Lemma 9.15.6. Let $E/F$ be an algebraic extension of fields. If $E$ is generated by $\alpha _ i \in E$, $i \in I$ over $F$ and if for each $i$ the minimal polynomial of $\alpha _ i$ over $F$ splits completely in $E$, then $E/F$ is normal.
Proof. Let $P_ i$ be the minimal polynomial of $\alpha _ i$ over $F$. Let $\alpha _ i = \alpha _{i, 1}, \alpha _{i, 2}, \ldots , \alpha _{i, d_ i}$ be the roots of $P_ i$ over $E$. Given two embeddings $\sigma , \sigma ' : E \to \overline{F}$ over $F$ we see that
\[ \{ \sigma (\alpha _{i, 1}), \ldots , \sigma (\alpha _{i, d_ i})\} = \{ \sigma '(\alpha _{i, 1}), \ldots , \sigma '(\alpha _{i, d_ i})\} \]
because both sides are equal to the set of roots of $P_ i$ in $\overline{F}$. The elements $\alpha _{i, j}$ generate $E$ over $F$ and we find that $\sigma (E) = \sigma '(E)$. Hence $E/F$ is normal by Lemma 9.15.5. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: