Proving that $[F(x_1 , x_2) : F] = p_1 p_2$, where $p_1 = deg min(x_1 , F)$ and $p_2 = deg min(x_2 , F)$ are coprime.
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Let $K / F$ be a extension of fields and let $x_1 , x_2 in K$ be two algebraic elements over $F$ and let $p_1 = deg min(x_1 , F)$ and $p_2 = deg min(x_2 , F)$ be two coprime natural numbers. I want to show that $[F(x_1 , x_2) : F] = p_1 p_2$. By multiplicity of degree formula, we know that
$$
[F(x_1 , x_2) : F] = p_1 pmbox,
$$
where $p = deg min(x_2 , F(x_1))$. Then I need we need to prove that $min(x_2 , F(x_1))(X) in F[X]$, using that $gcd(p_1 , p_2) = 1$. Thank you very much in advance.
field-theory
asked Aug 26 at 22:34
joseabp91
1,100411
add a comment |Â
up vote
0
down vote
favorite
Let $K / F$ be a extension of fields and let $x_1 , x_2 in K$ be two algebraic elements over $F$ and let $p_1 = deg min(x_1 , F)$ and $p_2 = deg min(x_2 , F)$ be two coprime natural numbers. I want to show that $[F(x_1 , x_2) : F] = p_1 p_2$. By multiplicity of degree formula, we know that
$$
[F(x_1 , x_2) : F] = p_1 pmbox,
$$
where $p = deg min(x_2 , F(x_1))$. Then I need we need to prove that $min(x_2 , F(x_1))(X) in F[X]$, using that $gcd(p_1 , p_2) = 1$. Thank you very much in advance.
field-theory
asked Aug 26 at 22:34
joseabp91
1,100411
Use the fact that degree multiplies in a tower of field extensions $E/K/F$.
– John Brevik
Aug 26 at 22:39
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $K / F$ be a extension of fields and let $x_1 , x_2 in K$ be two algebraic elements over $F$ and let $p_1 = deg min(x_1 , F)$ and $p_2 = deg min(x_2 , F)$ be two coprime natural numbers. I want to show that $[F(x_1 , x_2) : F] = p_1 p_2$. By multiplicity of degree formula, we know that
$$
[F(x_1 , x_2) : F] = p_1 pmbox,
$$
where $p = deg min(x_2 , F(x_1))$. Then I need we need to prove that $min(x_2 , F(x_1))(X) in F[X]$, using that $gcd(p_1 , p_2) = 1$. Thank you very much in advance.
field-theory
asked Aug 26 at 22:34
joseabp91
1,100411
Let $K / F$ be a extension of fields and let $x_1 , x_2 in K$ be two algebraic elements over $F$ and let $p_1 = deg min(x_1 , F)$ and $p_2 = deg min(x_2 , F)$ be two coprime natural numbers. I want to show that $[F(x_1 , x_2) : F] = p_1 p_2$. By multiplicity of degree formula, we know that
$$
[F(x_1 , x_2) : F] = p_1 pmbox,
$$
where $p = deg min(x_2 , F(x_1))$. Then I need we need to prove that $min(x_2 , F(x_1))(X) in F[X]$, using that $gcd(p_1 , p_2) = 1$. Thank you very much in advance.
field-theory
asked Aug 26 at 22:34
joseabp91
1,100411
asked Aug 26 at 22:34
joseabp91
1,100411
asked Aug 26 at 22:34
joseabp91
1,100411
asked Aug 26 at 22:34
joseabp91
1,100411
1,100411
Use the fact that degree multiplies in a tower of field extensions $E/K/F$.
– John Brevik
Aug 26 at 22:39
add a comment |Â
Use the fact that degree multiplies in a tower of field extensions $E/K/F$.
– John Brevik
Aug 26 at 22:39
Use the fact that degree multiplies in a tower of field extensions $E/K/F$.
– John Brevik
Aug 26 at 22:39
Use the fact that degree multiplies in a tower of field extensions $E/K/F$.
– John Brevik
Aug 26 at 22:39
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
HINT:
$$p_1 = [F(x_1):F]; big|; [F(x_1,x_2):F(x_1)][F(x_1):F] = [F(x_1,x_2):F]$$
$$p_2 = [F(x_2):F]; big|; [F(x_1,x_2):F(x_2)][F(x_2):F] = [F(x_1,x_2):F]$$
Also:
$$[F(x_1,x_2):F] le p_1p_2$$
answered Aug 26 at 22:49
Stefan4024
29.6k53377
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
accepted
HINT:
$$p_1 = [F(x_1):F]; big|; [F(x_1,x_2):F(x_1)][F(x_1):F] = [F(x_1,x_2):F]$$
$$p_2 = [F(x_2):F]; big|; [F(x_1,x_2):F(x_2)][F(x_2):F] = [F(x_1,x_2):F]$$
Also:
$$[F(x_1,x_2):F] le p_1p_2$$
answered Aug 26 at 22:49
Stefan4024
29.6k53377
add a comment |Â
up vote
1
down vote
accepted
HINT:
$$p_1 = [F(x_1):F]; big|; [F(x_1,x_2):F(x_1)][F(x_1):F] = [F(x_1,x_2):F]$$
$$p_2 = [F(x_2):F]; big|; [F(x_1,x_2):F(x_2)][F(x_2):F] = [F(x_1,x_2):F]$$
Also:
$$[F(x_1,x_2):F] le p_1p_2$$
answered Aug 26 at 22:49
Stefan4024
29.6k53377
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
HINT:
$$p_1 = [F(x_1):F]; big|; [F(x_1,x_2):F(x_1)][F(x_1):F] = [F(x_1,x_2):F]$$
$$p_2 = [F(x_2):F]; big|; [F(x_1,x_2):F(x_2)][F(x_2):F] = [F(x_1,x_2):F]$$
Also:
$$[F(x_1,x_2):F] le p_1p_2$$
answered Aug 26 at 22:49
Stefan4024
29.6k53377
HINT:
$$p_1 = [F(x_1):F]; big|; [F(x_1,x_2):F(x_1)][F(x_1):F] = [F(x_1,x_2):F]$$
$$p_2 = [F(x_2):F]; big|; [F(x_1,x_2):F(x_2)][F(x_2):F] = [F(x_1,x_2):F]$$
Also:
$$[F(x_1,x_2):F] le p_1p_2$$
answered Aug 26 at 22:49
Stefan4024
29.6k53377
answered Aug 26 at 22:49
Stefan4024
29.6k53377
answered Aug 26 at 22:49
Stefan4024
29.6k53377
answered Aug 26 at 22:49
Stefan4024
29.6k53377
29.6k53377
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%2f2895597%2fproving-that-fx-1-x-2-f-p-1-p-2-where-p-1-deg-minx-1-f-an%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
Use the fact that degree multiplies in a tower of field extensions $E/K/F$.
– John Brevik
Aug 26 at 22:39