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Let $ L/K $ be a finite extension of fields and $ L_{1},L_{2} $ two intermediate fields that are Galois over $ K $. Is the composite field $ L_{1}L_{2} $ (i.e. the smallest subfield of $ L $ that contains both $ L_{1} $ and $ L_{2}$) Galois over $ K $?

My thought would be that this indeed is true since as $ L_{1} $ is Galois over $ K $, it is the splitting field of a family of separable polynomials $ \{f_{i} \} _{i \in I} $ over $ K $ and therefore $ L_{1}L_{2} $ is the splitting field of the same family of polynomials over $ L_{2} $. On the other hand, as $ L_{2} $ is Galois over $ K $, $ L_{2} $ is also the splitting field of a family $ \{g_{j}\}_{j \in J} $ of separable polynomials over $ K $ so $ L_{1}L_{2} $ is the splitting field of $ \{f_{i} \} _{i \in I} \cup \{g_{j}\}_{j \in J} $ over $ K $ thus the composite is Galois over $ K $.

Is there anything wrong with my answer? Thank you in advance for any help!

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  • $\begingroup$ Your answer assumes that $L_1$ and $L_2$ are Galois over not just $K$, but also over $L_1\cap L_2$. $\endgroup$ – user416426 Apr 13 '17 at 6:24
  • $\begingroup$ What if I write $ L_{1}=K(\alpha_{1}, \dots , \alpha_{r}) $ and $ L_{2}=K(\alpha_{r+1}, \dots ,\alpha_{n}) $ where $ \alpha_{1}, \dots ,\alpha_{r} $ are the roots of a separable polynomial $ f_{1} $ over $ K $ and $\alpha_{r+1}, \dots \alpha_{n} $ are the roots of another separable polynomial $ f_{2} $ over $ K $. Then $ L_{1}L_{2}=K(\alpha_{1}, \dots , \alpha_{n}) $ will be the splitting field of the separable polynomial $ \frac{f_{1}f_{2}}{gcd_{K[X]}(f_{1},f_{2})^{2}} $, hence Galois. Would this be a better approach? $\endgroup$ – Raizen Apr 13 '17 at 8:08
  • $\begingroup$ In the denominator above there should be just the GCD, not GCD squared $\endgroup$ – Raizen Apr 13 '17 at 8:15
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    $\begingroup$ But, @NoahRiggenbach, isn’t it automatically the case that if $K\subset M\subset L$ and $L$ is finite Galois over $K$, that then $L$ is Galois over $M$? After all, $L$ has been gotten by adjoining all roots of a $K$-polynomial, $f$, and $f$ is all the more so an $M$-polynomial as well. $\endgroup$ – Lubin Apr 13 '17 at 21:57
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    $\begingroup$ Oh, @NoahRiggenbach, I always mentioned it when I taught Galois Theory. $\endgroup$ – Lubin Apr 13 '17 at 22:38

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