$E := \mathbb{Q}( \sqrt{3}, \sqrt{5}, \sqrt[3]{7})/ \mathbb{Q}$. Determine a $ϑ ∈ E$ with $E = \mathbb{Q}(ϑ)$.

Question

Hey I have this exercise where I don't know how to proceed.

Let $E := \mathbb{Q}( \sqrt{3}, \sqrt{5}, \sqrt[3]{7})$ be a field extension of $\mathbb{Q}$. Determine a $ϑ ∈ E$ with the property $E = \mathbb{Q}(ϑ)$.

I was thinking of proving $\mathbb{Q}( \sqrt{3}, \sqrt{5}, \sqrt[3]{7})=\mathbb{Q}( \sqrt{3} \cdot \sqrt{5}\cdot \sqrt[3]{7})$

But I don't know if it's right and I don't know how to proceed. I should prove that $\sqrt{3}, \sqrt{5}, \sqrt[3]{7} \in \mathbb{Q}( \sqrt{3} \cdot \sqrt{5}\cdot \sqrt[3 ]{7})$ and that $\sqrt{3} \cdot \sqrt{5}\cdot \sqrt[3]{7} \in \mathbb{Q}( \sqrt{3}, \sqrt{5} , \sqrt[3]{7})$?

Can someone help me?

Answer 1

Let us consider the numbers $a=\sqrt 3$, $b=\sqrt 5$, $c=\sqrt[3]7$, and
$$ t = (a+b)c = (\sqrt 3+\sqrt 5)\sqrt[3]7\ . $$ We want to show that the inclusion $K:=\Bbb Q(t)\subseteq \Bbb Q(a,b,c)$ is in fact an equality. So we need algebraic expressions in $t$ reproducing each of $a,b,c$.

For this, we compute: $$ \begin{aligned} t^3 &=(a+b)^3\cdot 7=(\sqrt 3+\sqrt 5)^3\cdot 7=(18\sqrt 3+14\sqrt 5)\cdot 7\in K\ ,\\ t^9 &=(a+b)^9\cdot 7^3= (\sqrt 3+\sqrt 5)^9\cdot 7^3 =(70416\sqrt3 + 54544\sqrt5)\cdot 7^3\in K\ ,\\ \end{aligned} $$ and because the determinant $$ \begin{vmatrix} 18 & 14\\ 70416 & 54544 \end{vmatrix} = 2^5 \begin{vmatrix} 9 & 7\\ 4401 & 3409 \end{vmatrix} $$ does not vanish (as it does not modulo ten). So we can extract $a,b$ linearly from $t^3,t^9$. This gives first $a,b\in K$, then $c=(a+b)/t\in K$.

$\square$

Answer 2

The existing answer is great. I'd like to add that there is a general theorem, which you surely know, but also that this theorem's proof provides us with an algorithm for finding the primitive element.

Say $F$ is a field and $K=F(\alpha_1,\alpha_2,\cdots,\alpha_n)$ where $\alpha_2,\cdots,\alpha_n$ are separable over $F$ and all $\alpha_\bullet$ are algebraic over $F$. Then there exists $\gamma\in K,\,F(\gamma)=K$.

If $F$ is a finite field then so is $K$ and $K^{\times}$ is cyclic; we may take $\gamma$ to be any generator of $K^{\times}$. If $F$ is infinite, as is the case for $\Bbb Q=F$, we need to work harder.

The $n>2$ case follows from induction on $n=2$, the $n=1$ case being immaterial. I won't write the whole proof but I'll follow the algorithm. Let's start with $K'=\Bbb Q(\sqrt{3},\sqrt{5})$. Let $f(x)=x^2-3,g(x)=x^2-5$ the minimal polynomials. All we have to do is find $\lambda\in\Bbb Q$ such that $\lambda$ is not of the form $\frac{\alpha'-\sqrt{3}}{\sqrt{5}-\beta'}$ where $\alpha',\beta'\neq\sqrt{5}$ are roots of $f$, $g$ (! in general, this question is considered for roots in a normal closure $L'$ of $K'/F$) and then declare $\gamma:=\sqrt{3}+\lambda\sqrt{5}$.

Ok, so what are the possible values of $\lambda$ which are excluded? $\lambda=0$ and $\lambda=\frac{-\sqrt{3}-\sqrt{3}}{\sqrt{5}-(-\sqrt{5})}=-2\sqrt{\frac{3}{5}}$, but the last one is not in $F=\Bbb Q$ so in fact any nonzero $\lambda$ will work. Let's take $\lambda=1$ then.

Now we consider $K$ as an extension of $K'=\Bbb Q(\sqrt{3}+\sqrt{5})$, $K=K'(\sqrt[3]{7})=\Bbb Q(\sqrt{3}+\sqrt{5},\sqrt[3]{7})$. Now we have to fix a normal closure $L$ of $K/\Bbb Q$, of course $L=K(\zeta_3)$. The relevant $f,g$ are now $f(x)=x^4-16x^2+4$, $g(x)=x^3-7$ (use Vieta's formulae to obtain the coefficients of $f$). The excluded $\lambda$ are $\lambda=0$ and also $\lambda=\frac{-2\sqrt{5}\text{ or }-2\sqrt{3}\text{ or }-2\sqrt{3}-2\sqrt{5}}{\sqrt[3]{7}(1-\zeta_3)\text{ or }\sqrt[3]{7}(1-\zeta_3^2)}$; it's fairly clear none of these will be rational. We can surely take $\lambda=1$ again.

It follows $\sqrt{3}+\sqrt{5}+\sqrt[3]{7}$ is a primitive element $\vartheta$ for $K$ over $\Bbb Q$. You can use Dan's answer - the determinant trick - to explicitly express $\sqrt{3},\sqrt{5}$ and $\sqrt[3]{7}$ in terms of $\vartheta$. In general that's a computational nightmare. By the abstract proof, I already know that this will be possible - with zero guesswork!