# Primitive element of a field extension

Hi, I want to determine a primitive element for root expressions of the extension $\mathbb{Q}\left( m^{\frac{1}{a}}, n^{\frac{1}{b}} \right)$ over $\mathbb{Q}$.

Question 1: In the case $m$ and $n$ are coprime and squarefree, I think that the degree of field extension is $ab$. It is right?

Question 2: I want to determine a primitive element. My assumption is, that tt's always the element $m^{\frac{1}{a}}+n^{\frac{1}{b}}$.

For that I would have to show that $\frac{m^{\frac{1}{a}}}{n^{\frac{1}{b}}}\frac{\zeta_{a}^{i}-\zeta_{a}^{j}}{1-\zeta_{b}^{k}}$ is not 1 for $i, j \in \left\lbrace 0, \ldots, a-1\right\rbrace$, $k \in \left\lbrace 1, \ldots, a-1\right\rbrace$.

Thanks and best regards Florian M.

• You should change the tag to "number theory". – Dimitrije Kostic Nov 13 '11 at 19:00
• I retagged. (The problem is that "number theory" was in all likelihood intended. Yet this doesn't work as there must not be a space in a tag so it becomes 'number' and 'theory'.) – user9072 Nov 13 '11 at 19:25
• I think this question needs a precise definition of $\mathbf{Q}(m^{1/a},n^{1/b})$ before it is meaningful. The issues here are a little thornier than one might think at first. – Kevin Buzzard Nov 13 '11 at 21:31
• I'm not sure about the degree, but at least it's not difficult to prove that $m^{1/a}+n^{1/b}$ is a primitive element. Namely, if an element of the Galois group fixes this sum then it has to fix both terms, as can be seen by looking at the real parts. (Strictly speaking this works when $m,n$ are positive, are you assuming this ?) – François Brunault Nov 13 '11 at 21:49
• @Florian: Let $L$ be the Galois closure of $\mathbf{Q}(m^{1/a},n^{1/b})$ and $\sigma \in \mathrm{Gal}(L/\mathbf{Q})$ fixing $m^{1/a}+n^{1/b}$. Since $\sigma(m^{1/a})$ is conjugate to $m^{1/a}$ we must have $\sigma(m^{1/a})=\zeta \cdot m^{1/a}$ for some $a$-th root of unity $\zeta$, and similarly $\sigma(n^{1/b})=\zeta' \cdot n^{1/b}$. Now we have $\zeta \cdot m^{1/a}+\zeta' \cdot n^{1/b}=m^{1/a}+n^{1/b}$ and looking at the real parts we get $\zeta=\zeta'=1$ so $\sigma$ fixes both $m^{1/a}$ and $n^{1/b}$. This shows $\mathbf{Q}(m^{1/a},n^{1/b})=\mathbf{Q}(m^{1/a}+n^{1/b})$. – François Brunault Nov 24 '11 at 18:59

The answer to Q1 is yes, here is a somewhat pedestrian proof.

I'll assume $m,n \geq 2$ are coprime and squarefree, and that $m^{1/a}$ (resp. $n^{1/b}$) denotes the unique real $a$-th root of $m$ (resp. real $b$-th root of $n$).

Recall that $K=\mathbf{Q}(n^{1/b})$ has degree $b$ over $\mathbf{Q}$ (use Eisenstein's criterion). Let us prove directly that $X^a-m$ is irreducible over $K$. Over $\mathbf{C}$ we have

\begin{equation*} X^a-m = \prod_{k=0}^{a-1} X-\zeta_a^k m^{1/a} \end{equation*} If $P$ is a nontrivial factor of $X^a-m$ over $K$, of degree $1 \leq d \leq a-1$, then the constant term of $P$ is of the form $\zeta \cdot m^{d/a}$ for some root of unity $\zeta$. Since $K \subset \mathbf{R}$, we must have $\zeta=\pm 1$, so that $m^{d/a} \in K$. Since $d/a$ is not an integer, we deduce that there must exist a prime $p$ (dividing $a$) such that $m^{1/p} \in K$. In particular $p$ divides $b$.

Now using the Galois correspondence, it is not too hard to show that the unique subfield of degree $p$ of $K$ is $\mathbf{Q}(n^{1/p})$. So we get $\mathbf{Q}(m^{1/p})=\mathbf{Q}(n^{1/p})$. It remains to compare the discriminants of these number fields to get a contradiction. The prime $p$ divides at most one of the numbers $m$ and $n$, say $p$ doesn't divide $m$. Then any prime divisor $q$ of $m$ ramifies in $\mathbf{Q}(m^{1/p})$ but not in $\mathbf{Q}(n^{1/p})$, whence a contradiction.

As noted in my comment, the answer to Q2 is always yes, and one doesn't need the assumption "$m,n$ coprime and squarefree" for that.

• Hi, but how do you use Galois correspondence? There is no Galois extension!? Thanks and best greetings Florian – Florian Modler Nov 22 '11 at 10:31
• @Florian : One uses the Galois correspondence for the Galois closure $L=\mathbf{Q}(n^{1/b},\zeta_b)$ of $K$. In fact the Galois correspondence is not needed here. Since $\mathbf{Q}(m^{1/p}) \subset \mathbf{Q}(n^{1/b})$ we have that $n^{1/b}$ has degree $b/p$ over $\mathbf{Q}(m^{1/p})$. Its minimal polynomial is a product of $b/p$ factors of the form $X-\n^{1/b} \zeta_b^{k}$ with $0 \leq k \leq b-1$. Looking at the constant term yields $\zeta \cdot n^{1/p} \in \mathbf{Q}(m^{1/p})$ with $\zeta =\pm 1$ like before so that $\mathbf{Q}(m^{1/p})=\mathbf{Q}(n^{1/p})$. – François Brunault Nov 22 '11 at 12:03