The following proof is from Robert Ash, Abstract Algebra, 6.8.2, and I have difficulty following it:

Let $E/F$ be a radical extension, and let $N$ be a normal closure of $E$ over $F$. Then $N/F$ is also a radical extension.

Proof. $E$ is obtained from $F$ by successively adjoining $\alpha_1, \ldots, \alpha_r$, where $\alpha_i$ is the $n$-th root of an element in $F_{i-1}$. On the other hand, $N$ is obtained from $F$ by adjoining not only the $\alpha_i$, but their conjugates $\alpha_{i1},\ldots, \alpha_{im(i)}$. For any fixed $i$ and $j$, there is an automorphism $\sigma \in \mbox{Gal}(N/F)$ such that $\sigma(\alpha_i) = \alpha_{ij}$ (see (3.2.3), (3.5.5) and (3.5.6)). Thus $$ \alpha_{ij}^n = \sigma(\alpha_i)^n = \alpha(\alpha_i^n) $$ and since $\alpha_i^n$ belongs to $F(\alpha_1, \ldots, \alpha_{i-1})$, it follows from (3.5.1) that $\sigma(\alpha_i^n)$ belongs to the splitting field $K_i$ of $\prod_{j=1}^{i-1} \mbox{min}(\alpha_j, F)$ over $F$. [Take $K_1 = F$, and note that since $\alpha_i^n = b_i \in F$, we have $\sigma(\alpha_i^n) = \sigma(b_i) = b_i \in F$. Alternatively, observe that by (3.5.1), $\sigma$ must take a root of $X^n - b_1$ to another root of this polynomial.] Thus we can display $N$ as a radical extension of $F$ by successively adjoining $$ \alpha_{11},\ldots, \alpha_{1m(1)}, \ldots, \alpha_{r1},\ldots, \alpha_{rm(r)}. \quad \square $$

I do not understand the 2nd paragraph. If we set $b_i := \alpha_i^n \in F(\alpha_1, \ldots, \alpha_{i-1})$, then $\alpha_i$ is a root of $X^n - b_i$. But $\sigma$ is an $F$-monomorphism, and $X^n - b_i$ is a polynomial over the larger field $F(\alpha_1, \ldots, \alpha_{i-1})$, so (3.5.1) does not gives that $\sigma(\alpha_i)$ is also a root of this polynomial. If it would be, then as $K_i$ is normal over $F$, and hence also over $F(\alpha_1, \ldots, \alpha_{i-1})$, this would give $\sigma(\alpha_i) \in K_i$. But other polynomials then $X^n - b_i$ does not come to my mind to somehow use normality of $K_i$ (as a splitting field, but it must be a polynomial over $F$ so that we can use (3.5.1)).

The cited parts:

(3.5.1) Let $\sigma : E \to E$ be an $F$-monomorphism, and assume that the polynomial $f \in F[X]$ splits over $E$. If $\alpha$ is a root of $f$ in $E$, then so is $\sigma(\alpha)$.

(3.2.3) If $\alpha$ and $\beta$ are roots of the irreducible polynomial $f \in F[X]$ in an extension $E$ of $F$, then $F(\alpha)$ is isomorphic to $F(\beta)$ via an isomorphism that carries $\alpha$ into $\beta$ and is the identity on $F$.

(3.5.5) The finite extension $E/F$ is normal iff every $F$-monomorphism of $E$ into an algebraic closure $C$ is actually an $F$-automorphism of $E$.

(3.5.6.) In the above, the algebraic closure could be replaced by any fixed normal extension of $F$ containing $E$.

  • $\begingroup$ @reuns Sorry, I do not understand what this has to do with the arguments given in the proof? $\endgroup$ – StefanH Oct 2 '17 at 15:52
  • $\begingroup$ I meant if $\alpha$ is a root of $x^m-a$ and $\sigma \in \text{Gal}(N/F)$ then $\sigma(\alpha)$ is a root of $x^m-\sigma(a)$. Thus the $F$-conjugates of $\sqrt[m]{a}$ are $\zeta_d^l \sqrt[m]{a_j}$ where $d= [F(\sqrt[m]{a}):F]$ and $a_j$ are the cojugates of $a$. $\endgroup$ – reuns Oct 2 '17 at 16:26

As i understand it, it works the following way:

You look at the polynomial $f_i$ defined by multiplying all the minimal polynomials of the $\alpha$'s up to $\alpha_{i-1}$. Now you know that his splitting field $K_i$ contains $F(\alpha_1,\dots,\alpha_{i-1})$. Now, since $\alpha_i^n$ is contained in $F(\alpha_1,\dots,\alpha_{i-1})$, it is contained in $K_i$. But $K_i$ is normal over $F$, so the minimal polynomial of $\alpha_i^n$ in $F$ splits in $K_i$ and therefore the monomorphism $\sigma$ sends $\alpha_i^n$ to another root of this polynomial in $K_i$.

  • $\begingroup$ Ok. Thank you. May I ask you concerning one of the first sentences of the proof: "$N$ is obtained from $F$ by adjoining not only the $\alpha_i$, but their conjugates $\alpha_{i1},\ldots, \alpha_{im(i)}$" - So isn't everything already said here, that we adjoin roots and hence it could be written as a radical extension? Why still arguing that those conjugates are in $N$ then? $\endgroup$ – StefanH Oct 2 '17 at 18:21
  • 1
    $\begingroup$ @StefanH We are not arguing that the conjugates are in $N$. Instead we are arguing, that they are actually roots of elements in $F_i$. Thats why we are writing any conjugates $\alpha_{ij}$ $n$-power as an element in $F_i$. $\endgroup$ – Verdruss Oct 2 '17 at 18:49

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