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!

  • $\begingroup$ You might check through linear algebra that the minimal polynomial of $\sqrt[6]{7}+\sqrt[4]{5}$ over $\mathbb{Q}$ has degree $24$. Since $\mathbb{Q}(\sqrt[6]{7}+\sqrt[4]{5})\subseteq\mathbb{Q}(\sqrt[6]{7},\sqrt[4]{5})$ the extensions through $\sqrt[6]{7}$ and $\sqrt[4]{5}$ are independent. $\endgroup$ Apr 13, 2017 at 12:20

3 Answers 3


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.

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.

  • $\begingroup$ Thanks! But may I please ask how can we see that the factorization over $\Bbb Q(\sqrt[6]{7})$ is either $(x−\sqrt[4]{5})(x+\sqrt[4]{5})(x^2+\sqrt{5})$ or $(x^2-\sqrt{5})(x^2+\sqrt{5})$, which can allow us to conclude that it it factorize then $\sqrt{5}\in \Bbb Q(\sqrt[6]{7})$? $\endgroup$ Apr 14, 2017 at 3:33
  • $\begingroup$ @ymxu0809 If a factorization exist over $\mathbb{Q}(\sqrt[6]{7}$, this is also a factorization over the reals. We know what the irreducible factors are over the reals, so a factorization over the smaller field must be obtained from one over the larger field. This is used also in the final argument. $\endgroup$
    – egreg
    Apr 14, 2017 at 8:09
  • $\begingroup$ Thanks! I do know that the factorization comes from $\Bbb R$, but how can I see that the only possible ways we can gain the factorization over $\Bbb R$ must involves the existence of $\sqrt{5}$? That is, how to see that there is not possible that we can have $5=abcd$ or $5=abc$ or $5=ab$ where $a,b,c,d\in\Bbb R$ but none of them involves $\sqrt{5}$? $\endgroup$ Apr 14, 2017 at 23:32
  • $\begingroup$ @ymxu0809 If $x^4-5$ factors in some number field $K\subseteq\mathbb{R}$, then its factorization contains one among $x-\sqrt[4]{5}$, $x+\sqrt[4]{5}$ or $x^2-\sqrt{5}$; thus $\sqrt{5}$ must belong to $K$. Remember that factorization in $\mathbb{R}[x]$ is unique. $\endgroup$
    – egreg
    Apr 14, 2017 at 23:42
  • $\begingroup$ Yes it intuitively makes sence to me... But how can I see that 5 can never be a product of transendental elements or some product of other algebraic number where $\sqrt{5}$ does not appear? Is that the fact that as the factorization is unique, once we have a factorization already, we cannot have another factorization? $\endgroup$ Apr 14, 2017 at 23:48

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}) $.

  • $\begingroup$ It seems to be a quite general argument. But as I have not learnt about the terminology ramified, could you please give some explaination on this word? And for the set $LK$, do you mean the set $\lbrace lk: l\in L, k\in K\rbrace$? $\endgroup$ Apr 14, 2017 at 3:37

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.


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