Avoiding normal closures in proving a characterization of normal extensions

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The following is Exercise V.3.21 from Hungerford's Algebra (GTM 73).
Let $F$ be algebraic over $K$. $F$ is normal over $K$ if and only if for every $K$-monomorphism of fields $sigma:Fto N$, where $N$ is any normal extension of $K$ containing $F$, $sigma(F)=F$ so that $sigma$ is a $K$-automorphism of $F$. [Hint: Adapt the proof of Theorem
3.14, using Theorem 3.16.]
Theorem 3.14 is the usual characterization of normal extensions. Theorem 3.16 states the properties of normal closures. However, I don't think we need normal closures here.
We shall prove that the above condition implies the following, which will prove that $F$ is normal over $K$: If $overlineK$ is any algebraic closure of $K$ containing $F$, then for any $K$-monomorphism of fields $sigma:FtooverlineK$, $operatornameimsigma=F$ so that $sigma$ is actually a $K$-automorphism of $F$.
Proof. Extend $sigma:FtooverlineK$ to a $K$-monomorphism $overlinesigma:NtooverlineK$ (this is possible since $N$ is algebraic over $F$). Since $N$ is normal over $K$, we have $operatornameimoverlinesigma=N$, so $sigma$ actually maps $F$ into $N$. By hypothesis $operatornameimsigma=F$, so we're done.
The other direction is rather obvious, and does not need normal closures either. So I have not made any use of normal closures here. But why is the hint asking me to do so? Is my proof correct?
proof-verification field-theory galois-theory
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The following is Exercise V.3.21 from Hungerford's Algebra (GTM 73).
Let $F$ be algebraic over $K$. $F$ is normal over $K$ if and only if for every $K$-monomorphism of fields $sigma:Fto N$, where $N$ is any normal extension of $K$ containing $F$, $sigma(F)=F$ so that $sigma$ is a $K$-automorphism of $F$. [Hint: Adapt the proof of Theorem
3.14, using Theorem 3.16.]
Theorem 3.14 is the usual characterization of normal extensions. Theorem 3.16 states the properties of normal closures. However, I don't think we need normal closures here.
We shall prove that the above condition implies the following, which will prove that $F$ is normal over $K$: If $overlineK$ is any algebraic closure of $K$ containing $F$, then for any $K$-monomorphism of fields $sigma:FtooverlineK$, $operatornameimsigma=F$ so that $sigma$ is actually a $K$-automorphism of $F$.
Proof. Extend $sigma:FtooverlineK$ to a $K$-monomorphism $overlinesigma:NtooverlineK$ (this is possible since $N$ is algebraic over $F$). Since $N$ is normal over $K$, we have $operatornameimoverlinesigma=N$, so $sigma$ actually maps $F$ into $N$. By hypothesis $operatornameimsigma=F$, so we're done.
The other direction is rather obvious, and does not need normal closures either. So I have not made any use of normal closures here. But why is the hint asking me to do so? Is my proof correct?
proof-verification field-theory galois-theory
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The following is Exercise V.3.21 from Hungerford's Algebra (GTM 73).
Let $F$ be algebraic over $K$. $F$ is normal over $K$ if and only if for every $K$-monomorphism of fields $sigma:Fto N$, where $N$ is any normal extension of $K$ containing $F$, $sigma(F)=F$ so that $sigma$ is a $K$-automorphism of $F$. [Hint: Adapt the proof of Theorem
3.14, using Theorem 3.16.]
Theorem 3.14 is the usual characterization of normal extensions. Theorem 3.16 states the properties of normal closures. However, I don't think we need normal closures here.
We shall prove that the above condition implies the following, which will prove that $F$ is normal over $K$: If $overlineK$ is any algebraic closure of $K$ containing $F$, then for any $K$-monomorphism of fields $sigma:FtooverlineK$, $operatornameimsigma=F$ so that $sigma$ is actually a $K$-automorphism of $F$.
Proof. Extend $sigma:FtooverlineK$ to a $K$-monomorphism $overlinesigma:NtooverlineK$ (this is possible since $N$ is algebraic over $F$). Since $N$ is normal over $K$, we have $operatornameimoverlinesigma=N$, so $sigma$ actually maps $F$ into $N$. By hypothesis $operatornameimsigma=F$, so we're done.
The other direction is rather obvious, and does not need normal closures either. So I have not made any use of normal closures here. But why is the hint asking me to do so? Is my proof correct?
proof-verification field-theory galois-theory
The following is Exercise V.3.21 from Hungerford's Algebra (GTM 73).
Let $F$ be algebraic over $K$. $F$ is normal over $K$ if and only if for every $K$-monomorphism of fields $sigma:Fto N$, where $N$ is any normal extension of $K$ containing $F$, $sigma(F)=F$ so that $sigma$ is a $K$-automorphism of $F$. [Hint: Adapt the proof of Theorem
3.14, using Theorem 3.16.]
Theorem 3.14 is the usual characterization of normal extensions. Theorem 3.16 states the properties of normal closures. However, I don't think we need normal closures here.
We shall prove that the above condition implies the following, which will prove that $F$ is normal over $K$: If $overlineK$ is any algebraic closure of $K$ containing $F$, then for any $K$-monomorphism of fields $sigma:FtooverlineK$, $operatornameimsigma=F$ so that $sigma$ is actually a $K$-automorphism of $F$.
Proof. Extend $sigma:FtooverlineK$ to a $K$-monomorphism $overlinesigma:NtooverlineK$ (this is possible since $N$ is algebraic over $F$). Since $N$ is normal over $K$, we have $operatornameimoverlinesigma=N$, so $sigma$ actually maps $F$ into $N$. By hypothesis $operatornameimsigma=F$, so we're done.
The other direction is rather obvious, and does not need normal closures either. So I have not made any use of normal closures here. But why is the hint asking me to do so? Is my proof correct?
proof-verification field-theory galois-theory
asked Aug 17 at 0:55
Colescu
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2,8131733
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Everything you say seems completely correct, but I think our problem is that the exercise is sloppily stated.
I think that the exercise should be stated as follows: Let $Fsupset K$ be an algebraic extension and $N$ a normal extension of $K$ containing $F$. Then $F$ is normal over $K$ if and only if for every $K$-morphism $sigma:Fto N$, $sigma(F)=F$.
What he seems to want is that to check for normality of the extension $Fsupset K$, you need only embed the situation in some (perhaps well-chosen) normal extension of $K$, not (for instance) an algebraically closed field containing $K$. Then this would be a useful theorem.
Notice that our author has violated the principle of putting all the quantifiers at the beginning. Let this be a lesson to all of us: youâÂÂre asking for trouble if you embed quantifiers in the middle of a proposition.
Thanks for answering! Yes, that's what I thought what he meant, i.e., the $N$ is arbitrarily chosen. So my proof is correct? Which means we don't really need normal closures here?
â Colescu
Aug 17 at 3:41
Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, IâÂÂd prefer to say that the given situation is $Nsupset Fsupset K$, with $N$ normal over $K$. To apply the proposition, you choose whatever $N$ you like, subject to those conditions.
â Lubin
Aug 18 at 2:52
Thank you for helping!
â Colescu
Aug 18 at 7:49
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Everything you say seems completely correct, but I think our problem is that the exercise is sloppily stated.
I think that the exercise should be stated as follows: Let $Fsupset K$ be an algebraic extension and $N$ a normal extension of $K$ containing $F$. Then $F$ is normal over $K$ if and only if for every $K$-morphism $sigma:Fto N$, $sigma(F)=F$.
What he seems to want is that to check for normality of the extension $Fsupset K$, you need only embed the situation in some (perhaps well-chosen) normal extension of $K$, not (for instance) an algebraically closed field containing $K$. Then this would be a useful theorem.
Notice that our author has violated the principle of putting all the quantifiers at the beginning. Let this be a lesson to all of us: youâÂÂre asking for trouble if you embed quantifiers in the middle of a proposition.
Thanks for answering! Yes, that's what I thought what he meant, i.e., the $N$ is arbitrarily chosen. So my proof is correct? Which means we don't really need normal closures here?
â Colescu
Aug 17 at 3:41
Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, IâÂÂd prefer to say that the given situation is $Nsupset Fsupset K$, with $N$ normal over $K$. To apply the proposition, you choose whatever $N$ you like, subject to those conditions.
â Lubin
Aug 18 at 2:52
Thank you for helping!
â Colescu
Aug 18 at 7:49
add a comment |Â
up vote
2
down vote
accepted
Everything you say seems completely correct, but I think our problem is that the exercise is sloppily stated.
I think that the exercise should be stated as follows: Let $Fsupset K$ be an algebraic extension and $N$ a normal extension of $K$ containing $F$. Then $F$ is normal over $K$ if and only if for every $K$-morphism $sigma:Fto N$, $sigma(F)=F$.
What he seems to want is that to check for normality of the extension $Fsupset K$, you need only embed the situation in some (perhaps well-chosen) normal extension of $K$, not (for instance) an algebraically closed field containing $K$. Then this would be a useful theorem.
Notice that our author has violated the principle of putting all the quantifiers at the beginning. Let this be a lesson to all of us: youâÂÂre asking for trouble if you embed quantifiers in the middle of a proposition.
Thanks for answering! Yes, that's what I thought what he meant, i.e., the $N$ is arbitrarily chosen. So my proof is correct? Which means we don't really need normal closures here?
â Colescu
Aug 17 at 3:41
Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, IâÂÂd prefer to say that the given situation is $Nsupset Fsupset K$, with $N$ normal over $K$. To apply the proposition, you choose whatever $N$ you like, subject to those conditions.
â Lubin
Aug 18 at 2:52
Thank you for helping!
â Colescu
Aug 18 at 7:49
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Everything you say seems completely correct, but I think our problem is that the exercise is sloppily stated.
I think that the exercise should be stated as follows: Let $Fsupset K$ be an algebraic extension and $N$ a normal extension of $K$ containing $F$. Then $F$ is normal over $K$ if and only if for every $K$-morphism $sigma:Fto N$, $sigma(F)=F$.
What he seems to want is that to check for normality of the extension $Fsupset K$, you need only embed the situation in some (perhaps well-chosen) normal extension of $K$, not (for instance) an algebraically closed field containing $K$. Then this would be a useful theorem.
Notice that our author has violated the principle of putting all the quantifiers at the beginning. Let this be a lesson to all of us: youâÂÂre asking for trouble if you embed quantifiers in the middle of a proposition.
Everything you say seems completely correct, but I think our problem is that the exercise is sloppily stated.
I think that the exercise should be stated as follows: Let $Fsupset K$ be an algebraic extension and $N$ a normal extension of $K$ containing $F$. Then $F$ is normal over $K$ if and only if for every $K$-morphism $sigma:Fto N$, $sigma(F)=F$.
What he seems to want is that to check for normality of the extension $Fsupset K$, you need only embed the situation in some (perhaps well-chosen) normal extension of $K$, not (for instance) an algebraically closed field containing $K$. Then this would be a useful theorem.
Notice that our author has violated the principle of putting all the quantifiers at the beginning. Let this be a lesson to all of us: youâÂÂre asking for trouble if you embed quantifiers in the middle of a proposition.
answered Aug 17 at 3:27
Lubin
41.2k34184
41.2k34184
Thanks for answering! Yes, that's what I thought what he meant, i.e., the $N$ is arbitrarily chosen. So my proof is correct? Which means we don't really need normal closures here?
â Colescu
Aug 17 at 3:41
Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, IâÂÂd prefer to say that the given situation is $Nsupset Fsupset K$, with $N$ normal over $K$. To apply the proposition, you choose whatever $N$ you like, subject to those conditions.
â Lubin
Aug 18 at 2:52
Thank you for helping!
â Colescu
Aug 18 at 7:49
add a comment |Â
Thanks for answering! Yes, that's what I thought what he meant, i.e., the $N$ is arbitrarily chosen. So my proof is correct? Which means we don't really need normal closures here?
â Colescu
Aug 17 at 3:41
Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, IâÂÂd prefer to say that the given situation is $Nsupset Fsupset K$, with $N$ normal over $K$. To apply the proposition, you choose whatever $N$ you like, subject to those conditions.
â Lubin
Aug 18 at 2:52
Thank you for helping!
â Colescu
Aug 18 at 7:49
Thanks for answering! Yes, that's what I thought what he meant, i.e., the $N$ is arbitrarily chosen. So my proof is correct? Which means we don't really need normal closures here?
â Colescu
Aug 17 at 3:41
Thanks for answering! Yes, that's what I thought what he meant, i.e., the $N$ is arbitrarily chosen. So my proof is correct? Which means we don't really need normal closures here?
â Colescu
Aug 17 at 3:41
Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, IâÂÂd prefer to say that the given situation is $Nsupset Fsupset K$, with $N$ normal over $K$. To apply the proposition, you choose whatever $N$ you like, subject to those conditions.
â Lubin
Aug 18 at 2:52
Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, IâÂÂd prefer to say that the given situation is $Nsupset Fsupset K$, with $N$ normal over $K$. To apply the proposition, you choose whatever $N$ you like, subject to those conditions.
â Lubin
Aug 18 at 2:52
Thank you for helping!
â Colescu
Aug 18 at 7:49
Thank you for helping!
â Colescu
Aug 18 at 7:49
add a comment |Â
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