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Show that $mathbb{Q}(sqrt{5}+sqrt[3]{2})=mathbb{Q}(sqrt{5},sqrt[3]{2})$.

I've got that $big[mathbb{Q}(sqrt{5}+sqrt[3]{2}):mathbb{Q}big] in {1,2,3,6}$ because it's going to divide $big[mathbb{Q}(sqrt{5},sqrt[3]{2}):mathbb{Q}big]=6$. Clearly it is not $1$. I want to show that it is not $2$ or $3$. So I'm saying that if it is $2$ then $alpha = sqrt{5}+sqrt[3]{2}$ satisfies the relation
$$alpha^2 + balpha = k$$ for $b,k in mathbb{Q}$ since $alpha$ will be the root of a monic irreducible polynomial of degree $2$. How can I obtain a contradiction from this? I also need to do the degree $3$ case somehow.

Also if there is a better way to do this than what I'm doing I'd be excited to learn about it.

Since you know that $[mathbb Q[sqrt[3]{2},sqrt 5]:mathbb Q] = 6$, you know that each of the six values $$1,sqrt{5},\sqrt[3]{2},sqrt[3]{2}sqrt{5},\sqrt[3]{4},sqrt[3]{4}sqrt 5tag{1}$$
are linearly independent over $mathbb Q$.

Now $$(sqrt[3]2+sqrt 5)^2=5cdot 1 + 2cdot sqrt[3]2sqrt 5 + 1cdotsqrt[3]4$$

Is it possible for $1,sqrt[3]2+sqrt5,(sqrt[3]2+sqrt 5)^2$ to be linearly dependent over $mathbb Q$?

Do the same with by adding the cube $(sqrt[3]2+sqrt5)^3$.

Another way to look at it use (1) as a basis, and write elements of the field as:

$$(a,b,c,d,e,f)to acdot 1 + bcdot sqrt5+csqrt[3]2+dsqrt[3]2sqrt5+esqrt[3]4+fsqrt[3]4sqrt5$$

Then $$begin{align}(1,0,0,0,0,0)&leftrightarrow 1\(0,1,1,0,0,0)&leftrightarrow sqrt 5+ sqrt[3]2\(5,0,0,2,1,0)&leftrightarrow (sqrt5+sqrt[3]2)^2\
(2,5,15,0,0,3)&leftrightarrow (sqrt5+sqrt[3]2)^3
end{align}$$

And those four vectors are "obviously" linearly independent.

Let's show that $s=sum sqrt[n_i]{d_i}$ ($d_i >0$ rationals ) generates $mathbb{Q}(sqrt[n_i]{d_i})$. Consider all in larger Galois extension $K supset mathbb{Q}(sqrt[n_i]{d_i})supset mathbb{Q}$. Now, to show that $s$ generates all the $sqrt[n_i]{d_i}$ it's enough to show that whenever a Galois transformation $phi$ of $K$ preserves $s$ it must preserve all the $sqrt[n_i]{d_i}$. Now any Galois transformation $phi$ takes $sqrt[n_i]{d_i}$ to some $omega_i sqrt[n_i]{d_i}$ where $omega_i^{n_i} =1$, so $phi(s) = sum omega_i sqrt[n_i]{d_i}$.
Note that
$$left|sum_i omega_i sqrt[n_i]{d_i}right| le sum_i sqrt[n_i]{d_i}$$ with equality
if and only if $omega_i$ have the same argument. Therefore, if $sum_i omega_i sqrt[n_i]{d_i}= sum_i sqrt[n_i]{d_i}$ then all the $omega_i$ must be $1$.

Conclusion: $phi(s) = s implies phi( sqrt[n_i]{d_i}) = sqrt[n_i]{d_i}$ for all $i$.

$bf{Added:}$ A solution that uses mostly linear algebra, inspired from @
Thomas Andrews: 's solution.

We have the equalities
begin{eqnarray}
left(begin{array} {c}
1\sqrt[3]{2}+sqrt{5}\ (sqrt[3]{2}+sqrt{5})^2\ (sqrt[3]{2}+sqrt{5})^3\ (sqrt[3]{2}+sqrt{5})^4\ (sqrt[3]{2}+sqrt{5})^5
end{array}right) = left(
begin{array}{cccccc}
1 & 0 & 0 & 0 & 0 & 0 \
0 & 1 & 0 & 1 & 0 & 0 \
5 & 0 & 1 & 0 & 2 & 0 \
2 & 15 & 0 & 5 & 0 & 3 \
25 & 2 & 30 & 8 & 20 & 0 \
100 & 125 & 2 & 25 & 10 & 50 \
end{array}
right) cdot left(begin{array} {c} 1 \ sqrt[3]{2}\ sqrt[3]{4} \ sqrt{5} \ sqrt[3]{2} sqrt{5} \sqrt[3]{4} sqrt{5}
end{array}
right)
end{eqnarray}
hence
begin{eqnarray}
left(begin{array} {c} 1 \ sqrt[3]{2}\ sqrt[3]{4} \ sqrt{5} \ sqrt[3]{2} sqrt{5} \sqrt[3]{4} sqrt{5}
end{array}
right)
= left(
begin{array}{cccccc}
1 & 0 & 0 & 0 & 0 & 0 \
frac{2275}{4054} & -frac{1714}{2027} & frac{195}{2027} & frac{500}{2027} & -frac{9}{4054} & -frac{30}{2027} \
frac{2875}{2027} & -frac{1325}{2027} & -frac{955}{2027} & frac{150}{2027} & frac{100}{2027} & -frac{9}{2027} \
-frac{2275}{4054} & frac{3741}{2027} & -frac{195}{2027} & -frac{500}{2027} & frac{9}{4054} & frac{30}{2027} \
-frac{6505}{2027} & frac{1325}{4054} & frac{1491}{2027} & -frac{75}{2027} & -frac{50}{2027} & frac{9}{4054} \
-frac{5143}{2027} & frac{2335}{2027} & -frac{650}{2027} & -frac{991}{2027} & frac{15}{2027} & frac{100}{2027} \
end{array}
right) cdot left(begin{array} {c}
1\sqrt[3]{2}+sqrt{5}\ (sqrt[3]{2}+sqrt{5})^2\ (sqrt[3]{2}+sqrt{5})^3\ (sqrt[3]{2}+sqrt{5})^4\ (sqrt[3]{2}+sqrt{5})^5
end{array}right)
end{eqnarray}

For instance, we have

$
tiny
sqrt{5} = -frac{2275}{4054} + frac{3741}{2027} (sqrt[3]{2}+sqrt{5}) -frac{195}{2027}(sqrt[3]{2}+sqrt{5})^2 -frac{500}{2027} (sqrt[3]{2}+sqrt{5})^3+ frac{9}{4054} (sqrt[3]{2}+sqrt{5})^4+ frac{30}{2027} (sqrt[3]{2}+sqrt{5})^5$

Brute-Force Method

Let $alpha:=sqrt{5}+sqrt[3]{2}$. Then, $alpha^3-3sqrt{5}alpha^2+15alpha-5sqrt{5}=(alpha-sqrt{5})^3=2$, whence $$left(alpha^3+15alpha-2right)^3=5left(3alpha^2+5right)^2,,$$ or $alpha$ is a root of the polynomial $$f(x):=x^6-15x^4-4x^3+75x^2-60x-121 in mathbb{Q}[x],.$$ If this polynomial is reducible, then consider it modulo $3$, so $$f(x)=x^6+2x^3+2=left(x^2+2x+2right)^3$$ in $mathbb{F}_3$, and $x^2+2x+2$ is an irreducible element of $mathbb{F}_3[x]$. That is, $f(x)$ must have a monic quadratic factor $g(x)$ in $mathbb{Q}[x]$, and by Gauss's Lemma, $g(x)inmathbb{Z}[x]$. Hence, $g(x)=x^2+ax+b$ with $a,binmathbb{Z}$ and $bequiv 2 equiv -1pmod{3}$. Let $f(x)=g(x),h(x)$ for some $h(x)inmathbb{Q}[x]$ (which again yields $h(x)inmathbb{Z}[x]$). Since the coefficient of $x^5$ in $f(x)$ is $0$, $h(x)=x^4-ax^3+cx^2+dx+e$ for some $c,d,einmathbb{Z}$.

Now, $$f(x)=x^6-4x^3+4=left(x^3-2right)^2=left(x^3+2^3right)^2=(x+2)^2left(x^2-2x+4right)^2$$ in $mathbb{F}_5$, where both $x+2$ and $x^2-2x+4$ are irreducible elements of $mathbb{F}_5[x]$. Hence, in $mathbb{F}_5$, either $$g(x)=(x+2)^2=x^2+4x+4text{ or }g(x)=x^2-2x+4,.$$ In either case, $bequiv 4equiv -1pmod{5}$. Together with $bequiv -1pmod{3}$, we deduce that $b equiv -1pmod{15}$. However, $b$ must divide the constant term $-121=-11^2$ of $f(x)$. This means $b=-1$ or $b=-121$.

If $b=-1$, then $g(x)=x^2+ax-1$ and $h(x)=x^4-ax^3+cx^2+dx+121$. Equating the coefficients of $f(x)$ and $g(x),h(x)$, we have $$c-a^2-1=-15,,,, ac+a+d=-4,,$$ $$ad-c+121=75,,text{ and }121a-d=-60,.$$ Consequently, $c=a^2-14$ and $d=121a+60$, so that $$begin{align}0&=(ad-c+121)-75=ad-c+46\&=a(121a+60)-left(a^2-14right)+46=60left(2a^2+a+1right),.end{align}$$ However, $2a^2+a+1=0$ does not have an integer solution.

If $b=-121$, $g(x)=x^2+ax-121$ and $h(x)=x^4-ax^3+cx^2+dx+1$. Equating the coefficients of $f(x)$ and $g(x),h(x)$, we have $$c-a^2-121=-15,,,, ac+121a+d=-4,,$$ $$ad-121c+1=75,,text{ and }a-121d=-60,.$$ Consequently, $c=a^2+106$ and $d=frac{a+60}{121}$, so that $$begin{align}0&=(ad-121c+1)-75=ad-121c-74\&=frac{a}{121}(a+60)-121left(a^2+106right)-75,,end{align}$$ or $$0=-frac{1}{121}left(14640a^2-60a+1561021right),.$$ However, $14640a^2-60a+1561021=0$ does not have an integer solution.

We have derived a contradiction from the hypothesis that $f(x)$ is reducible over $mathbb{Q}$, so $f(x)$ must be irreducible in $mathbb{Q}[x]$. Hence, $mathbb{Q}(alpha)cong mathbb{Q}[x]/big(f(x)big)$ is a field extension of index $deg(f)=6$ over $mathbb{Q}$.

Clearly $LHSsubseteq RHS$. Now it suffices to write down the minimal polynomial of $sqrt{5}+sqrt[3]{2}$ and note it has degree $6$.