# Conditions required for Galois Correspondence

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:

$$\text{Gal}(M/K)\simeq\frac{\text{Gal}(P/K)}{\text{Gal}(P/M)}$$

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 $$ℚ(\sqrt 2)/ℚ$$ with its intermediate field $$ℚ(\sqrt 2)$$.

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