Exact Sequence of Galois Groups











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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.






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    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







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      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







      share|cite|improve this question













      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






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          $(sigma,tau) mapsto (sigma - tau)|_{E_1cap E_2}$ 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).






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            $(sigma,tau) mapsto (sigma - tau)|_{E_1cap E_2}$ 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).






            share|cite|improve this answer

























              up vote
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              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).






              share|cite|improve this answer























                up vote
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                accepted







                up vote
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                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).






                share|cite|improve this answer












                $(sigma,tau) mapsto (sigma - tau)|_{E_1cap E_2}$ 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).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                user3482749

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