I first read about the following problem.

Prove: $|\mathbb{Q}(\sqrt[4]{2}, \sqrt{3}) : \mathbb{Q}| =8$

So I could prove this statement by showing $\sqrt[4]{2} \notin \mathbb{Q}(\sqrt{3})$ using this trick. I am wondering if there is a more generalized proof - if the problem were to solve for

$|\mathbb{Q}(\sqrt[m]{p}, \sqrt[n]{q}):\mathbb{Q}|$ (Say with $p,q$ prime).

How does one go about?

  • 1
    $\begingroup$ Who told you that it is $4$? $\endgroup$
    – Kenny Lau
    Jan 7, 2018 at 7:31
  • $\begingroup$ Sorry, it should be $8$, typo $\endgroup$
    – Bryan Shih
    Jan 7, 2018 at 7:31
  • $\begingroup$ If the goal is something like $m \cdot n$ then you will need more by way of restrictions, e.g., consider $16^{1/4}$ and $4^{1/2}$. (You may also wish to specify the sets to which $p, q, m, n$ can belong. E.g., can $p = 2 - \sqrt{2}$? Etc.) $\endgroup$ Jan 7, 2018 at 7:39

2 Answers 2


When $p\neq q$, the extension has degree $nm$ as you might expect. When $p=q$, the extension has degree $\mathsf{lcm}(n,m)$. To see this, use Karpilovsky's theorem (see this answer for more details)


For the case $p\ne q$ the order is $mn$ and we prove it here. For primes $p,q$ and positive integers $m,n$ we take the following basis for the extended field: $$\Large v_{ij}={p^{i\over m}}{q^{j\over n}}\qquad i\in\lbrace1,...,m\rbrace,j\in\lbrace1,...,n\rbrace$$ so we can represent $x\in\mathbb{Q}(\sqrt[4]{2}, \sqrt{3}) : \mathbb{Q}$ as a linear combination of $v_{ij}$'s as below: $$\Large x=\Sigma_{i,j}{a_{ij}}{v_{ij}}$$ which can also be written as following: $$\Large x=\Sigma_{i}\Sigma_{j}{a_{ij}}{v_{ij}}=\Sigma_{i}{p^{i\over m}}\Sigma_{j}{a_{ij}}{q^{j\over n}}$$ Now to prove $v_{ij}$'s are really basis must show that: $$\Large x=0\to a_{ij}=0\qquad\forall i,j$$

Before that we try to prove a lemma. First define $b_i=\Sigma_{j}{a_{ij}}{q^{j\over n}}$.

Lemma: $\large\Sigma_{i}{b_i}{p^{i\over m}}=0$ yields to $\large b_i=0\qquad\forall j$.

Proof of Lemma: Let $v={p^{1\over m}}$. Then we have $\large\Sigma_{i=1}^{m}{b_i}{v^i}=0$ where $b_i$'s are quotient and $\large{{v^j}\over{v^k}}\notin\Bbb Q\qquad \forall j,k\in\lbrace1,...,n\rbrace$. Multiplying $v$ to both side of the equation leads us to: $$\large\Sigma_{i}{b_i}{v^{i+1}}=0$$ from $\large v^n=q$ by substitution and two equations we arrived at we can finally obtain: $$(b_2b_np-b_1^2)v^2+(b_3b_np-b_1b_2)v^3+...+(b_n^2p-b_1b_{n-1})v^n=0$$ or $$(b_2b_np-b_1^2)v+(b_3b_np-b_1b_2)v^2+...+(b_n^2p-b_1b_{n-1})v^{n-1}=0$$ where new coefficients are quotient either and the order of equation has reduced to $n-1$. By iterating this process we can show that all of those coefficients are zero $\bullet$

Since we have shown $\large b_i=\Sigma_{j}{a_{ij}}{q^{j\over n}}=0\qquad \forall i$ then we can apply this lemma again on ${a_{ij}}$'s for any arbitrary $i\in\lbrace1,...,m\rbrace$ and deduce $a_{ij}=0\qquad \forall i,j$

This is what we wanted show so we prove that the basis is linearly independent and the order of extended field is then $\large mn$.

If $p=q$ let's define $m=m^{'}d$ and $n=n^{'}d$ where $d=g.c.d.(m,n)$ and $g.c.d.(m^{'},n^{'})=1$. Therefore we have:

$$\Large v_{ij}={p^{i\over {m^{'}d}}}{p^{j\over {n^{'}d}}}={p^{{in^{'}+jm^{'}}\over{m^{'}n^{'}d}}}\qquad i\in\lbrace1,...,m\rbrace,j\in\lbrace1,...,n\rbrace$$ In this case we easily see that the order is $\large {{mn}\over{g.c.d.(m,n)}}$.

  • $\begingroup$ I do not understand the "where $a_{ij}$'s are quotient" in your lemma. You seem to use the same name $a_{ij}$ for two different things $\endgroup$ Jan 7, 2018 at 16:04
  • $\begingroup$ Thanks for the feedback!... it is incomplete and imprecise... i'm gonna fix it in moment.... $\endgroup$ Jan 7, 2018 at 16:07
  • $\begingroup$ This is still incorrect. In the statement of the lemma, you probably mean $\large b_i=0\qquad\forall i$ rather than $\large b_i=0\qquad\forall j$. In the proof of the lemma, all your $n$'s should be $m$'s, right ? $\endgroup$ Jan 7, 2018 at 16:26
  • $\begingroup$ The last claim in the proof of the lemma, "by iterating this process we can show all the coefficients are zero" is false. Counterexample : consider the sum $\sum_{i=1}^{m+1} b_i p^{\frac{i}{m}}$ where $b_1=-p,b_{m+1}=1$ and the other $b_i$ are zero. You can indeed "iterate the process", but that doesn't mean all the coefficients are zero $\endgroup$ Jan 7, 2018 at 16:29
  • $\begingroup$ Oops! Did I miss something during modification? $\endgroup$ Jan 7, 2018 at 16:30

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