From Robert Ash (Basic Abstract Algebra)

Suppose that $E=F(\alpha)$ is a finite Galois extension of $F$, where $\alpha$ is a root of the irreducible polynomial $f \in F[x]$. Assume that the roots of $f$ are $\alpha_1 = \alpha, \alpha_2, ... , \alpha_n$. Describe, as best you can from the given information, the Galois group of $E/F$.

Answer: $G = \{\sigma_1, ... ,\sigma_n\}$ where $\sigma_i$ is the unique $F$-automorphism of $E$ that takes $\alpha$ to $\alpha_i$.

There are two things that confuse me about this answer:

1) What if some of the $\alpha_i$'s are in F? Then how would we have an $F$-automorphism that takes $\alpha$ to $\alpha_i$? Because we know that any $F$-automorhpism must fix all elements of $F$, right?

2) Don't we need $E$ to be a normal extension? Because how else would we know that the other roots of $f$ are also contained in $E$?

Thank you in advance


1) If any of the $\alpha_i$ were in $F$, then $f$ has a factor $X-\alpha_i$. Since $f$ is irreducible, we would have $f = x-\alpha_i$ and so $\alpha_i$ is the only root.

Still, you have the identity that takes $\alpha_i$ to itself, and this is the entire Galois group.

2) By definition, a Galois extension is normal and separable. If this wasn't your definition then you must have shown it to be equivalent earlier.


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