Let $K=\mathbb{Q}(\sqrt[n]a)$ where $a\in\mathbb{Q}$, $a>0$ and suppose $[K:\mathbb{Q}]=n$. Let $E$ be any subfield of $K$ and let $[E:\mathbb{Q}]=d.$ Prove that $E=\mathbb{Q}(\sqrt[d]a)$.

It's quite clear that $\mathbb{Q}(\sqrt[d]a)$ must have $x^d-a$ as its minimal polynomial in order to have degree $d$. And also $d|n$. We also have $x^n-a$ is irreduicible. How does it follow that $x^d-a$ must also be irreducible? Also I'm having trouble showing $\mathbb{Q}(\sqrt[d]a)$ is the unique subfield of degree $d$. If we consider $N_{K/E}(\mathbb{Q}(\sqrt[d]a))$, which is in $E$, what are all the automorphisms? If $n$ is even the conjugation $\sqrt[n]a\mapsto-\sqrt[n]a$ is an automorphism.

For the second part, I have to show that if $n$ is odd then $K$ has no nontrivial subfields which are Galois over $\mathbb{Q}$ and if $n$ is even then the only nontrivial subfield of $K$ which is Galois over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{a})$.

Any help is appreciated!

• Hint: Look at $[K:E]$ too. – user14972 Apr 2 '13 at 2:47
• Also, keep in mind that $K/\mathbb{Q}$ usually isn't Galois, so you have to take care when reasoning about automorphisms. Its normal closure is $K(\zeta_n)$, where $\zeta_n$ is a primitive $n$-th root of unity. – user14972 Apr 2 '13 at 2:49
• Well $[K:E]=n/d$ since degrees are multiplicative. How does that help? Yeah I agree that $K$ needs roots of unity to be Galois. – user70520 Apr 2 '13 at 2:50
• Over $\mathbb Q(a^{1/d})$, $a^{1/n}$ is a root of $x^{n/d} - a^{1/d}$, so has degree at most $n/d$ over this field. Thus $[K :\mathbb Q(a^{1/d})] \leq n/d$, so from $[K :\mathbb Q(a^{1/d})] [ \mathbb Q(a^{1/d}) : \mathbb Q ] = [ K : \mathbb Q] = n$, we must have $[ \mathbb Q(a^{1/d}) : \mathbb Q ] = d$. Since $x^{1/d}$ is a root of $x^d - a$, $x^d - a$ must be irreducible. Showing that these subfields are unique is the hard part, and I am not quite sure how to do this. – user15464 Apr 2 '13 at 2:54

You are right to consider the norm of $\sqrt[n]{a}$ down to $E$. It is a product of elements of the form $\sigma(\sqrt[n]{a})$ where $\sigma$ is an embedding of $K$ into $\mathbf{C}$ which fixes $E$. Since $\sigma$ fixes $E$, in particular it fixes $\mathbf{Q}$, so it sends $\sqrt[n]{a}$ to another root of $X^n-a$ in $\mathbf{C}$. Thus it must have the form $\zeta\sqrt[n]{a}$ for some $n$-th root of unity $\zeta$ (I'm not saying the same root of unity appears for every $\sigma$, just that for each $\sigma$, there is some $n$-th root of unity). The number of embeddings of $K$ into $\mathbf{C}$ which fix $E$ is equal to the number of roots of the minimal polynomial for $\sqrt[n]{a}$ over $E$ (think about how you define a homomorphism from the simple extension $K=E(\sqrt[n]{a})$ to some other extension of $E$, using the description of $K$ as $E[X]/(g(X))$ where $g(X)$ is the minimal polynomial for $\sqrt[n]{a}$). Because we are in characteristic zero, $g(X)$ has precisely $[K:E]=n/d$ roots. That is to say, you get one embedding for each of the $n/d$ roots of the minimal polynomial. So, when you multiply all these elements together, you're going to get a root of unity $\zeta$ of some order times $\sqrt[n]{a}^{n/d}$. Since $a$ is positive (this is crucial), $K\subseteq\mathbf{R}$ (I'm assuming you think of $K$ as a subfield of $\mathbf{C}$ and $\sqrt[n]{a}$ means the unique positive $n$-th root of $a$). So $\zeta\sqrt[n]{a}^{n/d}\in\mathbf{R}$ and so is $\sqrt[n]{a}^{n/d}$, which means that $\zeta\in\mathbf{R}$. The only roots of unity in $\mathbf{R}$ are $\pm 1$, so $\zeta=\pm 1$, and thus $\sqrt[n]{a}^{n/d}=\sqrt[d]{a}$ is in $E$ (I mean, again, the unique positive $d$-th root of $a$ in $\mathbf{R}$). Now you have $\mathbf{Q}(\sqrt[d]{a})\subseteq E$, and by degree considerations, using that $X^n-a$ is irreducible over $\mathbf{Q}$, you can conclude that this inclusion is an equality.

For the second part of the question, think about when an extension $K$ of the form $\mathbf{Q}(\sqrt[n]{a})$ with $a$ a root of $X^n-a$, $a>0$, and $X^n-a$ irreducible (equivalently $[K:\mathbf{Q}]=n$) can contain all the roots of $X^n-a$ (because this is what is necessary and sufficient for the extension to be Galois).

Here is a more general situation. Let $F$ be a field, $a \in F^\times$, and assume $X^n - a$ is irreducible over $F$.

(1) We want to show for each $d|n$ that $X^d - a$ is irreducible over $F$.

(2) Writing $\sqrt[n]{a}$ as notation for a root of $X^n - a$, assume any $n$th roots of unity in $F(\sqrt[n]{a})$ in fact lie in $F$. (Example: $F = {\mathbf Q}$ and $a > 0$, so ${\mathbf Q}(\sqrt[n]{a})$ is isomorphic to a subfield of ${\mathbf R}$, which makes it clear that the only roots of unity at all in ${\mathbf Q}(\sqrt[n]{a})$ are $\pm 1$, which lie in ${\mathbf Q}$.) We want to show, for each $d|n$, that the only field between $F$ and $F(\sqrt[n]{a})$ of degree $d$ is $F(\sqrt[d]{a})$, where $\sqrt[d]{a} := \sqrt[n]{a}^{n/d}$.

Proof of (1): Write $\sqrt[d]{a}$ for $\sqrt[n]{a}^{n/d}$, so $\sqrt[d]{a}$ is a root of $X^d - a$. In the tower $F \subset F(\sqrt[d]{a}) \subset F(\sqrt[n]{a})$, we have $[F(\sqrt[d]{a}):F] \leq d$ and $[F(\sqrt[n]{a}):F(\sqrt[d]{a})] \leq n/d$, since $\sqrt[d]{a}$ is a root of $X^d - a \in F[X]$ and $\sqrt[n]{a}$ is a root of $X^{n/d} - \sqrt[d]{a} \in F(\sqrt[d]{a})[X]$. Because $$[F(\sqrt[n]{a}):F] = [F(\sqrt[n]{a}):F(\sqrt[d]{a})][F(\sqrt[d]{a}):F]$$ and we assume the left side is $n$, it follows that our upper bounds for the terms on the right must be equalities. In particular, $[F(\sqrt[d]{a}):F] = d$, so $X^d - a$ must be irreducible over $F$ (it has a root with degree $d$ over $F$).

Proof of (2): Let $d|n$ and suppose $E$ is a field with $F \subset E \subset F(\sqrt[n]{a})$ and $[E:F] = d$. To prove $E = F(\sqrt[d]{a})$, it suffices to show $\sqrt[d]{a} \in E$, since that would give us $F(\sqrt[d]{a}) \subset E$ and we already saw in (1) that $F(\sqrt[d]{a})$ has degree $d$ over $F$, so the containment $F(\sqrt[d]{a}) \subset E$ would have to be an equality.

Let $f(X)$ be the minimal polynomial of $\sqrt[n]{a}$ over $E$, so $f(X)|(X^n - a)$ and $\deg f = n/d$. Any two roots of $f(X)$ are $n$th roots of $a$, and thus have a ratio that is an $n$th root of unity, so in terms of the one root $\sqrt[n]{a}$ we can write any other root of $f(X)$ as $\zeta\sqrt[n]{a}$ for some $n$th root of unity $\zeta$. (I am not making any assumptions about the $n$th roots of unity being distinct, in case $F$ has positive characteristic, and these individual $n$th roots of unity need not lie in $F(\sqrt[n]{a})$.) In a splitting field, the factorization of $f(X)$ is $\prod_{i \in I} (X - \zeta_i\sqrt[n]{a})$ for some $n$th roots of unity $\zeta_i$ ($I$ is just an index set). The constant term of $f(X)$ is in $E$, so $(\prod_{i \in I} \zeta_i)\sqrt[n]{a}^{n/d} \in E$. Therefore $(\prod_{i \in I} \zeta_i)\sqrt[n]{a}^{n/d} \in F(\sqrt[n]{a})$, so $\prod_{i \in I} \zeta_i \in F(\sqrt[n]{a})$. The only $n$th roots of unity in $F(\sqrt[n]{a})$ are, by hypothesis, in $F$, so $\prod_{i \in I} \zeta_i \in F \subset E$. Therefore $\sqrt[n]{a}^{n/d} = \sqrt[d]{a}$ is in $E$, so we're done.

To see an example of this not involving the rational numbers, let $k$ be a field and $F = k(t)$, the rational functions over $k$ in one indeterminate. The polynomial $X^n - t$ is irreducible over $k(t)$, since it is Eisenstein at $t$. We let $\sqrt[n]{t}$ denote one root of $X^n - t$, so $F(\sqrt[n]{t}) = k(\sqrt[n]{t})$ has degree $n$ over $k(t)$. All roots of unity in $k(\sqrt[n]{t})$ -- not just $n$th roots of unity -- are in $k$, because $k(\sqrt[n]{t})$ is a rational function field in one indeterminate over $k$ (it's $k$-isomorphic to $k(t)$) and there's a general theorem that any roots of unity in a rational function field over $k$ are in the constant field $k$. By the above work, the only fields between $k(t)$ and $k(\sqrt[n]{t})$ are $k(\sqrt[d]{t})$ for $d|n$.

An example where the hypothesis that all $n$th roots of unity in $F(\sqrt[n]{a})$ are in $F$ is false, and the conclusion is nevertheless true, is $F = {\mathbf Q}(i)$, $a = 2$, and $n = 8$: $[{\mathbf Q}(i,\sqrt{2}):{\mathbf Q}(i)] = 8$. The extension ${\mathbf Q}(i,\sqrt{2})/{\mathbf Q}(i)$ is Galois with a cyclic Galois group, but not all 8th roots of unity are in ${\mathbf Q}(i)$.