Show that $[K:F]_s = [K:L]_s [L:F]_s$ and $[K:F]_i = [K:L]_i [L:F]_i$.












1












$begingroup$


This is a problem from Patrick Morandi's Field and Galois Theory: Chapter I.4, exercise 15.



Let $K$ be a finite extension of $F$. If $S={xin K mid x text{ is separable over } F }$ and $I={xin K mid x text{ is purely inseparable over } F }$ are the separable and purely inseparable closures of $F$ in $K$, respectively, we define the separable degree $[K:F]_s = [S:F]$ and the purely inseparable degree $[K:F]_i = [K:S]$.



Now using these definitions,




Prove the following product formulas for separability and inseparability degree: If $F subseteq L subseteq K$ are fields, then show that $[K:F]_s = [K:L]_s [L:F]_s$ and $[K:F]_i = [K:L]_i [L:F]_i$.




Proving just one of the equality will be enough (thanks to Tower law and the property that $[K:F]_s [K:F]_i = [K:F]$).



I started proving:



$[K:F]_s=[K:L]_s[L:F]_s$. Suppose $[K:L]_s=[S_1:F]$ and $[L:F]_s=[S_2:F]$
where $S_1$ and $S_2$ are separable closures of $K/L$ and $L/F$ respectively. And if we take ${a_i mid i=1,dots,m}$ and ${b_i mid i=1,dots,n}$ are the basis of $K/L$ and $L/F$ respectively. Is it true that ${a_ib_j mid i=1,dots,m, j=1,dots,n}$ is a basis of $S/F$? Then for $x in S$ how do we proceed?






field-theory galois-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If you can show that $[Kcolon F]_s$ is the number of distinct $F$-morphisms of $K$ into an algebraically closed extension of $K$, the multiplicativity of the separable degree drops out relatively easily (no bases involved).
    $endgroup$
    – Lubin
    Oct 15 '15 at 1:03












  • $begingroup$
    Yeah I know then I have to prove the equivalency b/w these two defn then how to show that? Otherwise I have to go directly.
    $endgroup$
    – user152715
    Oct 15 '15 at 5:34










  • $begingroup$
    As a matter of fact, a few months ago I was feeling unsure about these concepts, and wrote up, for my own benefit, a treatment that turned out to bounce back and forth between counting morphisms and looking at separable polynomials and their roots. It took more space than I expected.
    $endgroup$
    – Lubin
    Oct 15 '15 at 14:09










  • $begingroup$
    @Lubin, could you please just write down your answer here, so that others can learn
    $endgroup$
    – 王李远
    May 5 '17 at 9:16






  • 1




    $begingroup$
    @王李远 , instead of writing up a whole thing, I’ve put my private note on my Brown web site. It’s at math.brown.edu/~lubinj/sep.pdf . I hope this suffices; if not let me know.
    $endgroup$
    – Lubin
    May 5 '17 at 15:21
















1












$begingroup$


This is a problem from Patrick Morandi's Field and Galois Theory: Chapter I.4, exercise 15.



Let $K$ be a finite extension of $F$. If $S={xin K mid x text{ is separable over } F }$ and $I={xin K mid x text{ is purely inseparable over } F }$ are the separable and purely inseparable closures of $F$ in $K$, respectively, we define the separable degree $[K:F]_s = [S:F]$ and the purely inseparable degree $[K:F]_i = [K:S]$.



Now using these definitions,




Prove the following product formulas for separability and inseparability degree: If $F subseteq L subseteq K$ are fields, then show that $[K:F]_s = [K:L]_s [L:F]_s$ and $[K:F]_i = [K:L]_i [L:F]_i$.




Proving just one of the equality will be enough (thanks to Tower law and the property that $[K:F]_s [K:F]_i = [K:F]$).



I started proving:



$[K:F]_s=[K:L]_s[L:F]_s$. Suppose $[K:L]_s=[S_1:F]$ and $[L:F]_s=[S_2:F]$
where $S_1$ and $S_2$ are separable closures of $K/L$ and $L/F$ respectively. And if we take ${a_i mid i=1,dots,m}$ and ${b_i mid i=1,dots,n}$ are the basis of $K/L$ and $L/F$ respectively. Is it true that ${a_ib_j mid i=1,dots,m, j=1,dots,n}$ is a basis of $S/F$? Then for $x in S$ how do we proceed?






field-theory galois-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If you can show that $[Kcolon F]_s$ is the number of distinct $F$-morphisms of $K$ into an algebraically closed extension of $K$, the multiplicativity of the separable degree drops out relatively easily (no bases involved).
    $endgroup$
    – Lubin
    Oct 15 '15 at 1:03












  • $begingroup$
    Yeah I know then I have to prove the equivalency b/w these two defn then how to show that? Otherwise I have to go directly.
    $endgroup$
    – user152715
    Oct 15 '15 at 5:34










  • $begingroup$
    As a matter of fact, a few months ago I was feeling unsure about these concepts, and wrote up, for my own benefit, a treatment that turned out to bounce back and forth between counting morphisms and looking at separable polynomials and their roots. It took more space than I expected.
    $endgroup$
    – Lubin
    Oct 15 '15 at 14:09










  • $begingroup$
    @Lubin, could you please just write down your answer here, so that others can learn
    $endgroup$
    – 王李远
    May 5 '17 at 9:16






  • 1




    $begingroup$
    @王李远 , instead of writing up a whole thing, I’ve put my private note on my Brown web site. It’s at math.brown.edu/~lubinj/sep.pdf . I hope this suffices; if not let me know.
    $endgroup$
    – Lubin
    May 5 '17 at 15:21














1












1








1


1



$begingroup$


This is a problem from Patrick Morandi's Field and Galois Theory: Chapter I.4, exercise 15.



Let $K$ be a finite extension of $F$. If $S={xin K mid x text{ is separable over } F }$ and $I={xin K mid x text{ is purely inseparable over } F }$ are the separable and purely inseparable closures of $F$ in $K$, respectively, we define the separable degree $[K:F]_s = [S:F]$ and the purely inseparable degree $[K:F]_i = [K:S]$.



Now using these definitions,




Prove the following product formulas for separability and inseparability degree: If $F subseteq L subseteq K$ are fields, then show that $[K:F]_s = [K:L]_s [L:F]_s$ and $[K:F]_i = [K:L]_i [L:F]_i$.




Proving just one of the equality will be enough (thanks to Tower law and the property that $[K:F]_s [K:F]_i = [K:F]$).



I started proving:



$[K:F]_s=[K:L]_s[L:F]_s$. Suppose $[K:L]_s=[S_1:F]$ and $[L:F]_s=[S_2:F]$
where $S_1$ and $S_2$ are separable closures of $K/L$ and $L/F$ respectively. And if we take ${a_i mid i=1,dots,m}$ and ${b_i mid i=1,dots,n}$ are the basis of $K/L$ and $L/F$ respectively. Is it true that ${a_ib_j mid i=1,dots,m, j=1,dots,n}$ is a basis of $S/F$? Then for $x in S$ how do we proceed?






field-theory galois-theory







share|cite|improve this question











$endgroup$




This is a problem from Patrick Morandi's Field and Galois Theory: Chapter I.4, exercise 15.



Let $K$ be a finite extension of $F$. If $S={xin K mid x text{ is separable over } F }$ and $I={xin K mid x text{ is purely inseparable over } F }$ are the separable and purely inseparable closures of $F$ in $K$, respectively, we define the separable degree $[K:F]_s = [S:F]$ and the purely inseparable degree $[K:F]_i = [K:S]$.



Now using these definitions,




Prove the following product formulas for separability and inseparability degree: If $F subseteq L subseteq K$ are fields, then show that $[K:F]_s = [K:L]_s [L:F]_s$ and $[K:F]_i = [K:L]_i [L:F]_i$.




Proving just one of the equality will be enough (thanks to Tower law and the property that $[K:F]_s [K:F]_i = [K:F]$).



I started proving:



$[K:F]_s=[K:L]_s[L:F]_s$. Suppose $[K:L]_s=[S_1:F]$ and $[L:F]_s=[S_2:F]$
where $S_1$ and $S_2$ are separable closures of $K/L$ and $L/F$ respectively. And if we take ${a_i mid i=1,dots,m}$ and ${b_i mid i=1,dots,n}$ are the basis of $K/L$ and $L/F$ respectively. Is it true that ${a_ib_j mid i=1,dots,m, j=1,dots,n}$ is a basis of $S/F$? Then for $x in S$ how do we proceed?






field-theory galois-theory




field-theory galois-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 17:57









Brahadeesh

6,42442363




6,42442363










asked Oct 14 '15 at 18:06









user152715user152715

1




1








  • 1




    $begingroup$
    If you can show that $[Kcolon F]_s$ is the number of distinct $F$-morphisms of $K$ into an algebraically closed extension of $K$, the multiplicativity of the separable degree drops out relatively easily (no bases involved).
    $endgroup$
    – Lubin
    Oct 15 '15 at 1:03












  • $begingroup$
    Yeah I know then I have to prove the equivalency b/w these two defn then how to show that? Otherwise I have to go directly.
    $endgroup$
    – user152715
    Oct 15 '15 at 5:34










  • $begingroup$
    As a matter of fact, a few months ago I was feeling unsure about these concepts, and wrote up, for my own benefit, a treatment that turned out to bounce back and forth between counting morphisms and looking at separable polynomials and their roots. It took more space than I expected.
    $endgroup$
    – Lubin
    Oct 15 '15 at 14:09










  • $begingroup$
    @Lubin, could you please just write down your answer here, so that others can learn
    $endgroup$
    – 王李远
    May 5 '17 at 9:16






  • 1




    $begingroup$
    @王李远 , instead of writing up a whole thing, I’ve put my private note on my Brown web site. It’s at math.brown.edu/~lubinj/sep.pdf . I hope this suffices; if not let me know.
    $endgroup$
    – Lubin
    May 5 '17 at 15:21














  • 1




    $begingroup$
    If you can show that $[Kcolon F]_s$ is the number of distinct $F$-morphisms of $K$ into an algebraically closed extension of $K$, the multiplicativity of the separable degree drops out relatively easily (no bases involved).
    $endgroup$
    – Lubin
    Oct 15 '15 at 1:03












  • $begingroup$
    Yeah I know then I have to prove the equivalency b/w these two defn then how to show that? Otherwise I have to go directly.
    $endgroup$
    – user152715
    Oct 15 '15 at 5:34










  • $begingroup$
    As a matter of fact, a few months ago I was feeling unsure about these concepts, and wrote up, for my own benefit, a treatment that turned out to bounce back and forth between counting morphisms and looking at separable polynomials and their roots. It took more space than I expected.
    $endgroup$
    – Lubin
    Oct 15 '15 at 14:09










  • $begingroup$
    @Lubin, could you please just write down your answer here, so that others can learn
    $endgroup$
    – 王李远
    May 5 '17 at 9:16






  • 1




    $begingroup$
    @王李远 , instead of writing up a whole thing, I’ve put my private note on my Brown web site. It’s at math.brown.edu/~lubinj/sep.pdf . I hope this suffices; if not let me know.
    $endgroup$
    – Lubin
    May 5 '17 at 15:21








1




1




$begingroup$
If you can show that $[Kcolon F]_s$ is the number of distinct $F$-morphisms of $K$ into an algebraically closed extension of $K$, the multiplicativity of the separable degree drops out relatively easily (no bases involved).
$endgroup$
– Lubin
Oct 15 '15 at 1:03






$begingroup$
If you can show that $[Kcolon F]_s$ is the number of distinct $F$-morphisms of $K$ into an algebraically closed extension of $K$, the multiplicativity of the separable degree drops out relatively easily (no bases involved).
$endgroup$
– Lubin
Oct 15 '15 at 1:03














$begingroup$
Yeah I know then I have to prove the equivalency b/w these two defn then how to show that? Otherwise I have to go directly.
$endgroup$
– user152715
Oct 15 '15 at 5:34




$begingroup$
Yeah I know then I have to prove the equivalency b/w these two defn then how to show that? Otherwise I have to go directly.
$endgroup$
– user152715
Oct 15 '15 at 5:34












$begingroup$
As a matter of fact, a few months ago I was feeling unsure about these concepts, and wrote up, for my own benefit, a treatment that turned out to bounce back and forth between counting morphisms and looking at separable polynomials and their roots. It took more space than I expected.
$endgroup$
– Lubin
Oct 15 '15 at 14:09




$begingroup$
As a matter of fact, a few months ago I was feeling unsure about these concepts, and wrote up, for my own benefit, a treatment that turned out to bounce back and forth between counting morphisms and looking at separable polynomials and their roots. It took more space than I expected.
$endgroup$
– Lubin
Oct 15 '15 at 14:09












$begingroup$
@Lubin, could you please just write down your answer here, so that others can learn
$endgroup$
– 王李远
May 5 '17 at 9:16




$begingroup$
@Lubin, could you please just write down your answer here, so that others can learn
$endgroup$
– 王李远
May 5 '17 at 9:16




1




1




$begingroup$
@王李远 , instead of writing up a whole thing, I’ve put my private note on my Brown web site. It’s at math.brown.edu/~lubinj/sep.pdf . I hope this suffices; if not let me know.
$endgroup$
– Lubin
May 5 '17 at 15:21




$begingroup$
@王李远 , instead of writing up a whole thing, I’ve put my private note on my Brown web site. It’s at math.brown.edu/~lubinj/sep.pdf . I hope this suffices; if not let me know.
$endgroup$
– Lubin
May 5 '17 at 15:21










1 Answer
1






active

oldest

votes


















1












$begingroup$

Patrick Morandi sometimes poses exercises in a section earlier than you are expected to solve them. This seems to be one of those exercises. You can read Lemmas 8.9 and 8.11 on pages 82-83, which will help answer the problem thoroughly. I am giving their statements below:




Lemma 8.9. Let $K$ be a finite extension of $F$, and let $S$ be the separable closure of $F$ in $K$. Then $[S:F]$ is equal to the number of $F$-homomorphisms from $K$ to an algebraic closure of $F$.



Lemma 8.11. Suppose that $F subseteq L subseteq K$ are fields with $[K:F] < infty$. Then $[K:F]_i = [K:L]_i [L:F]_i$.




To be honest, I find that the treatment in Lang's Algebra is far more intuitive, so do try to read the relevant sections from Lang to see how he discusses separable and purely inseparable extensions.



Prof. Lubin has shared his succint notes on this problem in the comments. The link is



www.math.brown.edu/~lubinj/sep.pdf



Another answer by Prof. Lubin (using Lang's definition of separable degree) is given here: $[K : F]_s = [K : L]_s [L : F]_s $ and $[K : F]_i = [K : L]_i [L : F]_i $






share|cite|improve this answer









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    $begingroup$

    Patrick Morandi sometimes poses exercises in a section earlier than you are expected to solve them. This seems to be one of those exercises. You can read Lemmas 8.9 and 8.11 on pages 82-83, which will help answer the problem thoroughly. I am giving their statements below:




    Lemma 8.9. Let $K$ be a finite extension of $F$, and let $S$ be the separable closure of $F$ in $K$. Then $[S:F]$ is equal to the number of $F$-homomorphisms from $K$ to an algebraic closure of $F$.



    Lemma 8.11. Suppose that $F subseteq L subseteq K$ are fields with $[K:F] < infty$. Then $[K:F]_i = [K:L]_i [L:F]_i$.




    To be honest, I find that the treatment in Lang's Algebra is far more intuitive, so do try to read the relevant sections from Lang to see how he discusses separable and purely inseparable extensions.



    Prof. Lubin has shared his succint notes on this problem in the comments. The link is



    www.math.brown.edu/~lubinj/sep.pdf



    Another answer by Prof. Lubin (using Lang's definition of separable degree) is given here: $[K : F]_s = [K : L]_s [L : F]_s $ and $[K : F]_i = [K : L]_i [L : F]_i $






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Patrick Morandi sometimes poses exercises in a section earlier than you are expected to solve them. This seems to be one of those exercises. You can read Lemmas 8.9 and 8.11 on pages 82-83, which will help answer the problem thoroughly. I am giving their statements below:




      Lemma 8.9. Let $K$ be a finite extension of $F$, and let $S$ be the separable closure of $F$ in $K$. Then $[S:F]$ is equal to the number of $F$-homomorphisms from $K$ to an algebraic closure of $F$.



      Lemma 8.11. Suppose that $F subseteq L subseteq K$ are fields with $[K:F] < infty$. Then $[K:F]_i = [K:L]_i [L:F]_i$.




      To be honest, I find that the treatment in Lang's Algebra is far more intuitive, so do try to read the relevant sections from Lang to see how he discusses separable and purely inseparable extensions.



      Prof. Lubin has shared his succint notes on this problem in the comments. The link is



      www.math.brown.edu/~lubinj/sep.pdf



      Another answer by Prof. Lubin (using Lang's definition of separable degree) is given here: $[K : F]_s = [K : L]_s [L : F]_s $ and $[K : F]_i = [K : L]_i [L : F]_i $






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Patrick Morandi sometimes poses exercises in a section earlier than you are expected to solve them. This seems to be one of those exercises. You can read Lemmas 8.9 and 8.11 on pages 82-83, which will help answer the problem thoroughly. I am giving their statements below:




        Lemma 8.9. Let $K$ be a finite extension of $F$, and let $S$ be the separable closure of $F$ in $K$. Then $[S:F]$ is equal to the number of $F$-homomorphisms from $K$ to an algebraic closure of $F$.



        Lemma 8.11. Suppose that $F subseteq L subseteq K$ are fields with $[K:F] < infty$. Then $[K:F]_i = [K:L]_i [L:F]_i$.




        To be honest, I find that the treatment in Lang's Algebra is far more intuitive, so do try to read the relevant sections from Lang to see how he discusses separable and purely inseparable extensions.



        Prof. Lubin has shared his succint notes on this problem in the comments. The link is



        www.math.brown.edu/~lubinj/sep.pdf



        Another answer by Prof. Lubin (using Lang's definition of separable degree) is given here: $[K : F]_s = [K : L]_s [L : F]_s $ and $[K : F]_i = [K : L]_i [L : F]_i $






        share|cite|improve this answer









        $endgroup$



        Patrick Morandi sometimes poses exercises in a section earlier than you are expected to solve them. This seems to be one of those exercises. You can read Lemmas 8.9 and 8.11 on pages 82-83, which will help answer the problem thoroughly. I am giving their statements below:




        Lemma 8.9. Let $K$ be a finite extension of $F$, and let $S$ be the separable closure of $F$ in $K$. Then $[S:F]$ is equal to the number of $F$-homomorphisms from $K$ to an algebraic closure of $F$.



        Lemma 8.11. Suppose that $F subseteq L subseteq K$ are fields with $[K:F] < infty$. Then $[K:F]_i = [K:L]_i [L:F]_i$.




        To be honest, I find that the treatment in Lang's Algebra is far more intuitive, so do try to read the relevant sections from Lang to see how he discusses separable and purely inseparable extensions.



        Prof. Lubin has shared his succint notes on this problem in the comments. The link is



        www.math.brown.edu/~lubinj/sep.pdf



        Another answer by Prof. Lubin (using Lang's definition of separable degree) is given here: $[K : F]_s = [K : L]_s [L : F]_s $ and $[K : F]_i = [K : L]_i [L : F]_i $







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 17:44









        BrahadeeshBrahadeesh

        6,42442363




        6,42442363






























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