When I'm self-studying Parick Morandi's book Field and Galois Theory,I came across some problems,which I can't work out fully. Let $K$, $L$, be two extension fields of base field $F$. If $K/F$ and $L/F$ are both Galois extension ,how to proof that their composite $KL$ is also Galois over $F$. And why we have the same conclusion when "Galois "is replaced by "normal","separable","purely inseparable",or "algebraic"? I worked on these problems and searched this website for two days but could only proof the cases for "separable","purely inseparable",and "algebraic".I have no idea how to proof the "normal" and "Galois" cases. In addition ,why the degree inequality holds for $$ [KL:F]\leq[K:F][L:F]$$ If you know how to proof these ,please tell me.I am here waiting for the answers.Thank you!

  • $\begingroup$ Actually if the "normal" case is right,then the “Galois " case follows since Galois extension is precisely normal and separable extension and the "separable "case I have proven. $\endgroup$ – 王李远 May 11 '16 at 13:45

You already know how to prove the separable case, and a field extension is galois if it is finite, separable, and normal, so it remains to prove the normal case.
First notice that, if a field extension $L/F$ is generated by elements $x_i$ such that all conjugates of $x_i$ are inside $L,$ then $L/F$ is normal by linearity. Then observe that, if $B_1:=\{x_1,\cdots,x_n\}$ is a basis for $L/F,$ and $B_2:=\{y_1,\cdots,y_m\}$ a basis for $K/F,$ then $\{x_iy_j|i=1,\cdots,n, j=1,\cdots,m\}$ generates $KL/F.$
For the degree inequality, it is also explained by the fact that $\{x_iy_j|i=1,\cdots,n, j=1,\cdots,m\}$ generates $LK/F$ if $B_i$ is a generating set for $i=1,2.$
I think this is enough to prove the proposition, if something is confusing, inappropriate, or wrong, please inform me, thanks.

  • $\begingroup$ I'm not so sure that : $\{x_iy_j|i=1,\cdots,n, j=1,\cdots,m\}$ is a basis for $KL/F.$@awllower $\endgroup$ – 王李远 May 11 '16 at 13:59
  • $\begingroup$ @王李远 Thanks, that set only generates $KL/F,$ which suffices for our purpose here. $\endgroup$ – awllower May 11 '16 at 14:19
  • $\begingroup$ I know that $\{x_iy_j|i=1,\cdots,n, j=1,\cdots,m\}$ generates $KL/F$, but this doesn't mean that they are a basis. For example ,for the extension $Q( \sqrt[3]{2},\omega)/Q$, $\sqrt[3]{2}$,$\omega $ generates the field ,but not a basis. $\endgroup$ – 王李远 May 11 '16 at 14:20
  • $\begingroup$ Yes ,it is suffices.but how about the degree inequality@awllower $\endgroup$ – 王李远 May 11 '16 at 14:22
  • $\begingroup$ In fact, if an extension $F/K$ is generated by $n$ elements, then $\left[F:K\right]\le n,$ so the generation of $KL/F$ by $nm$ elements suffices for that inequality. Also, I did not claim the set being a basis now. $\endgroup$ – awllower May 11 '16 at 14:34

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