2c - c = c notin {a^p-a mid a in K}$.
How is this compatible with the Artin-Schreier correspondence? I am grateful for any help!
EDIT 1: Note that the Artin-Schreier correspondence is usually proved by constructing an inverse map, as it was done e.g. in this answer.
field-theory galois-theory splitting-field galois-extensions
field-theory galois-theory splitting-field galois-extensions
asked 2 days ago
Algebrus
407210
asked 2 days ago
Algebrus
407210
edited 2 days ago
asked 2 days ago
Algebrus
407210
asked 2 days ago
Algebrus
407210
asked 2 days ago
Algebrus
407210
407210
This question has an open bounty worth +50
reputation from Algebrus ending in 6 days.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from Algebrus ending in 6 days.
This question has not received enough attention.
2
The bijection is rather between the Galois extensions of $K$ of order $p$ and $bigl(Ksetminus{a^p-a|ain K}bigr)/Bbb F_p^*$. (If $lambdainBbb F_p^*$ then $L_{lambda c}=L_c$.)
– user8268
2 days ago
@user8268: Yes, this seems to be the case, but why is the AS-correspondence always stated as I did it above?
– Algebrus
2 days ago
add a comment |
2
The bijection is rather between the Galois extensions of $K$ of order $p$ and $bigl(Ksetminus{a^p-a|ain K}bigr)/Bbb F_p^*$. (If $lambdainBbb F_p^*$ then $L_{lambda c}=L_c$.)
– user8268
2 days ago
@user8268: Yes, this seems to be the case, but why is the AS-correspondence always stated as I did it above?
– Algebrus
2 days ago
2
2
The bijection is rather between the Galois extensions of $K$ of order $p$ and $bigl(Ksetminus{a^p-a|ain K}bigr)/Bbb F_p^*$. (If $lambdainBbb F_p^*$ then $L_{lambda c}=L_c$.)
– user8268
2 days ago
The bijection is rather between the Galois extensions of $K$ of order $p$ and $bigl(Ksetminus{a^p-a|ain K}bigr)/Bbb F_p^*$. (If $lambdainBbb F_p^*$ then $L_{lambda c}=L_c$.)
– user8268
2 days ago
@user8268: Yes, this seems to be the case, but why is the AS-correspondence always stated as I did it above?
– Algebrus
2 days ago
@user8268: Yes, this seems to be the case, but why is the AS-correspondence always stated as I did it above?
– Algebrus
2 days ago
add a comment |
1 Answer
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I should really check out a definite source, but I think the Artin-Schreier correspondence means the following. To summarize, the problem observed by user8268 can be resolved by insisting that the Galois groups should have a preferred generator.
So something like the following.
Let $L$ and $L'$ be two cyclic degree $p$ extensions of $K$, and let $sigma$ (resp. $sigma'$) be the respective preferred generators. We call $(L,sigma)$ and $(L',sigma')$ equivalent, if there exists a $K$-isomorphism $psi:Lto L'$ such that $$psicircsigma=sigma'circpsi.$$
Then the AS-correspondence is a bijection between the non-trivial cosets of the subgroup $A={x^p-xmid xin K}le(K,+)$ and the equivalence classes of pairs $(L,sigma)$. If $c+A$ is a non-trivial coset of $A$ then it corresponds to an extension $L=K(beta)$ with $beta^p-beta+c=0$ together with the preferred automorphism $sigma:Lto L$ uniquely determined by $sigma(beta)=beta+1$.
A consequence of this formulation is that while the splitting fields of $x^p-x+c$ and $x^p-x+2c$ are both equal to $L=K(beta)$, the above correspondence associates a different generator of the Galois group with the latter polynomial. The automorphism $sigma$ that maps $betamapsto beta+1$ will map $2betamapsto 2beta+2$, so we should associate $x^p-x+2c$ with the pair $(K(beta),sigma^2)$ instead of $(K(beta),sigma)$. Observe that those two pairs cannot be equivalent according to the above definition because $sigma'=sigma^2$ (and here every possible $psi$ commutes with $sigma$).
A few closing remarks:
- I may be out of my depth here, but I hazard a guess that the interpretation of this result in terms of Galois cohomology makes it necessary to specify the action of a fixed generator of the cyclic group.
- An analogous problem is present in Kummer theory also. For example, with $K=Bbb{Q}(omega)$, $omega=e^{2pi i/3},$ the cyclic cubic extensions $K(root3of2)$ and $K(root3of4)$ are clearly equal (as subsets of $Bbb{C}$) even though $2$ and $4$ belong to distinct (multiplicative) cosets of the subgroup of cubes of $K^*$. Here we also resolve the problem by observing that the automorphisms of $K(root3of2)$ that map $root3of2mapstoomegaroot3of2$ and $root3of4mapsto omegaroot3of4$ are similarly squares of each other, i.e. distinct generators of the Galois group.
answered 16 hours ago
Jyrki Lahtonen
107k12166364
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I should really check out a definite source, but I think the Artin-Schreier correspondence means the following. To summarize, the problem observed by user8268 can be resolved by insisting that the Galois groups should have a preferred generator.
So something like the following.
Let $L$ and $L'$ be two cyclic degree $p$ extensions of $K$, and let $sigma$ (resp. $sigma'$) be the respective preferred generators. We call $(L,sigma)$ and $(L',sigma')$ equivalent, if there exists a $K$-isomorphism $psi:Lto L'$ such that $$psicircsigma=sigma'circpsi.$$
Then the AS-correspondence is a bijection between the non-trivial cosets of the subgroup $A={x^p-xmid xin K}le(K,+)$ and the equivalence classes of pairs $(L,sigma)$. If $c+A$ is a non-trivial coset of $A$ then it corresponds to an extension $L=K(beta)$ with $beta^p-beta+c=0$ together with the preferred automorphism $sigma:Lto L$ uniquely determined by $sigma(beta)=beta+1$.
A consequence of this formulation is that while the splitting fields of $x^p-x+c$ and $x^p-x+2c$ are both equal to $L=K(beta)$, the above correspondence associates a different generator of the Galois group with the latter polynomial. The automorphism $sigma$ that maps $betamapsto beta+1$ will map $2betamapsto 2beta+2$, so we should associate $x^p-x+2c$ with the pair $(K(beta),sigma^2)$ instead of $(K(beta),sigma)$. Observe that those two pairs cannot be equivalent according to the above definition because $sigma'=sigma^2$ (and here every possible $psi$ commutes with $sigma$).
A few closing remarks:
- I may be out of my depth here, but I hazard a guess that the interpretation of this result in terms of Galois cohomology makes it necessary to specify the action of a fixed generator of the cyclic group.
- An analogous problem is present in Kummer theory also. For example, with $K=Bbb{Q}(omega)$, $omega=e^{2pi i/3},$ the cyclic cubic extensions $K(root3of2)$ and $K(root3of4)$ are clearly equal (as subsets of $Bbb{C}$) even though $2$ and $4$ belong to distinct (multiplicative) cosets of the subgroup of cubes of $K^*$. Here we also resolve the problem by observing that the automorphisms of $K(root3of2)$ that map $root3of2mapstoomegaroot3of2$ and $root3of4mapsto omegaroot3of4$ are similarly squares of each other, i.e. distinct generators of the Galois group.
answered 16 hours ago
Jyrki Lahtonen
107k12166364
add a comment |
up vote
1
down vote
accepted
I should really check out a definite source, but I think the Artin-Schreier correspondence means the following. To summarize, the problem observed by user8268 can be resolved by insisting that the Galois groups should have a preferred generator.
So something like the following.
Let $L$ and $L'$ be two cyclic degree $p$ extensions of $K$, and let $sigma$ (resp. $sigma'$) be the respective preferred generators. We call $(L,sigma)$ and $(L',sigma')$ equivalent, if there exists a $K$-isomorphism $psi:Lto L'$ such that $$psicircsigma=sigma'circpsi.$$
Then the AS-correspondence is a bijection between the non-trivial cosets of the subgroup $A={x^p-xmid xin K}le(K,+)$ and the equivalence classes of pairs $(L,sigma)$. If $c+A$ is a non-trivial coset of $A$ then it corresponds to an extension $L=K(beta)$ with $beta^p-beta+c=0$ together with the preferred automorphism $sigma:Lto L$ uniquely determined by $sigma(beta)=beta+1$.
A consequence of this formulation is that while the splitting fields of $x^p-x+c$ and $x^p-x+2c$ are both equal to $L=K(beta)$, the above correspondence associates a different generator of the Galois group with the latter polynomial. The automorphism $sigma$ that maps $betamapsto beta+1$ will map $2betamapsto 2beta+2$, so we should associate $x^p-x+2c$ with the pair $(K(beta),sigma^2)$ instead of $(K(beta),sigma)$. Observe that those two pairs cannot be equivalent according to the above definition because $sigma'=sigma^2$ (and here every possible $psi$ commutes with $sigma$).
A few closing remarks:
- I may be out of my depth here, but I hazard a guess that the interpretation of this result in terms of Galois cohomology makes it necessary to specify the action of a fixed generator of the cyclic group.
- An analogous problem is present in Kummer theory also. For example, with $K=Bbb{Q}(omega)$, $omega=e^{2pi i/3},$ the cyclic cubic extensions $K(root3of2)$ and $K(root3of4)$ are clearly equal (as subsets of $Bbb{C}$) even though $2$ and $4$ belong to distinct (multiplicative) cosets of the subgroup of cubes of $K^*$. Here we also resolve the problem by observing that the automorphisms of $K(root3of2)$ that map $root3of2mapstoomegaroot3of2$ and $root3of4mapsto omegaroot3of4$ are similarly squares of each other, i.e. distinct generators of the Galois group.
answered 16 hours ago
Jyrki Lahtonen
107k12166364
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I should really check out a definite source, but I think the Artin-Schreier correspondence means the following. To summarize, the problem observed by user8268 can be resolved by insisting that the Galois groups should have a preferred generator.
So something like the following.
Let $L$ and $L'$ be two cyclic degree $p$ extensions of $K$, and let $sigma$ (resp. $sigma'$) be the respective preferred generators. We call $(L,sigma)$ and $(L',sigma')$ equivalent, if there exists a $K$-isomorphism $psi:Lto L'$ such that $$psicircsigma=sigma'circpsi.$$
Then the AS-correspondence is a bijection between the non-trivial cosets of the subgroup $A={x^p-xmid xin K}le(K,+)$ and the equivalence classes of pairs $(L,sigma)$. If $c+A$ is a non-trivial coset of $A$ then it corresponds to an extension $L=K(beta)$ with $beta^p-beta+c=0$ together with the preferred automorphism $sigma:Lto L$ uniquely determined by $sigma(beta)=beta+1$.
A consequence of this formulation is that while the splitting fields of $x^p-x+c$ and $x^p-x+2c$ are both equal to $L=K(beta)$, the above correspondence associates a different generator of the Galois group with the latter polynomial. The automorphism $sigma$ that maps $betamapsto beta+1$ will map $2betamapsto 2beta+2$, so we should associate $x^p-x+2c$ with the pair $(K(beta),sigma^2)$ instead of $(K(beta),sigma)$. Observe that those two pairs cannot be equivalent according to the above definition because $sigma'=sigma^2$ (and here every possible $psi$ commutes with $sigma$).
A few closing remarks:
- I may be out of my depth here, but I hazard a guess that the interpretation of this result in terms of Galois cohomology makes it necessary to specify the action of a fixed generator of the cyclic group.
- An analogous problem is present in Kummer theory also. For example, with $K=Bbb{Q}(omega)$, $omega=e^{2pi i/3},$ the cyclic cubic extensions $K(root3of2)$ and $K(root3of4)$ are clearly equal (as subsets of $Bbb{C}$) even though $2$ and $4$ belong to distinct (multiplicative) cosets of the subgroup of cubes of $K^*$. Here we also resolve the problem by observing that the automorphisms of $K(root3of2)$ that map $root3of2mapstoomegaroot3of2$ and $root3of4mapsto omegaroot3of4$ are similarly squares of each other, i.e. distinct generators of the Galois group.
answered 16 hours ago
Jyrki Lahtonen
107k12166364
I should really check out a definite source, but I think the Artin-Schreier correspondence means the following. To summarize, the problem observed by user8268 can be resolved by insisting that the Galois groups should have a preferred generator.
So something like the following.
Let $L$ and $L'$ be two cyclic degree $p$ extensions of $K$, and let $sigma$ (resp. $sigma'$) be the respective preferred generators. We call $(L,sigma)$ and $(L',sigma')$ equivalent, if there exists a $K$-isomorphism $psi:Lto L'$ such that $$psicircsigma=sigma'circpsi.$$
Then the AS-correspondence is a bijection between the non-trivial cosets of the subgroup $A={x^p-xmid xin K}le(K,+)$ and the equivalence classes of pairs $(L,sigma)$. If $c+A$ is a non-trivial coset of $A$ then it corresponds to an extension $L=K(beta)$ with $beta^p-beta+c=0$ together with the preferred automorphism $sigma:Lto L$ uniquely determined by $sigma(beta)=beta+1$.
A consequence of this formulation is that while the splitting fields of $x^p-x+c$ and $x^p-x+2c$ are both equal to $L=K(beta)$, the above correspondence associates a different generator of the Galois group with the latter polynomial. The automorphism $sigma$ that maps $betamapsto beta+1$ will map $2betamapsto 2beta+2$, so we should associate $x^p-x+2c$ with the pair $(K(beta),sigma^2)$ instead of $(K(beta),sigma)$. Observe that those two pairs cannot be equivalent according to the above definition because $sigma'=sigma^2$ (and here every possible $psi$ commutes with $sigma$).
A few closing remarks:
- I may be out of my depth here, but I hazard a guess that the interpretation of this result in terms of Galois cohomology makes it necessary to specify the action of a fixed generator of the cyclic group.
- An analogous problem is present in Kummer theory also. For example, with $K=Bbb{Q}(omega)$, $omega=e^{2pi i/3},$ the cyclic cubic extensions $K(root3of2)$ and $K(root3of4)$ are clearly equal (as subsets of $Bbb{C}$) even though $2$ and $4$ belong to distinct (multiplicative) cosets of the subgroup of cubes of $K^*$. Here we also resolve the problem by observing that the automorphisms of $K(root3of2)$ that map $root3of2mapstoomegaroot3of2$ and $root3of4mapsto omegaroot3of4$ are similarly squares of each other, i.e. distinct generators of the Galois group.
answered 16 hours ago
Jyrki Lahtonen
107k12166364
edited 10 hours ago
answered 16 hours ago
Jyrki Lahtonen
107k12166364
answered 16 hours ago
Jyrki Lahtonen
107k12166364
answered 16 hours ago
Jyrki Lahtonen
107k12166364
107k12166364
add a comment |
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The bijection is rather between the Galois extensions of $K$ of order $p$ and $bigl(Ksetminus{a^p-a|ain K}bigr)/Bbb F_p^*$. (If $lambdainBbb F_p^*$ then $L_{lambda c}=L_c$.)
– user8268
2 days ago
@user8268: Yes, this seems to be the case, but why is the AS-correspondence always stated as I did it above?
– Algebrus
2 days ago