Proving $\sqrt[3]{5} \notin \mathbb{Q}(\sqrt[3]{2})$ with solvability of system in $\mathbb{Q}$

Question

We were asked to show that $\sqrt[3]{5} \notin \mathbb{Q}(\sqrt[3]{2})$ by reducing the problem to the solvability of system in $\mathbb{Q}$. There was also a hint with it that a system does not need to be linear. I don't really know what exactly they're expecting. Intuitively I would start by $\sqrt[3]{5} \in \mathbb{Q}(\sqrt[3]{2})$ such that $\sqrt[3]{5}= a + b\sqrt[3]{2} + c\sqrt[3]{4}$ for some $a,b,c \in \mathbb{Q}$ and try searching for a contradiction as I have seen in similar posts on this sort of problems. However when I cube it, it gets quite large and I get lost in the calculation. The answers on other posts on this all use concepts we haven't seen yet so I assume there is another way to solve this using the hint.

Answer 1

The cube is not that bad, and that is surely what you are expected to do if they tell you to "reduce to a system over $\mathbb{Q}$." You just need to do the algebra carefully.

(Corrected)

Use the formula $$(x+y+z)^3 = x^3+y^3 + z^3 + 3x^2y+3x^2z + 3xy^2 + 3y^2z + 3xz^2 + 3yz^2 + 6xyz.$$ We have: $$\begin{align*} (a+b2^{1/3}+c2^{2/3})^3 &= a^3 + 2b^3 + 4c^4\\ &\qquad+ 3a^2b2^{1/3} + 3a^2c2^{2/3} + 3ab^22^{2/3} + 6b^2c2^{1/3}\\ &\qquad + 6ac^22^{1/3} + 6bc^22^{2/3} + 12abc\\ &= \bigl( a^3+2b^3+4c^3+12abc\bigr)\\ &\qquad + \bigl(3a^2b + 6b^2c + 6ac^2)2^{1/3}\\ &\qquad + \bigl(3a^2c + 3ab^2 + 6bc^2)2^{2/3} \end{align*}$$ For $a+b2^{1/3}+c2^{2/3}$ to equal $\sqrt[3]{5}$, you need this to equal $5$, so you have the system $$\begin{align*} a^3+2b^3+4c^3+12abc &=5\\ 3a^2b+6b^2c+6ac^2 &= 0\\ 3a^2c+3ab^2+6bc^2 &=0 \end{align*}$$ We can divide the last two equations by $3$ to get $$\begin{align*} a^3+2b^3+4c^3+12abc &=5\\ a^2b+2b^2c+2ac^2 &= 0\\ a^2c+ab^2+2bc^2 &=0 \end{align*}$$ If any of $a$, $b$, or $c$ are zero, then the second or third equation give that at least two of them are zero. Then the first equation would ask us to solve $a^3=5$, $2b^3=5$, or $4c^3=5$ in rational numbers, which is impossible. Thus, all of $a,b,c$ would necessarily be nonzero. So we may assume they are all nonzero.

Rewrite the last equation as $$a(ac + b^2) + 2bc^2 = 0,$$ which gives $(ac+b^2) = -\frac{2bc^2}{a}$. Now rewrite the second equation as $$a^2b+2c(b^2+ac) = 0,$$ so plugging in the value of $ac+b^2$ we get: $$\begin{align*} a^2b+2c(b^2+ac) &=0\\ a^2b + 2c\left(-\frac{2bc^2}{a}\right) &= 0\\ a^2b - \frac{4bc^3}{a} &= 0\\ a^3b - 4bc^3 &=0\\ b(a^3-4c^3)&=0\\ a^3-4c^3 &=0\\ a^3 &= 4c^3. \end{align*}$$ This is impossible for rationals $a$ and $c$: the power of $2$ that appears on the left is a multiple of $3$ (positive or negative). But the power of $2$ that appears on the right hand side is of the form $3m+2$, with $m$ an integer.

Thus, there are no solutions to this system in rationals, hence $\sqrt[3]{5}$ is not an element of $\mathbb{Q}[\sqrt[3]{2}]$.

Answer 2

If you keep your calm and do algebra, it all works out.

Let us write $$\sqrt[3]{5} = a + b\sqrt[3]{2} + c\sqrt[3]{4}.$$

Squaring, $$\sqrt[3]{25} = (a^2 + 4bc) + (2c^2+2ab)\sqrt[3]{2} + (b^2+2ac)\sqrt[3]{4}.$$

Thus $$5 = (a^3+4abc+2b^3+4abc+4c^3+4abc) + (2ac^2+2a^2b+2b^2c+4ac^2)\sqrt[3]{2} + (ab^2+2a^2c+2bc^2+2ab^2)\sqrt[3]{4}.$$

This in turn reduces to the system $$2a^2c+2bc^2+3ab^2 = 0,$$ $$2a^2b+2b^2c+6ac^2=0,$$ $$a^3+2b^3+4c^3+12abc=5$$ over the rationals. We wish to explain why this has no solutions.

Let us first pick the smallest integer $N$ so that each of $aN, bN, cN$ is an integer. Write these as $A=aN, B = bN, C = cN$ and multiply each equation by $N^3$ to find $$2A^2C+2BC^2+3AB^2=0,$$ $$A^2B+B^2C+3AC^2=0,$$ $$A^3+2B^3+4C^3+12ABC=5N^3.$$

If $N$ is odd, then we get a contradiction as follows.

From the last equation, we can deduce that $A$ is odd. From the first equation, we then deduce that $B$ is even. The quantity $A^2B+B^2C$ is then even, and so from the second equation we deduce that $3AC^2$ is even. This As $A$ is odd, we find that $C$ is even. Say that $B = 2^xB', C = 2^yC'$ where $B', C'$ are odd. Then our second equation can be rewritten $$2^xA^2B'+2^{2x+y}(B')^2(C') + 3\cdot 2^{2y}A(C')^2 = 0.$$

If $2y > x,$ then we could divide by $2^x$ to get a contradiction that a sum of an odd number plus two even numbers is even. Thus we must have $x \geq 2y.$ The first equation reads $$2^{y+1}A^2C' + 2^{x+2y+1}(B')(C')^2 + 3\cdot 2^{2x}A(B')^2 = 0.$$

If $y+1 < 2x,$ we could divide by it and get a similar contradiction to before, so we now know $y+1\geq 2x.$ But we also know $x\geq 2y.$ Both of these inequalities at once are bad for positive integers $x,y$: note that $$y+1 \geq 2x \geq 2(2y) = 4y,$$ but $y$ needs to be a positive integer and no positive integer obeys $y+1 \geq 4y.$ Thus there are no solutions.

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Therefore $N$ is even. But if $N$ is even, then our last equation implies $A$ is even. If $B$ and $C$ were both even, then we could divide $N$ by 2 to get a smaller $N$ while still clearing denominators, contradicting its minimality; if $B, C$ were both odd, then our second equation would find a contradiction, as the outer two terms (multiples of $A$) would be even but the middle term $B^2C$ would be odd, yet they sum to the even number zero. Hence one of $B,C$ is even and the other is odd.

Take our last equation modulo $8.$ The terms $A^3$ and $5N^3$ both become zero, since $A, N$ are even and $2^3 = 8.$ The term $12ABC$ is also 0, since $12$ gives a factor of 4 and then $A$ gives another factor of $2.$

Thus the equation becomes $8 \mid (2B^3+4C^3)$ after reducing modulo 8. But this is impossible: if $C$ is the even one of $B,C,$ then $4C%3$ is a multiple of $8$ and so $8 \mid 2B^3,$ absurd since $B$ odd; the same problem happens if $B$ is the even one, and we conclude.