Degree of extension $\Bbb Q(\sqrt[4]{5},\sqrt[6]{7})$.

Question

I tried to consider the tower of extension $\Bbb Q\subset \Bbb Q(\sqrt[6]{7})\subset\Bbb Q(\sqrt[4]{5},\sqrt[6]{7})$.

The minimal polynomial of $\Bbb Q(\sqrt[6]{7})$ over $\Bbb Q$ is $x^6-7$ by Eisenstein. But although it is easy to see that it has no root in $\Bbb Q(\sqrt[6]{7})$, how can I formally conclude that $x^4-5$ is irreducible over $\Bbb Q(\sqrt[6]{7})$ and thus we can see the basis of $\Bbb Q(\sqrt[6]{7}, \sqrt[4]{5})$？

I know that for if we have the degree of $\Bbb Q(\sqrt[6]{7})$ and $\Bbb Q(\sqrt[4]{5})$ are coprime, then it can be much simpler. But how to deal with that in this case where they are not coprime. Any help would be appreciate. Thanks so much！

Answer 1

The polynomial $x^4-5$ can be factored over $\mathbb{R}$, which contains $\mathbb{Q}(\sqrt[6]{7})$, as $$ x^4-5=(x-\sqrt[4]{5})(x+\sqrt[4]{5})(x^2+\sqrt{5}) $$ Thus a factorization over $\mathbb{Q}(\sqrt[6]{7})$ can only be the one above or $(x^2-\sqrt{5})(x^2+\sqrt{5})$. So you just need to show that $\sqrt{5}\notin\mathbb{Q}(\sqrt[6]{7})$.

Suppose $\mathbb{Q}(\sqrt{5})\subseteq\mathbb{Q}(\sqrt[6]{7})$. Then the degree of $\sqrt[6]{7}$ over $\mathbb{Q}(\sqrt{5})$ is $3$ by the dimension formula. The factorization of $x^6-7$ over $\mathbb{R}$ is $$ (x^3-\sqrt{7})(x^3+\sqrt{7})= (x-\sqrt[6]{7})(x^2+\sqrt[6]{7}\,x+\sqrt[3]{7}) (x+\sqrt[6]{7})(x^2-\sqrt[6]{7}\,x+\sqrt[3]{7}) $$ Since $\sqrt{7}\notin\mathbb{Q}(\sqrt{5})$, you can only get degree three factors as $$ (x-\sqrt[6]{7})(x^2-\sqrt[6]{7}\,x+\sqrt[3]{7}) $$ or $$ (x+\sqrt[6]{7})(x^2+\sqrt[6]{7}\,x+\sqrt[3]{7}) $$ and in both cases you'd conclude that $\sqrt[6]{7}\in\mathbb{Q}(\sqrt{5})$, which is impossible.

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Suppose $x^4-5$ can be factored as $f(x)g(x)$ over some extension $K$ of $\mathbb{Q}$, $K\subseteq\mathbb{R}$; suppose also that $f(x)$ and $g(x)$ are non constant. Since $x^4-5$ is monic, also $f$ and $g$ can be assumed monic. Continuing like this, we can assume that $x^4-5$ is factored into monic factors, irreducible over $K[x]$.

Let $h(x)\in K[x]$ be one of these factors; its factorization in $\mathbb{R}[x]$ must consist of polynomials in the set $\{x-\sqrt[4]{5},x+\sqrt[4]{5},x+\sqrt{5}\}$, which are the irreducible factors of $x^4-5$ in $\mathbb{R}[x]$, because of uniqueness of factorization in $F[x]$ (for $F$ any field).

Now it's just a matter of checking the various possibilities. A factorization of $x^4-5$ can only be with degrees

• $1$, $1$ and $2$
• $2$ and $2$
• $1$ and $3$

If a degree $1$ factor appears, then $\sqrt[4]{5}\in K$; if a degree $2$ factor appears, then $\sqrt{5}\in K$. In both cases, $\sqrt{5}\in K$.

The same argument applies for the second part of the proof.

Answer 2

In general, we have the following: Let $ K/\mathbf Q $ and $ L/\mathbf Q $ be two number fields such that there exists a rational prime $ p $ which is totally ramified in $ K/\mathbf Q $ and unramified in $ L/\mathbf Q $. Then, the extensions $ K/\mathbf Q $ and $ L/\mathbf Q $ are linearly disjoint, that is, $ [KL : \mathbf Q] = [K : \mathbf Q][L : \mathbf Q] $.

Proof. Let $ \mathfrak q $ be a prime of $ LK $ lying over the rational prime $ p $. Then, $ e_{\mathfrak q | p} \geq [K : \mathbf Q] $, since ramification indices are multiplicative across towers and $ p $ is totally ramified in $ K/\mathbf Q $. On the other hand, if we let $ \mathfrak p $ be the prime of $ L $ lying below $ \mathfrak q $; then we have that

$$ [K : \mathbf Q] \leq e_{\mathfrak q | p} = e_{\mathfrak q | \mathfrak p} e_{\mathfrak p | p} = e_{\mathfrak q | \mathfrak p} \leq [LK : L] $$

However, we obviously have that $ [K : \mathbf Q] \geq [LK : L] $; thus it follows that $ [K : \mathbf Q] = [LK : L] $, and multiplication by $ [L : \mathbf Q] $ on both sides of the equality gives the result.

Now, notice that we have exactly the situation of this claim with the rational prime $ p = 5 $, which is totally ramified in $ \mathbf Q(\sqrt[4]{5}) $ but unramified in $ \mathbf Q(\sqrt[6]{7}) $.

Answer 3

To show that the polynomial $X^4-5$ is irreducible over $\mathbb{Q}(\sqrt[6]{7})$ you must assume that it is not irreducible,thus it can be factorized to polynomials of degree:$1* 1* 1* 1$ or $2*2$ or $3*1$ or $2*1*1$.

You have to work by cases but its difficult because you have complicated elements in $\mathbb{Q}(\sqrt[6]{7})$ thus consider the opposite tower.

If you prove it then a basis for the final extension will be:

$A=\{\sqrt[6]{7}^j*\sqrt[4]{5}^i|j=0,1...5 ,i=0,1,2,3,\}$

This is one way but it has many calculations.