Mixing
Exact Sequence of Galois Groups
up vote
0
down vote
favorite
Let $E_1/F$, $E_2/F$ be Galois extensions. Then $E_1E_2/F$ and $E_1cap E_2/F$ are Galois extensions. Supposedly there is a short exact sequence
$$1to mathrm{Gal}(E_1E_2/F) xrightarrow{varphi} mathrm{Gal}(E_1/F)times mathrm{Gal}(E_2/F) to mathrm{Gal}(E_1cap E_2/F) to 1$$
where $varphi(sigma) = (sigma|_{E_1},sigma|_{E_2})$. However, I cannot figure out what the map $mathrm{Gal}(E_1/F)times mathrm{Gal}(E_2/F) to mathrm{Gal}(E_1cap E_2/F)$ should be.
field-theory galois-theory
asked 2 days ago
Rdrr
12210
add a comment |
up vote
0
down vote
favorite
Let $E_1/F$, $E_2/F$ be Galois extensions. Then $E_1E_2/F$ and $E_1cap E_2/F$ are Galois extensions. Supposedly there is a short exact sequence
$$1to mathrm{Gal}(E_1E_2/F) xrightarrow{varphi} mathrm{Gal}(E_1/F)times mathrm{Gal}(E_2/F) to mathrm{Gal}(E_1cap E_2/F) to 1$$
where $varphi(sigma) = (sigma|_{E_1},sigma|_{E_2})$. However, I cannot figure out what the map $mathrm{Gal}(E_1/F)times mathrm{Gal}(E_2/F) to mathrm{Gal}(E_1cap E_2/F)$ should be.
field-theory galois-theory
asked 2 days ago
Rdrr
12210
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $E_1/F$, $E_2/F$ be Galois extensions. Then $E_1E_2/F$ and $E_1cap E_2/F$ are Galois extensions. Supposedly there is a short exact sequence
$$1to mathrm{Gal}(E_1E_2/F) xrightarrow{varphi} mathrm{Gal}(E_1/F)times mathrm{Gal}(E_2/F) to mathrm{Gal}(E_1cap E_2/F) to 1$$
where $varphi(sigma) = (sigma|_{E_1},sigma|_{E_2})$. However, I cannot figure out what the map $mathrm{Gal}(E_1/F)times mathrm{Gal}(E_2/F) to mathrm{Gal}(E_1cap E_2/F)$ should be.
field-theory galois-theory
asked 2 days ago
Rdrr
12210
Let $E_1/F$, $E_2/F$ be Galois extensions. Then $E_1E_2/F$ and $E_1cap E_2/F$ are Galois extensions. Supposedly there is a short exact sequence
$$1to mathrm{Gal}(E_1E_2/F) xrightarrow{varphi} mathrm{Gal}(E_1/F)times mathrm{Gal}(E_2/F) to mathrm{Gal}(E_1cap E_2/F) to 1$$
where $varphi(sigma) = (sigma|_{E_1},sigma|_{E_2})$. However, I cannot figure out what the map $mathrm{Gal}(E_1/F)times mathrm{Gal}(E_2/F) to mathrm{Gal}(E_1cap E_2/F)$ should be.
field-theory galois-theory
field-theory galois-theory
asked 2 days ago
Rdrr
12210
asked 2 days ago
Rdrr
12210
asked 2 days ago
Rdrr
12210
asked 2 days ago
Rdrr
12210
asked 2 days ago
Rdrr
12210
12210
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
$(sigma,tau) mapsto (sigma - tau)|_{E_1cap E_2}$ works: it's surjective (take any $sigma$ in the target, extend it to some $bar{sigma}$ on $E_1$ any way you like, and then $(bar{sigma},0)mapsto sigma$), and its kernel is the set of all pairs of maps which agree on $E_1cap E_2$, which clearly includes the image of $phi$, and anything in the kernel is a pair of maps$(sigma,tau)$, defined on $E_1$ and $E_2$ respectively, and agreeing on $E_1cap E_2$, so there's an extension of them to $E_1E_2$ (take anything in $E_1E_2$, split it into a product of something in $E_1$ and something in $E_2$, and map the former by $sigma$ and the latter by $tau$, then multiply them - the fact that they agree on the intersection gives you that this is well-defined (any two such representations differ only by multiplying each side by something in the intersection and its inverse respectively, and $sigma$ and $tau$ send those differences to a pair of inverse elements, which cancel out at the end).
answered 2 days ago
user3482749
1,018411
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
$(sigma,tau) mapsto (sigma - tau)|_{E_1cap E_2}$ works: it's surjective (take any $sigma$ in the target, extend it to some $bar{sigma}$ on $E_1$ any way you like, and then $(bar{sigma},0)mapsto sigma$), and its kernel is the set of all pairs of maps which agree on $E_1cap E_2$, which clearly includes the image of $phi$, and anything in the kernel is a pair of maps$(sigma,tau)$, defined on $E_1$ and $E_2$ respectively, and agreeing on $E_1cap E_2$, so there's an extension of them to $E_1E_2$ (take anything in $E_1E_2$, split it into a product of something in $E_1$ and something in $E_2$, and map the former by $sigma$ and the latter by $tau$, then multiply them - the fact that they agree on the intersection gives you that this is well-defined (any two such representations differ only by multiplying each side by something in the intersection and its inverse respectively, and $sigma$ and $tau$ send those differences to a pair of inverse elements, which cancel out at the end).
answered 2 days ago
user3482749
1,018411
add a comment |
up vote
1
down vote
accepted
$(sigma,tau) mapsto (sigma - tau)|_{E_1cap E_2}$ works: it's surjective (take any $sigma$ in the target, extend it to some $bar{sigma}$ on $E_1$ any way you like, and then $(bar{sigma},0)mapsto sigma$), and its kernel is the set of all pairs of maps which agree on $E_1cap E_2$, which clearly includes the image of $phi$, and anything in the kernel is a pair of maps$(sigma,tau)$, defined on $E_1$ and $E_2$ respectively, and agreeing on $E_1cap E_2$, so there's an extension of them to $E_1E_2$ (take anything in $E_1E_2$, split it into a product of something in $E_1$ and something in $E_2$, and map the former by $sigma$ and the latter by $tau$, then multiply them - the fact that they agree on the intersection gives you that this is well-defined (any two such representations differ only by multiplying each side by something in the intersection and its inverse respectively, and $sigma$ and $tau$ send those differences to a pair of inverse elements, which cancel out at the end).
answered 2 days ago
user3482749
1,018411
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$(sigma,tau) mapsto (sigma - tau)|_{E_1cap E_2}$ works: it's surjective (take any $sigma$ in the target, extend it to some $bar{sigma}$ on $E_1$ any way you like, and then $(bar{sigma},0)mapsto sigma$), and its kernel is the set of all pairs of maps which agree on $E_1cap E_2$, which clearly includes the image of $phi$, and anything in the kernel is a pair of maps$(sigma,tau)$, defined on $E_1$ and $E_2$ respectively, and agreeing on $E_1cap E_2$, so there's an extension of them to $E_1E_2$ (take anything in $E_1E_2$, split it into a product of something in $E_1$ and something in $E_2$, and map the former by $sigma$ and the latter by $tau$, then multiply them - the fact that they agree on the intersection gives you that this is well-defined (any two such representations differ only by multiplying each side by something in the intersection and its inverse respectively, and $sigma$ and $tau$ send those differences to a pair of inverse elements, which cancel out at the end).
answered 2 days ago
user3482749
1,018411
$(sigma,tau) mapsto (sigma - tau)|_{E_1cap E_2}$ works: it's surjective (take any $sigma$ in the target, extend it to some $bar{sigma}$ on $E_1$ any way you like, and then $(bar{sigma},0)mapsto sigma$), and its kernel is the set of all pairs of maps which agree on $E_1cap E_2$, which clearly includes the image of $phi$, and anything in the kernel is a pair of maps$(sigma,tau)$, defined on $E_1$ and $E_2$ respectively, and agreeing on $E_1cap E_2$, so there's an extension of them to $E_1E_2$ (take anything in $E_1E_2$, split it into a product of something in $E_1$ and something in $E_2$, and map the former by $sigma$ and the latter by $tau$, then multiply them - the fact that they agree on the intersection gives you that this is well-defined (any two such representations differ only by multiplying each side by something in the intersection and its inverse respectively, and $sigma$ and $tau$ send those differences to a pair of inverse elements, which cancel out at the end).
answered 2 days ago
user3482749
1,018411
answered 2 days ago
user3482749
1,018411
answered 2 days ago
user3482749
1,018411
answered 2 days ago
user3482749
1,018411
1,018411
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005663%2fexact-sequence-of-galois-groups%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
