Mixing
Tensoring does not change the determinant (norm map)
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let $ K $ be a number field, and let $ L $ be a finite extension of $ K $. Let $ p $ be a prime ideal in $ mathcalO_K $, then we know that there is a canonical isomorphism
$$ L otimes_K K_p simeq prod_q mid p L_q. $$
($ q $ runs over all prime ideals in $ L $ lying over $ p $.)
Why does it follow from here that for any $ b in L $,
$$ N_L/K (b) = prod_q mid p N_L_q / K_p (b)? $$
More specifically, why is that tensoring over $ K $ with $ K_p $ does not change the determinant of the $K$-linear map $ x mapsto bx $?
galois-theory algebraic-number-theory tensor-products
asked Aug 1 at 23:28
kgs
1636
add a comment |Â
up vote
1
down vote
favorite
Let $ K $ be a number field, and let $ L $ be a finite extension of $ K $. Let $ p $ be a prime ideal in $ mathcalO_K $, then we know that there is a canonical isomorphism
$$ L otimes_K K_p simeq prod_q mid p L_q. $$
($ q $ runs over all prime ideals in $ L $ lying over $ p $.)
Why does it follow from here that for any $ b in L $,
$$ N_L/K (b) = prod_q mid p N_L_q / K_p (b)? $$
More specifically, why is that tensoring over $ K $ with $ K_p $ does not change the determinant of the $K$-linear map $ x mapsto bx $?
galois-theory algebraic-number-theory tensor-products
asked Aug 1 at 23:28
kgs
1636
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $ K $ be a number field, and let $ L $ be a finite extension of $ K $. Let $ p $ be a prime ideal in $ mathcalO_K $, then we know that there is a canonical isomorphism
$$ L otimes_K K_p simeq prod_q mid p L_q. $$
($ q $ runs over all prime ideals in $ L $ lying over $ p $.)
Why does it follow from here that for any $ b in L $,
$$ N_L/K (b) = prod_q mid p N_L_q / K_p (b)? $$
More specifically, why is that tensoring over $ K $ with $ K_p $ does not change the determinant of the $K$-linear map $ x mapsto bx $?
galois-theory algebraic-number-theory tensor-products
asked Aug 1 at 23:28
kgs
1636
Let $ K $ be a number field, and let $ L $ be a finite extension of $ K $. Let $ p $ be a prime ideal in $ mathcalO_K $, then we know that there is a canonical isomorphism
$$ L otimes_K K_p simeq prod_q mid p L_q. $$
($ q $ runs over all prime ideals in $ L $ lying over $ p $.)
Why does it follow from here that for any $ b in L $,
$$ N_L/K (b) = prod_q mid p N_L_q / K_p (b)? $$
More specifically, why is that tensoring over $ K $ with $ K_p $ does not change the determinant of the $K$-linear map $ x mapsto bx $?
galois-theory algebraic-number-theory tensor-products
asked Aug 1 at 23:28
kgs
1636
asked Aug 1 at 23:28
kgs
1636
asked Aug 1 at 23:28
kgs
1636
asked Aug 1 at 23:28
kgs
1636
1636
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
For $ain L$, you have two instances of the regular representation. You have $Lto L$ by $xmapsto ax$, and you have $Lotimes_KRto Lotimes_KR$ by $ximapsto(aotimes1)xi$, where $xi$ stands for an arbitrary element of the tensor product. Here, $R$ is any old $K$-algebra.
In the former case, the map is described completely by the result of multiplying $a$ to the elements $x_1,cdots,x_n$ of a $K$-basis of $L$. You get a matrix, and you take its determinant. And in the latter case, the map is decribed completely by the result of multiplying $aotimes1$ to the elements $x_1otimes1,cdots,x_notimes1$ of the corresponding $R$-basis of $Lotimes_KR$. You get the “same†matrix, and the determinant is $(det a)otimes1$. In the two cases, you have calculated the norm of $a$, respectively the norm of $aotimes1$.
In the case you (and I) are interested in, where $Lotimes K_p$ splits into a sum of $K_p$-algebras, one is tempted to take a basis of the tensor product that more nearly reflects the splitting. But taking a different basis doesn’t change the determinant.
I hope this addressed your worries; it certainly was not as efficient an explanation as I would have liked.
answered Aug 3 at 4:19
Lubin
40.7k34182
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
For $ain L$, you have two instances of the regular representation. You have $Lto L$ by $xmapsto ax$, and you have $Lotimes_KRto Lotimes_KR$ by $ximapsto(aotimes1)xi$, where $xi$ stands for an arbitrary element of the tensor product. Here, $R$ is any old $K$-algebra.
In the former case, the map is described completely by the result of multiplying $a$ to the elements $x_1,cdots,x_n$ of a $K$-basis of $L$. You get a matrix, and you take its determinant. And in the latter case, the map is decribed completely by the result of multiplying $aotimes1$ to the elements $x_1otimes1,cdots,x_notimes1$ of the corresponding $R$-basis of $Lotimes_KR$. You get the “same†matrix, and the determinant is $(det a)otimes1$. In the two cases, you have calculated the norm of $a$, respectively the norm of $aotimes1$.
In the case you (and I) are interested in, where $Lotimes K_p$ splits into a sum of $K_p$-algebras, one is tempted to take a basis of the tensor product that more nearly reflects the splitting. But taking a different basis doesn’t change the determinant.
I hope this addressed your worries; it certainly was not as efficient an explanation as I would have liked.
answered Aug 3 at 4:19
Lubin
40.7k34182
add a comment |Â
up vote
1
down vote
For $ain L$, you have two instances of the regular representation. You have $Lto L$ by $xmapsto ax$, and you have $Lotimes_KRto Lotimes_KR$ by $ximapsto(aotimes1)xi$, where $xi$ stands for an arbitrary element of the tensor product. Here, $R$ is any old $K$-algebra.
In the former case, the map is described completely by the result of multiplying $a$ to the elements $x_1,cdots,x_n$ of a $K$-basis of $L$. You get a matrix, and you take its determinant. And in the latter case, the map is decribed completely by the result of multiplying $aotimes1$ to the elements $x_1otimes1,cdots,x_notimes1$ of the corresponding $R$-basis of $Lotimes_KR$. You get the “same†matrix, and the determinant is $(det a)otimes1$. In the two cases, you have calculated the norm of $a$, respectively the norm of $aotimes1$.
In the case you (and I) are interested in, where $Lotimes K_p$ splits into a sum of $K_p$-algebras, one is tempted to take a basis of the tensor product that more nearly reflects the splitting. But taking a different basis doesn’t change the determinant.
I hope this addressed your worries; it certainly was not as efficient an explanation as I would have liked.
answered Aug 3 at 4:19
Lubin
40.7k34182
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For $ain L$, you have two instances of the regular representation. You have $Lto L$ by $xmapsto ax$, and you have $Lotimes_KRto Lotimes_KR$ by $ximapsto(aotimes1)xi$, where $xi$ stands for an arbitrary element of the tensor product. Here, $R$ is any old $K$-algebra.
In the former case, the map is described completely by the result of multiplying $a$ to the elements $x_1,cdots,x_n$ of a $K$-basis of $L$. You get a matrix, and you take its determinant. And in the latter case, the map is decribed completely by the result of multiplying $aotimes1$ to the elements $x_1otimes1,cdots,x_notimes1$ of the corresponding $R$-basis of $Lotimes_KR$. You get the “same†matrix, and the determinant is $(det a)otimes1$. In the two cases, you have calculated the norm of $a$, respectively the norm of $aotimes1$.
In the case you (and I) are interested in, where $Lotimes K_p$ splits into a sum of $K_p$-algebras, one is tempted to take a basis of the tensor product that more nearly reflects the splitting. But taking a different basis doesn’t change the determinant.
I hope this addressed your worries; it certainly was not as efficient an explanation as I would have liked.
answered Aug 3 at 4:19
Lubin
40.7k34182
For $ain L$, you have two instances of the regular representation. You have $Lto L$ by $xmapsto ax$, and you have $Lotimes_KRto Lotimes_KR$ by $ximapsto(aotimes1)xi$, where $xi$ stands for an arbitrary element of the tensor product. Here, $R$ is any old $K$-algebra.
In the former case, the map is described completely by the result of multiplying $a$ to the elements $x_1,cdots,x_n$ of a $K$-basis of $L$. You get a matrix, and you take its determinant. And in the latter case, the map is decribed completely by the result of multiplying $aotimes1$ to the elements $x_1otimes1,cdots,x_notimes1$ of the corresponding $R$-basis of $Lotimes_KR$. You get the “same†matrix, and the determinant is $(det a)otimes1$. In the two cases, you have calculated the norm of $a$, respectively the norm of $aotimes1$.
In the case you (and I) are interested in, where $Lotimes K_p$ splits into a sum of $K_p$-algebras, one is tempted to take a basis of the tensor product that more nearly reflects the splitting. But taking a different basis doesn’t change the determinant.
I hope this addressed your worries; it certainly was not as efficient an explanation as I would have liked.
answered Aug 3 at 4:19
Lubin
40.7k34182
answered Aug 3 at 4:19
Lubin
40.7k34182
answered Aug 3 at 4:19
Lubin
40.7k34182
answered Aug 3 at 4:19
Lubin
40.7k34182
40.7k34182
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2869599%2ftensoring-does-not-change-the-determinant-norm-map%23new-answer', 'question_page');
);
Post as a guest
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
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
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