Show that $\sqrt[3]{1+\sqrt{3}}$ isn't an element of the field $\mathbb{Q}(\sqrt{3} ,\sqrt[3]{2})$

Question

Setting $\alpha = \sqrt[3]{2}$ and $a+b\alpha+c\alpha^2=\sqrt[3]{1+\sqrt{3}}$ for $a,b,c$ in $\mathbb{Q}(\sqrt{3})$ (The minimal polynomial for $\sqrt[3]{2}$ in $\mathbb{Q}(\sqrt{3})$ is $x^3-2=0$ since if it were reducible that would imply the existence of $\sqrt[3]{2}$ in $\mathbb{Q}(\sqrt{3})$ which is easily shown not to be true) . This comes down to proving that the equations $$ a^3+9ab^2=1 $$ and $$ 3b^3+3a^2b=1$$ have no solutions $a,b$ in $\mathbb{Q}(\sqrt{3})$. This is obviously extremely difficult to prove so there should be a better approach.

This is too long for a comment.

Considering the the irreducible polynomials for $\sqrt[3]{2}$ and $\sqrt[3]{1+\sqrt{3}}$ over $\mathbb{Q}(\sqrt{3})$ we see they are $x^3-(1+\sqrt{3})$ and $x^3-2$ respectively. If the two extensions in the comments are equal that would imply that there is an automorphism of the field $\mathbb{Q}(\sqrt{3})$ which fixes $\mathbb{Q}$ and sends $1+\sqrt{3}$ to $2$ but there is only one nontrivial automrphism of this field that fixes $\mathbb{Q}$ and it sends $\sqrt{3}$ to $-\sqrt{3}$. So the two extensions can't be isomorphic, let alone equal. Is this correct?

Answer 1

Here is an alternative solution, which is essentially the same as Lubin's but with a more elementary presentation.

Lemma. Let $K$ be a subfield of $\mathbb C$ with $\sqrt[3]{2}\not\in K$ (whence it easily follows that $[K(\sqrt[3]{2}):K]=3$). Let $k\in K$. Then $k$ is a cube in $K(\sqrt[3]{2})$ iff one of $k,\frac{k}{2},\frac{k}{4}$ is already a cube in $K$.

Proof of lemma. The "if" direction is obvious, so let us concentrate on the "only if" part : suppose that $k$ is a cube in $K(\sqrt[3]{2})$, so that $k=(a+b\sqrt[3]{2}+c\sqrt[3]{4})^3$ for some $a,b,c\in K$. Putting $\theta=a+b\sqrt[3]{2}+c\sqrt[3]{4}$, one can compute that $$\theta^2=(2ac + b^2)\sqrt[3]{4} + 2(ab + c^2)\sqrt[3]{2} + (a^2 + 4bc)$$ and $$\theta^3=3(ab^2+a^2c+2bc^2)\sqrt[3]{4} + 3(a^2b+2b^2c+2ac^2)\sqrt[3]{2} + (a^3+2b^3+4c^3+12abc)$$

Since $\theta^3=k\in K$, we must have $z_1=ab^2+a^2c+2bc^2=0$ and $z_2=a^2b+2b^2c+2ac^2=0$. Now $az_1-bz_2=3c(a^3-2b^3)$, so either $a^3-2b^3=0$ (in which case $a=b=0$ because $\sqrt[3]{2}\not\in K$) or $c=0$. It easily follows that at most one of $a,b,c$ is nonzero and the lemma is proved.

Using the lemma above with $k=1+\sqrt{3}$ and $K={\mathbb Q}(k)$, we now have to see if $k,\frac{k}{2},\frac{k}{4}$ are cubes in $K$ or not.

If $k=(a+b\sqrt{3})^3$ with $a,b\in{\mathbb Q}$, we deduce $a^3+9ab^2=1$ and $3b^3+3a^2b=1$. We can write $a=\frac{u}{q},b=\frac{v}{q}$ where $u,v,q$ are integers. Then $u^3+9uv^2=q^3$ and $3u^3+3u^2b=q^3$. Reasoning modulo $5$, it can easily be checked that the only solution to this system in ${{\mathbb F}_5}^3$ is $(u,v,q)=(0,0,0)$. Returning to $\mathbb Z$, this means that each of $u,v,q$ must be divisible by $5$. But by infinite descent, this clearly implies that the only solution is $(u,v,q)=(0,0,0)$ in $\mathbb Z$ also, contradiction.

Similarly, in the two other cases we obtain a system where $(0,0,0)$ is the only solution modulo $5$. This finishes the proof.

Answer 2

This is a nice problem.

My argument is rather advanced, and it will be wonderful to see an elementary one. But let’s call the base field $K=\Bbb Q(\sqrt3\,)$. Later on, I’ll need to refer to its ring of integers $A=\Bbb Z[\sqrt3\,]$, the set of all $m+n\sqrt3$ for integers $m$ and $n$.

As I believe you’ve recognized, showing that $L=K(\sqrt[3]2\,)$ doesn’t contain $\sqrt[3]{1+\sqrt3\,}$ is equivalent to showing that $L$ is distinct from the extension $M=K\bigl(\sqrt[3]{1+\sqrt3\,}\bigr)$.

A powerful but rather advanced way of showing that two extensions are different is to show that the set of split primes in the one is different from the set of split primes in the other, and we can do that in this case. Let me explain what’s going on:

First, I should point out that $A$, the ring of algebraic integers of $K=\Bbb Q(\sqrt3\,)$, is a Euclidean domain, thus principal ideal domain. Except for the fact that $A$ has infinitely many elements whose reciprocals also are in $A$ ($2+\sqrt3$ is one of these, for instance), arithmetic in $A$ is much like that in $\Bbb Z$, with unique factorization, etc.

When we have an extension of $A$ of degree $n$, a prime $\pi$ of $A$ may split into several primes, or may remain prime. In the case of cubic extensions of $K$, as we have here, there are three possibilities for the behavior of the ideal $(\pi)$ in the ring $\mathcal O$ of algebraic integers of the extension:
(1) The ideal $(\pi)$ may no longer be prime in $\mathcal O$, but rather split into three primes $(\pi)=\mathfrak p_1\mathfrak p_2\mathfrak p_3$;
(2) $(\pi)$ may factorize as the product of two primes $(\pi)=\mathfrak p_1\mathfrak p_2$, with $\mathcal O/\mathfrak p_1$ isomorphic to $A/(\pi)$ and $\mathcal O/\mathfrak p_2$ the quadratic extension of $A/(\pi)$ (remember that these residue fields all are finite); or
(3) $(\pi)$ may remain prime in $\mathcal O$, but the residue field $\mathcal O/(\pi)$ will be the cubic extension of $A/(\pi)$.

I should say, in passing, that behavior (2) can happen only when the cubic extension is not normal. But at any rate, our aim is to find a prime $\pi$ of $A$ such that, perhaps, $(\pi)$ splits completely (case 1) in one of $L$, $M$, while $\pi$ remains prime in the other. And that’s just what happens when $\pi=5$; I’m about to demonstrate this.

First, what about $5$ as a number in $A=\Bbb Z[\sqrt3\,]$? Since $3$ is not a quadratic residue modulo $5$, this $\Bbb Z$-prime remains prime in $A$. But you see that the residue field $A/(5)$ is the quadratic extension of $\Bbb Z/(5)$, that is $A/(5)\cong\Bbb F_{25}$, whose multiplicative group is cyclic with $24$ elements. We want to look at $\Bbb F_{25}=\Bbb F_5(\sqrt3\,)$, that is, $\Bbb F_{25}\cong\Bbb F_5[x]/(x^2-3)$. We’re in luck here, because $1+x\sim1+\sqrt3$ is a generator of the multiplicative group: it has order $24$. Indeed, adjoining a cube root of it induces a cubic extension of $\Bbb F_{25}$. (You can see this by noticing that we’d be getting an element of order $3\cdot24=72$, and that $72$ does not divide $25^2-1$ but does divide $25^3-1$.)

In any event, $5$ is still prime in $\mathcal O_M$, the integers of $M$. Now let’s see that happens to $5$ in $\mathcal O_L$. Here, we have to consider $\sqrt[3]2$. But $2$ is a cube in $\Bbb F_5$, $2\equiv3^3\pmod5$, and of course in $\Bbb F_{25}$. Better yet, $\Bbb F_{25}$ has all cube roots of unity (check out $\>2+x\in\Bbb F_{25}$), so that it contains three cube roots of $2$, namely $3$, $3(2+x)$, and $3(2+x)^2$. Thus $5$ splits completely in $\mathcal O_L$.

This completes my argument that the two fields are distinct.

Answer 3

Hint

The proof below uses Theorem 1.6 of paper Cubic fields: a primer.

We have to prove that $L_1= K(\sqrt[3]{2})$ isn’t isomorphic to $L_2=K(\sqrt[3]{1+\sqrt{3}})$ where $K=\mathbb Q(\sqrt{3})$.

According to the theorem quoted above, that would imply that

$$\sqrt[3]{\frac{1+\sqrt{3}}{2}} \in K$$

We’re left to prove that it can’t be.

Answer 4

Keeping Lubin's notations $K=\mathbf Q(\sqrt 3),L=K(\sqrt [3]2),M=K(\sqrt [3] {1+\sqrt 3}) $, suppose that $L=M$. This implies that $L(j)=M(j)$, where $j$ is a primitive 3-rd root of unity. But both $L(j), M(j)$ are Kummer cubic extensions of $K(j)=\mathbf Q(j,\sqrt 3)=\mathbf Q(i,\sqrt 3)$, and Kummer theory tells us that $L(j)=M(j)$ iff there exists $c\in K(j)$ s.t. $1+\sqrt 3=2^r.c^3, r=1,2$. Norming down this equality from $K(j)$ to $K$ yields $(1+\sqrt 3)^2=4+2\sqrt 3=4^r.N(c)^3$; norming again from $K$ to $\mathbf Q$ produces an equation of the form $1=4^{r-1}b^3$, with $b\in \mathbf Q^*$. After clearing denominators, we get a contradiction with the unique factorization in $\mathbf Z$.

Answer 5

I want to use Dedekind's theorem on the cycle structure of elements of the Galois group when viewed as a group of permutations of roots

Everything happens inside the splitting field $F$ of the polynomial $$f(x)=(x^3-1)^2-3=x^6-2x^3-2$$ over $\Bbb{Q}$. The zeros of $f(x)$ are $(1\pm\sqrt3)^{1/3}\omega^j, j=0,1,2$, $\omega=(-1+i\sqrt3)/2$, so $$ F=\Bbb{Q}(\omega,\root3\of{1+\sqrt3},\root3\of{1-\sqrt3}). $$ As $(1+\sqrt3)(1-\sqrt3)=-2$ we see that $F$ contains the cube roots of $2$. For the same reason $F=\Bbb{Q}(\omega,\root3\of{1+\sqrt3},\root3\of2)$.

Let's view the Galois group $G=Gal(F/\Bbb{Q})$ as a subgroup of $S_6$ permuting those six roots.

• Lubin's calculations in the extensions of $\Bbb{F}_5$ imply that $f(x)$ remains irreducible modulo $5$. Therefore Dedekind's theorem implies that $G$ contains a 6-cycle $\tau$, and consequently also a product of two disjoint 3-cycles, namely $\tau^2$.
• Modulo $13$ the polynomial $f(x)$ has three linear factors and a single cubic factor. This is because in $\Bbb{F}_{13}$ we can choose $\sqrt{3}=4$. Consequently $1+\sqrt{3}=5=(-2)^3$ has a cube root in $\Bbb{F}_{13}$. As $3\mid(13-1)$ it actually has three distinct cube roots in there. But $1-\sqrt3=-3$ is not a cubic residue modulo $13$. Consequently $x^3+3$ is an irreducible cubic factor of $f(x)$ over $\Bbb{F}_{13}$. This means that $G$ contains a 3-cycle.
• The two previous bullets imply that the Sylow $3$-subgroup of $G$ cannot have order three. We can thus conclude that $9\mid [F:\Bbb{Q}]$. The main claim follows from this (and the tower law) immediately.