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?







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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?







    share|cite|improve this question






















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      1
      down vote

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      up vote
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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?







      share|cite|improve this question












      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?









      share|cite|improve this question











      share|cite|improve this question




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      asked Aug 17 at 0:55









      Colescu

      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: you’re asking for trouble if you embed quantifiers in the middle of a proposition.






          share|cite|improve this answer




















          • 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?
            – Colescu
            Aug 17 at 3:41










          • Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, I’d 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.
            – Lubin
            Aug 18 at 2:52










          • Thank you for helping!
            – Colescu
            Aug 18 at 7:49










          Your Answer




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          up vote
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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: you’re asking for trouble if you embed quantifiers in the middle of a proposition.






          share|cite|improve this answer




















          • 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?
            – Colescu
            Aug 17 at 3:41










          • Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, I’d 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.
            – Lubin
            Aug 18 at 2:52










          • Thank you for helping!
            – Colescu
            Aug 18 at 7:49














          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: you’re asking for trouble if you embed quantifiers in the middle of a proposition.






          share|cite|improve this answer




















          • 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?
            – Colescu
            Aug 17 at 3:41










          • Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, I’d 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.
            – Lubin
            Aug 18 at 2:52










          • Thank you for helping!
            – Colescu
            Aug 18 at 7:49












          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: you’re asking for trouble if you embed quantifiers in the middle of a proposition.






          share|cite|improve this answer












          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: you’re asking for trouble if you embed quantifiers in the middle of a proposition.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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?
            – Colescu
            Aug 17 at 3:41










          • Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, I’d 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.
            – Lubin
            Aug 18 at 2:52










          • Thank you for helping!
            – 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?
            – Colescu
            Aug 17 at 3:41










          • Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, I’d 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.
            – Lubin
            Aug 18 at 2:52










          • Thank you for helping!
            – 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?
          – 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?
          – Colescu
          Aug 17 at 3:41












          Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, I’d 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.
          – Lubin
          Aug 18 at 2:52




          Yes, your proof is correct. Rather than saying that $N$ is arbitrarily chosen, I’d 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.
          – Lubin
          Aug 18 at 2:52












          Thank you for helping!
          – Colescu
          Aug 18 at 7:49




          Thank you for helping!
          – Colescu
          Aug 18 at 7:49












           

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