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STATEMENT: Let $K$ be a splitting field for the family $\left\{f_i\right\}_{i\in I}$ and let $E$ be another splitting field. Any embedding of $E$ into $K^a$ inducing the identity on $k$ gives an isomorphism of $E$ onto $K$.

PROOF: Let the notation be as above. Note that $E$ contains a unique splitting field $E_i$ of $f_i$ and $K$ contains a unique splitting field $K_i$ of $f_i$. Any embeddinng $\sigma$ of $E$ into $K^a$ must map $E_i$ onto $K_i$ by $Theorem 3.1$, and hence maps $E$ into $K$.

QUESTION: Why can the author conclude that $E$ maps into $K$ after knowing that the unique splitting fields for each polynomial maps isomorphically into one another. Can't it be the case that it maps into a different field containing $K_i$?

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  • $\begingroup$ Just a rough idea, maybe I'm wrong. $K$ and $E$ are respectively the compositum of the $K_i$'s and the $E_i$'s. The statement show that any embedding of $E$ into $K^a$ should induce isomorphisms of $E_i$ and $K_i$ for each $i$. Every field containing all the $K_i$'s contains also $K$. So wherever you send $E$ its image will always be isomorphic to $K$. $\endgroup$ – Pgatti Dec 3 '14 at 17:47
  • $\begingroup$ But even if the image is isomorphic, it still doesn't justify how the author makes the conclusion that $E$ maps into $K$, right? $\endgroup$ – Enigma Dec 3 '14 at 18:50

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