This question is based on Theorem 10.2 of John Howie's 'Fields and Galois Theory', p.171. All fields have characteristic zero.

$M$ is a normal radical field extension of $K$, i.e. $M=K(\alpha_1,\alpha_2,...,\alpha_n)$ where $\alpha_i^{p_i}\in K(\alpha_1,...\alpha_{i-1})$ for some prime $p_i$.

We also define $P=M(\omega)$ where $\omega$ is a primitive $p_i$th root of unity. Howie points out that, as $P$ is a splitting field for $X^{p_i}-1$ over $M$, $P/M$ is a normal extension of $M$. So far so good.

As we have $M/K$ normal then, using the Fundamental Theorem of Galois Theory (FTGT), we have $\text{Gal}(P/M)\vartriangleleft \text{Gal}(P/K)$. Also by FTGT we have the following isomorphism:


However, in order for the FTGT to apply, I understand that we also require $P$ to be a normal extension of $K$. Howie does not seem to be explicit about why this condition holds. The only strategy I know would be to show that $P=K(\omega,\alpha_1,\alpha_2,...,\alpha_n)$ is a splitting field for a polynomial over $K$, however it is not obvious to me what that polynomial would be, or how I could show that its splitting fields was the one required.


You assume that $M/K$ is a normal extension. This means there is a polynomial $f\in K[x]$ such that $M$ is a splitting field of $f$ over $K$. But then note that $P$ is a splitting field of $(x^{p_i}-1)f(x)\in K[x]$, and hence $P/K$ is normal.

Note that in general if $P/M$ and $M/K$ are both normal extensions it doesn't imply that $P/K$ must be normal. Just in your case we are lucky that $x^{p_i}-1$ is also a polynomial in $K[x]$.


The vertical composition of normal field extensions isn’t normal, but the horizontal compositon is.

Let $E_1/F$ end $E_2/F$ be finite field extensions of a common base field $F$. If $E_1/F$ and $E_2/F$ are normal, then $E_1E_2/F$ is normal.

Proof. Let $f_1, f_2 ∈ F[X]$ be polynomials such that $E_1$ and $E_2$ are splitting fields of $f_1$ and $f_2$ respectively. Then $E_1E_2$ is a splitting field of $f_1f_2 ∈ F[X]$.

Here, $M(ω)/K$ is a horizontal composition of normal field extensions, by $M(ω) = K(ω)M$.

A counterexample for the failure of the vertical composition of normal field extensions to be normal is $ℚ(\sqrt[4] 2)/ℚ$ with its intermediate field $ℚ(\sqrt 2)$.

  • $\begingroup$ "a splitting field of" is a little ambiguous I'd say $\endgroup$
    – reuns
    Jul 26 '19 at 22:54
  • $\begingroup$ @reuns … how so? $\endgroup$
    – k.stm
    Jul 27 '19 at 5:24

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