## Modern Algebra (ECM3731) Part 2: Rings, fields and Galois theory

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Final remark about finding the Galois group of a finite normal extension $F\subseteq K$ in $\mathbb C$.
If $K$ is given as $K=F(\alpha_1,...,\alpha_n)$, then any $F$-linear automorphism is determined by what it does to the $\alpha_i$. Furthermore, an element of $K$ can only be moved to one of its conjugates over $F$. So if $\alpha_i$ has $m_i$ conjugates over $F$ (including itself), then that gives us $m_1\cdots m_n$ possible automorphisms. Just choose for each $\alpha_i$ a conjugate over $F$. If you can somehow show that the order of the Galois group is also $m_1\cdots m_n$ (usually by showing that $[K:F]=m_1\cdots m_n$), then you know that all above possibilities must occur and you have determined the Galois group as a set. Then you can pick out suitable automorphisms that move one of the $\alpha_i$ and fix all the others, and show that they generate the Galois group. This gives the Galois group as a set and in small examples you can determine the group structure by determining a commutation rule for the generators. The above cardinality argument avoids the application of the argument from the proof of Thm III.5 that I used in the notes and the solutions to the exercises. The latter is more insightful, but the former is shorter.