Is the composite of two simple radical extensions another simple radical extension?

Question

I encountered this (fact? Maybe I misunderstood) in a proof of a lemma in Dummit and Foote where the authors state something which (I think) implies this without proof, namely that the composition of two root extensions is another one. If what they state doesn't require that the composition of two simple radical extensions be another one, please explain why. I tried doing this in various ways, but (as the answer below suggests) this may not even be true. If so, there must be something wrong with my understanding or a way to avoid running into this supposed requirement.

They also didn't prove the statement that the composite of all fields $\sigma{K}$ for $\sigma\in Aut(L/F)$ is precisely $L$ (the Galois closure) and here's my attempt to prove it with $K$ an arbitrary extension of $F$.

Denote the composite of all such fields by $E$ and consider the action of any $\tau\in Aut(L/F)$ on it. We see that any such $\tau$ sends $E$ into itself (see the edit below). And so, any $\sigma\in Aut(L/F)$ is an automorphism of the field $E$ fixing $F$, But the number of embeddings $N$ of $E$ into an algebraic closure which contains $L$ that fix $F$ is the order of the set of cosets $\left|Aut\left(L/F\right)\right|/\left|Aut\left(L/E\right)\right|$ by the fundamental theorem. Combining these results we get, $$\frac{\left|Aut\left(L/F\right)\right|}{\left|Aut\left(L/E\right)\right|}\ge\left|Aut\left(E/F\right)\right|\ge\left|Aut\left(L/F\right)\right|$$ which (since $L/E$ is galois) leads to $$\left|Aut\left(L/E\right)\right|=\left[L:E\right]=1$$

Is this correct?

Edit: Let $B_i=\left\{b_{i,1},b_{i,2},...b_{i,m}\right\}$ denote any set of elements that span $\sigma_iK$ over $F$. Define $P=\left\{p_1,p_2,...p_s\right\}$ to be the set of all possible products of elements $b_{i,r}$ in the union $B_1 \cup B_2 \cup ...B_q$. The set $P$ spans $E$ over $F$. So, for arbitrary $\alpha \in E$ we have $$\alpha=\sum_{j=0}^sa_jp_j $$ $a_j \in F$ applying some $\tau \in Aut(L/F)$ and noting that $\tau$ fixes the $a_j$'s, we get $$\tau\left(\alpha\right)=\sum_{j=0}^sa_j\tau\left(p_j\right)$$

Each $p_j$ is a product of $b$'s in the union above and each $b_{i,r}$ is in some $\sigma_{i}K$ so $\tau$ sends it to an element of $\tau\sigma_{i}K$ which is an element of $E$ by definition. So, the image of a product of any number of $b$'s will still be in $E$ since all the images are in $E$. So, the image of $p_j$ is some $\beta$ again in $E$ which may or may not be in $P$, but this doesn't matter since $\tau\left(\alpha\right)$ is in $E$ anyway and we started with an arbitrary $\alpha$. Noting that $\tau$ can't send two elements to the same element since it's an automorphism of the larger field $L$, I conclude that any such $\tau$ is an automorphism of $E$ fixing $F$.

Answer 1

I don't think they ever state that composition of simple radical extension is also a radical extension. Moreover, I simply do not think it's true. For example, I don't think that $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ is a simple radical extension: its Galois group is not cyclic.

The only thing that they claim is that the composition of root extensions is also a root extensions, and that's easy. Suppose we have two root extensions: $F = F_0 \subset F_1 \subset \ldots \subset F_p$, and $F = G_0 \subset G_1 \subset \ldots \subset G_q$, such that $F_{i+1} = F_i(\sqrt[n_i]{a_i})$ for some $a_i \in F_i$, and same for $G_i$. WLOG we can assume that $p = q$.

Then, the composition of $F_p$ and $G_p$ is also a root extension: since $G_1 = G_0(\sqrt[k_0]{b_0})$, $b_0 \in G_0 = F = F_0$. Thus, we set $H_1 = F_1(\sqrt[k_0]{b_0}) = F_0(\sqrt[n_0]{a_0})(\sqrt[k_0]{b_0})$, and we see that $H_1$ is a root extension, and $F_1 \subset H_1$, $G_1 \subset H_1$ -- in fact, $H_1$ is the composition of $F_1$ and $G_1$. Similarly, since $a_1, b_1 \in H_1$, we set $H_2 = H_1(\sqrt[n_1]{a_1})(\sqrt[k_1]{b_1})$, and again $H_2$ is a root extension, and the composition of $F_2$ and $G_2$. Continuing this way, we obtain that $H_p$ is a root extension and the composition of $F_p$ and $G_p$.

Answer 2

You can take it as comment, you need to define the possible cases because the terms you are using are not standard

• Simple root extension $F(a^{1/n})/F$ with minimal polynomial $x^n-a$. In general a tower or compositum of such won't a simple root extension.

• Cyclotomic extension $F(\zeta_k)/F$

• A tower $F_l/F_0$ of such extensions : what we usually call a radical extension. Any element of a radical extension has an expression in term of nested sum/product/quotient of $n$-th roots of elements of $F$ and roots of unities.

• The normal closure of $F_l/F_0$ is obtained by adding to $F_l$ the roots of unities needed to make $x^{n_j}-a_j$ split completely as well as a root of $x^{n_j}-\sigma(a_j)$ for all $F$-homomorphism $\sigma : F_l \to \overline{F_l}$. By induction $\sigma(a)$ is in a radical extension so the normal closure is a radical extension. If I say normal instead of Galois it is because $\Bbb{F}_p(t)/\Bbb{F}_p(t^p)$ is a normal non-Galois radical extension.

• Cyclic extensions are radical extensions (add $\zeta_k$ to the base field so that $E(\zeta_k)/F(\zeta_k)$ is again cyclic, with the primitive element theorem $E(\zeta_k)=F(\zeta_k)(\beta)$, then $\gamma=\sum_{r=1}^m \sigma^r(\beta)\zeta_k^r$ satisfies $\sigma(\gamma) = \zeta_k^{-1}\gamma$ so it is a $m=[E(\zeta_k):F(\zeta_k)]$-th root)

• A tower of radical extensions such that $F_l/F_0$ is Galois (It is easy to show by induction adding iteratively the roots of $x^n-a$ to the base field that it decomposes as a tower of cyclic Galois radical extensions, this implies the whole Galois group is solvable (since $Gal(F_j/F_0) = Gal(F_l/F_0)/Gal(F_l/F_j)$) conversely any Galois extension with solvable Galois group decomposes as a tower of cyclic extensions, so we have the main theorem : radical extension = contained in a normal extension with solvable Galois group)