Norm transitivity in inseperable extensions confusion.
In the book I am reading, an exercise is to show that if a finite field extension $M/L$ is purely inseperable, then $N_{M/L}(alpha)=(-alpha)^{[M:L]}$. Then it asks to show that if $M/L$ is purely inseperable, and $L/K$ is seperable, then $N_{L/K}circ N_{M/L}=N_{M/K}$. The proof that I would like to employ is as follows:
$N_{M/K}(alpha)^{[M:L]}=N_{M/K}(alpha^{[M:L]})$
Since $alpha^{[M:L]}$ is in $L$, we then have this expression equals $N_{L/K}(alpha^{[M:L]})^{[M:L]}$, then since $[M:L]=p^s$ by pure inseperability, we may cancel the powers by injectivity of the frobenius morphism, giving:
$N_{M/K}(alpha)=N_{L/K}(alpha^{[M:L]})=N_{L/K}(-N_{M/L}(alpha))=N_{L/K}(-1)N_{L/K}(N_{M/L}(alpha))$
Since $N_{L/K}(-1)$ need not equal $1$, there is a mistake here somewhere, but I cant see where the argument fails.
abstract-algebra proof-verification field-theory
asked Dec 12 '18 at 7:53
user277182user277182
426212
add a comment |
In the book I am reading, an exercise is to show that if a finite field extension $M/L$ is purely inseperable, then $N_{M/L}(alpha)=(-alpha)^{[M:L]}$. Then it asks to show that if $M/L$ is purely inseperable, and $L/K$ is seperable, then $N_{L/K}circ N_{M/L}=N_{M/K}$. The proof that I would like to employ is as follows:
$N_{M/K}(alpha)^{[M:L]}=N_{M/K}(alpha^{[M:L]})$
Since $alpha^{[M:L]}$ is in $L$, we then have this expression equals $N_{L/K}(alpha^{[M:L]})^{[M:L]}$, then since $[M:L]=p^s$ by pure inseperability, we may cancel the powers by injectivity of the frobenius morphism, giving:
$N_{M/K}(alpha)=N_{L/K}(alpha^{[M:L]})=N_{L/K}(-N_{M/L}(alpha))=N_{L/K}(-1)N_{L/K}(N_{M/L}(alpha))$
Since $N_{L/K}(-1)$ need not equal $1$, there is a mistake here somewhere, but I cant see where the argument fails.
abstract-algebra proof-verification field-theory
asked Dec 12 '18 at 7:53
user277182user277182
426212
add a comment |
In the book I am reading, an exercise is to show that if a finite field extension $M/L$ is purely inseperable, then $N_{M/L}(alpha)=(-alpha)^{[M:L]}$. Then it asks to show that if $M/L$ is purely inseperable, and $L/K$ is seperable, then $N_{L/K}circ N_{M/L}=N_{M/K}$. The proof that I would like to employ is as follows:
$N_{M/K}(alpha)^{[M:L]}=N_{M/K}(alpha^{[M:L]})$
Since $alpha^{[M:L]}$ is in $L$, we then have this expression equals $N_{L/K}(alpha^{[M:L]})^{[M:L]}$, then since $[M:L]=p^s$ by pure inseperability, we may cancel the powers by injectivity of the frobenius morphism, giving:
$N_{M/K}(alpha)=N_{L/K}(alpha^{[M:L]})=N_{L/K}(-N_{M/L}(alpha))=N_{L/K}(-1)N_{L/K}(N_{M/L}(alpha))$
Since $N_{L/K}(-1)$ need not equal $1$, there is a mistake here somewhere, but I cant see where the argument fails.
abstract-algebra proof-verification field-theory
asked Dec 12 '18 at 7:53
user277182user277182
426212
In the book I am reading, an exercise is to show that if a finite field extension $M/L$ is purely inseperable, then $N_{M/L}(alpha)=(-alpha)^{[M:L]}$. Then it asks to show that if $M/L$ is purely inseperable, and $L/K$ is seperable, then $N_{L/K}circ N_{M/L}=N_{M/K}$. The proof that I would like to employ is as follows:
$N_{M/K}(alpha)^{[M:L]}=N_{M/K}(alpha^{[M:L]})$
Since $alpha^{[M:L]}$ is in $L$, we then have this expression equals $N_{L/K}(alpha^{[M:L]})^{[M:L]}$, then since $[M:L]=p^s$ by pure inseperability, we may cancel the powers by injectivity of the frobenius morphism, giving:
$N_{M/K}(alpha)=N_{L/K}(alpha^{[M:L]})=N_{L/K}(-N_{M/L}(alpha))=N_{L/K}(-1)N_{L/K}(N_{M/L}(alpha))$
Since $N_{L/K}(-1)$ need not equal $1$, there is a mistake here somewhere, but I cant see where the argument fails.
abstract-algebra proof-verification field-theory
abstract-algebra proof-verification field-theory
asked Dec 12 '18 at 7:53
user277182user277182
426212
asked Dec 12 '18 at 7:53
user277182user277182
426212
asked Dec 12 '18 at 7:53
user277182user277182
426212
asked Dec 12 '18 at 7:53
user277182user277182
426212
asked Dec 12 '18 at 7:53
user277182user277182
426212
426212
add a comment |
add a comment |
1 Answer
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There was a typo in the book, the previous part should state $N_{M/L}(alpha)=alpha^{[M:L]}$, without the minus. With the minus, its not multiplicative. This solves the issue.
answered Dec 12 '18 at 8:11
user277182user277182
426212
add a comment |
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There was a typo in the book, the previous part should state $N_{M/L}(alpha)=alpha^{[M:L]}$, without the minus. With the minus, its not multiplicative. This solves the issue.
answered Dec 12 '18 at 8:11
user277182user277182
426212
add a comment |
There was a typo in the book, the previous part should state $N_{M/L}(alpha)=alpha^{[M:L]}$, without the minus. With the minus, its not multiplicative. This solves the issue.
answered Dec 12 '18 at 8:11
user277182user277182
426212
add a comment |
There was a typo in the book, the previous part should state $N_{M/L}(alpha)=alpha^{[M:L]}$, without the minus. With the minus, its not multiplicative. This solves the issue.
answered Dec 12 '18 at 8:11
user277182user277182
426212
There was a typo in the book, the previous part should state $N_{M/L}(alpha)=alpha^{[M:L]}$, without the minus. With the minus, its not multiplicative. This solves the issue.
answered Dec 12 '18 at 8:11
user277182user277182
426212
answered Dec 12 '18 at 8:11
user277182user277182
426212
answered Dec 12 '18 at 8:11
user277182user277182
426212
answered Dec 12 '18 at 8:11
user277182user277182
426212
426212
add a comment |
add a comment |
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