A finite normal extension is also a splitting field
$begingroup$
Let $E/K $ be a finite normal extension, then there exists $p(x)in K[x] $ s.t. $E$ is the splitting field of $p(x) $.
The proof goes as follows:
$E=K(a_1,...,a_n) $ for some $a_1,...,a_nin Omega $ where $Omega $ is the algebraic closure of $K $.
Let $mu_i $ be the minimal polynomial of $a_i $ for $i=1,...,n $
We can send any $a_i $ onto another root of $mu _i $ through an homomorphism that fixes $K $.
Then the thesis follows.
I don't understand why we can send any $a_i $ onto another root of $mu _i $.
Note: the definition that we use of normal extension is:
$L/K$ is a normal extension if every homomorphism $phi :Krightarrow Omega $ that fixes $K $ sends $L $ onto itself.
field-theory galois-theory extension-field
edited Jan 6 at 11:22
Jyrki Lahtonen
110k13171386
asked Jan 5 at 17:31
Lucio TanziniLucio Tanzini
351114
$endgroup$
add a comment |
$begingroup$
Let $E/K $ be a finite normal extension, then there exists $p(x)in K[x] $ s.t. $E$ is the splitting field of $p(x) $.
The proof goes as follows:
$E=K(a_1,...,a_n) $ for some $a_1,...,a_nin Omega $ where $Omega $ is the algebraic closure of $K $.
Let $mu_i $ be the minimal polynomial of $a_i $ for $i=1,...,n $
We can send any $a_i $ onto another root of $mu _i $ through an homomorphism that fixes $K $.
Then the thesis follows.
I don't understand why we can send any $a_i $ onto another root of $mu _i $.
Note: the definition that we use of normal extension is:
$L/K$ is a normal extension if every homomorphism $phi :Krightarrow Omega $ that fixes $K $ sends $L $ onto itself.
field-theory galois-theory extension-field
edited Jan 6 at 11:22
Jyrki Lahtonen
110k13171386
asked Jan 5 at 17:31
Lucio TanziniLucio Tanzini
351114
$endgroup$
add a comment |
$begingroup$
Let $E/K $ be a finite normal extension, then there exists $p(x)in K[x] $ s.t. $E$ is the splitting field of $p(x) $.
The proof goes as follows:
$E=K(a_1,...,a_n) $ for some $a_1,...,a_nin Omega $ where $Omega $ is the algebraic closure of $K $.
Let $mu_i $ be the minimal polynomial of $a_i $ for $i=1,...,n $
We can send any $a_i $ onto another root of $mu _i $ through an homomorphism that fixes $K $.
Then the thesis follows.
I don't understand why we can send any $a_i $ onto another root of $mu _i $.
Note: the definition that we use of normal extension is:
$L/K$ is a normal extension if every homomorphism $phi :Krightarrow Omega $ that fixes $K $ sends $L $ onto itself.
field-theory galois-theory extension-field
edited Jan 6 at 11:22
Jyrki Lahtonen
110k13171386
asked Jan 5 at 17:31
Lucio TanziniLucio Tanzini
351114
$endgroup$
Let $E/K $ be a finite normal extension, then there exists $p(x)in K[x] $ s.t. $E$ is the splitting field of $p(x) $.
The proof goes as follows:
$E=K(a_1,...,a_n) $ for some $a_1,...,a_nin Omega $ where $Omega $ is the algebraic closure of $K $.
Let $mu_i $ be the minimal polynomial of $a_i $ for $i=1,...,n $
We can send any $a_i $ onto another root of $mu _i $ through an homomorphism that fixes $K $.
Then the thesis follows.
I don't understand why we can send any $a_i $ onto another root of $mu _i $.
Note: the definition that we use of normal extension is:
$L/K$ is a normal extension if every homomorphism $phi :Krightarrow Omega $ that fixes $K $ sends $L $ onto itself.
field-theory galois-theory extension-field
field-theory galois-theory extension-field
edited Jan 6 at 11:22
Jyrki Lahtonen
110k13171386
asked Jan 5 at 17:31
Lucio TanziniLucio Tanzini
351114
edited Jan 6 at 11:22
Jyrki Lahtonen
110k13171386
asked Jan 5 at 17:31
Lucio TanziniLucio Tanzini
351114
edited Jan 6 at 11:22
Jyrki Lahtonen
110k13171386
edited Jan 6 at 11:22
Jyrki Lahtonen
110k13171386
edited Jan 6 at 11:22
Jyrki Lahtonen
110k13171386
110k13171386
asked Jan 5 at 17:31
Lucio TanziniLucio Tanzini
351114
asked Jan 5 at 17:31
Lucio TanziniLucio Tanzini
351114
asked Jan 5 at 17:31
Lucio TanziniLucio Tanzini
351114
351114
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Here are the steps of a proof.
The $mu_{i}$ are irreducible.
Hence, given any two roots $a, b$ of $mu_{i}$, there is a $K$-isomorphism (meaning that it restricts to the identity on $K$) $$K(a) to K(b)$$ taking $a$ to $b$.
This isomorphism can be extended to an automorphism $phi$ of $Omega$.
Now since $L/K$ is normal, $phi$ has to send $L$ to $L$.
answered Jan 5 at 17:39
Andreas CarantiAndreas Caranti
56.8k34397
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
Here are the steps of a proof.
The $mu_{i}$ are irreducible.
Hence, given any two roots $a, b$ of $mu_{i}$, there is a $K$-isomorphism (meaning that it restricts to the identity on $K$) $$K(a) to K(b)$$ taking $a$ to $b$.
This isomorphism can be extended to an automorphism $phi$ of $Omega$.
Now since $L/K$ is normal, $phi$ has to send $L$ to $L$.
answered Jan 5 at 17:39
Andreas CarantiAndreas Caranti
56.8k34397
$endgroup$
add a comment |
$begingroup$
Here are the steps of a proof.
The $mu_{i}$ are irreducible.
Hence, given any two roots $a, b$ of $mu_{i}$, there is a $K$-isomorphism (meaning that it restricts to the identity on $K$) $$K(a) to K(b)$$ taking $a$ to $b$.
This isomorphism can be extended to an automorphism $phi$ of $Omega$.
Now since $L/K$ is normal, $phi$ has to send $L$ to $L$.
answered Jan 5 at 17:39
Andreas CarantiAndreas Caranti
56.8k34397
$endgroup$
add a comment |
$begingroup$
Here are the steps of a proof.
The $mu_{i}$ are irreducible.
Hence, given any two roots $a, b$ of $mu_{i}$, there is a $K$-isomorphism (meaning that it restricts to the identity on $K$) $$K(a) to K(b)$$ taking $a$ to $b$.
This isomorphism can be extended to an automorphism $phi$ of $Omega$.
Now since $L/K$ is normal, $phi$ has to send $L$ to $L$.
answered Jan 5 at 17:39
Andreas CarantiAndreas Caranti
56.8k34397
$endgroup$
Here are the steps of a proof.
The $mu_{i}$ are irreducible.
Hence, given any two roots $a, b$ of $mu_{i}$, there is a $K$-isomorphism (meaning that it restricts to the identity on $K$) $$K(a) to K(b)$$ taking $a$ to $b$.
This isomorphism can be extended to an automorphism $phi$ of $Omega$.
Now since $L/K$ is normal, $phi$ has to send $L$ to $L$.
answered Jan 5 at 17:39
Andreas CarantiAndreas Caranti
56.8k34397
answered Jan 5 at 17:39
Andreas CarantiAndreas Caranti
56.8k34397
answered Jan 5 at 17:39
Andreas CarantiAndreas Caranti
56.8k34397
answered Jan 5 at 17:39
Andreas CarantiAndreas Caranti
56.8k34397
56.8k34397
add a comment |
add a comment |
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