a problem on simple extension

Question

I have proved $\mathbb Q(\sqrt2,\sqrt3)=\mathbb Q(\sqrt 2+\sqrt 3)$ It was easy task as both were square root, now how to show $\mathbb Q(3^{1/3},\sqrt 2)=\mathbb Q (3^{1/3}+\sqrt 2) $

I tried cubing and squaring $3^{1/3}+\sqrt 2$ and ended up with $3^{1/3}(3^{1/3}+\sqrt 2)$ and $\sqrt 2+3\times[{3^{1/3}}/{\sqrt 2}](3^{1/3}+\sqrt 2)$, how to proceed after that

Answer 1

If $a=3^{1/3}+\sqrt{2}$ then $$ \sqrt{2}=\frac{1}{755}(48a^5+27a^4-320a^3-468a^2+879a-1092) $$ [How did I get this: By factoring $x^2-2$ over $\mathbb{Q}(3^{1/3}+\sqrt{2})$ on the computer]

This does not look like the solution that could be expected.

Thus the way to solve it is probably using more theory: Show that $\mathbb{Q}(3^{1/3},\sqrt{2}):\mathbb{Q}=6$ and (using the nicer basis $1,\sqrt{2},3^{1/3},3^{1/3}\sqrt{2},3^{2/3},3^{2/3}\sqrt{2}$) show that $1,a,a^2,a^3,a^4,a^5$ are linearly independent and that the minimal polynomial of $a$ thus must have degree 6.

This will show that the subfield $\mathbb{Q}(3^{1/3}+\sqrt{2})\le \mathbb{Q}(3^{1/3},\sqrt{2})$ is in fact equal.

Answer 2

Let $\alpha = \sqrt{2} + \sqrt[3]{3}$, so $\mathbf Q(\alpha) \subset \mathbf Q(\sqrt{2},\sqrt[3]{3})$. Because $\mathbf Q(\sqrt{2})$ and $\mathbf Q(\sqrt[3]{3})$ have relatively prime degrees, the field $\mathbf Q(\sqrt{2},\sqrt[3]{3})$ has degree $6$ over $\mathbf Q$ with basis $$ 1, \ \ \sqrt{2}, \ \ \sqrt[3]{3}, \ \ \sqrt{2}\sqrt[3]{3}, \ \ \sqrt[3]{9}, \ \ \sqrt{2}\sqrt[3]{9}. $$ We will show $\alpha$ is the root of an irreducible in $\mathbf Q[x]$ of degree $6$, so $\mathbf Q(\alpha) = \mathbf Q(\sqrt{2},\sqrt[3]{3})$.

Use the above basis to compute the $6 \times 6$ matrix for multiplication by $\alpha$ on $\mathbf Q(\sqrt{2},\sqrt[3]{3})$. The characteristic polynomial of that matrix is $$ x^6 - 6x^4 - 6x^3 + 12x^2 - 36x + 1 $$ and $\alpha$ is guaranteed to be a root of it. This polynomial is irreducible mod $13$ (I checked that with a computer, but in principle it is a finite hand calculation too), so it is irreducible over $\mathbf Q$.

One can bypass explicit calculations by proving a general result.

Theorem. If $K$ is a field of characteristic $0$ and $a$ and $b$ are algebraic over $K$ with $[K(a,b):K] = [K(a):K][K(b):K]$, then $K(a,b) = K(a+b)$. In particular, this is true when $[K(a):K]$ and $[K(b):K]$ are relatively prime.

Proof. See the answer by Pete Clark here, which includes a reference for a related result that also holds when $K$ has characteristic $p$.

Answer 3

Let $a=3^{1/3}+\sqrt{2}$. Consider $(a-\sqrt{2})^3=3$ and hence $a^3-3\sqrt{2}a^2+6a-2\sqrt{2}=3$. So $$\sqrt{2}=\frac{a^3+6a-3}{3a^2+2} \in \mathbb{Q}(a).$$ and $$3^{1/3}=a-\sqrt{2}\in \mathbb{Q}(a)$$ as well. Note: The nonzero element $3a^2+2$ is in $\mathbb{Q}(a)$ and $\mathbb{Q}(a)$ itself is a field. Hence $\frac {1}{3a^2+2} \in \mathbb{Q}(a)$ and to conclude $\frac{a^3+6a-3}{3a^2+2} \in \mathbb{Q}(a).$

Answer 4

$$(b_1,\ldots,b_6)=(1, \sqrt{2}, \sqrt[3]{3}, \sqrt{2}\sqrt[3]{3}, \sqrt[3]{9},\sqrt{2}\sqrt[3]{9})$$ is a $\Bbb{Q}$-basis of $\Bbb{Q}(\sqrt{2}, \sqrt[3]{3})$.

Let $a=\sqrt{2}+\sqrt[3]3$ and $A\in M_6(\Bbb{Q})$ be the matrix such that $a^i = \sum_{j=1}^6 A_{ij}b_j$.

Check that $\det(A)\ne 0$, which implies that $\Bbb{Q}(\sqrt{2}, \sqrt[3]{3})=\Bbb{Q}(a)$.

The inverse matrix $B=A^{-1}$ is such that $$b_i=\sum_{j=1}^6 B_{ij} a^j $$