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I suspect that $\mathbb{Q}[2^{1/3}]=\{x \in \mathbb{R}; ∃a,b \in \mathbb{Q}(x =a+b\cdot2^{1/3})\}$ is not a field. This is because $2^{1/3} \cdot 2^{1/3} = 2^{2/3}$ is not in $\mathbb{Q}[2^{1/3}]$. However, I am having difficulty proving the statement:

There does not exist $a,b \in \mathbb{Q}$ such that $a+b\cdot2^{1/3}=2^{2/3}$

I have not been able to find a contradiction assuming otherwise.

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    $\begingroup$ $x^3-2$ is irreducible $\endgroup$ – reuns Aug 11 '17 at 7:29
  • $\begingroup$ $2^{2/3} = ({2^{1/3}})^2$ $\endgroup$ – mathreadler Aug 11 '17 at 7:46
  • $\begingroup$ Historical note: if $2^{1/3}$ were a solution of $x^2=a+b\,x$ with rational $a,b$, the famous problem of doubling the cube with compass and straightedge would be solved. That was proved impossible in 1837 by Pierre Wantzel. $\endgroup$ – Professor Vector Aug 11 '17 at 8:09
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If $x^2=a+b\,x,$ we have $x^3=a\,x+b\,x^2=a\,x+b\,(a+b\,x)=a\,b+(a+b^2)\,x$. If $x=2^{1/3},$ we'd have $2^{1/3}=\frac{2-a\,b}{a+b^2},$ a rational number. Let $n$ be the smallest positive integer so that both $n\,2^{1/3}$ and $n\,2^{2/3}$ are integers. Then, $m=n(2^{1/3}-1)=n\,2^{1/3}-n$ is an integer $<n$, and $m\,2^{1/3}=n\,2^{2/3}-n\,2^{1/3}$ and $m\,2^{2/3}=2\,n-n\,2^{2/3}$ are integers, contradicting the minimality of $n$.

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  • $\begingroup$ Why $2^{\frac{1}{3}}$ is not rational number? See please my proof. $\endgroup$ – Michael Rozenberg Aug 11 '17 at 8:02
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    $\begingroup$ @MichaelRozenberg $2^{1/3}$ being irrational may be proven exactly the same way as $2^{1/2}$. $\endgroup$ – Arthur Aug 11 '17 at 8:14
  • $\begingroup$ @Arthur Which says that you have no a poof of this statement or at least it did not be written. See please my solution. It's a full solution. $\endgroup$ – Michael Rozenberg Aug 11 '17 at 8:17
  • $\begingroup$ @Michael Rozenberg To avoid boredom of spectators, I added a not so common proof for the irrationality of $2^{1/3}$. $\endgroup$ – Professor Vector Aug 11 '17 at 9:10
  • $\begingroup$ @Professor Vector It was nice! :) $\endgroup$ – Michael Rozenberg Aug 11 '17 at 9:15
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Let $\zeta=\sqrt[3]2$, and assume $\zeta^2=a+b\zeta$ with $a,b\in\mathbb Q$. Then $$ 2 = \zeta(a+b\zeta) = a\zeta+b\zeta^2 = a\zeta+b(a+b\zeta) = (a+b^2)\zeta + ba $$ and because $1$ and $\zeta$ are linearly independent over $\mathbb Q$ we must have $a+b^2=0$ and $ab=2$.

But then $a=-b^2$ and $ab=-b^3$. But there is no $b\in\mathbb Q$ such that $-b^3=2$.

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  • $\begingroup$ Why your last statement is true? See please my solution. $\endgroup$ – Michael Rozenberg Aug 11 '17 at 8:21
  • $\begingroup$ @MichaelRozenberg: Just as Professor Vector told you, the proof of that goes exactly like the extremely well-known proof that $\sqrt2$ is irrational. The OP does not seem to be in any doubt about that part. $\endgroup$ – Henning Makholm Aug 11 '17 at 8:23
  • $\begingroup$ If so the trick with the infinite descent is extremely well-known and my solution just more shorter than all solutions in this page. . What do you think? $\endgroup$ – Michael Rozenberg Aug 11 '17 at 8:26
  • $\begingroup$ @MichaelRozenberg: I do not think the word "shorter" means what you think it means. $\endgroup$ – Henning Makholm Aug 11 '17 at 8:27
  • $\begingroup$ See better please my solutions. It's two lines only! The rest it's just explanation because it seems that Professor Vector wanted to explain. $\endgroup$ – Michael Rozenberg Aug 11 '17 at 8:31
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Square brackets usually indicate a polynomial ring. For instance, $x^2\in \Bbb Q[x]$. The same is true for $2^{1/3}$ in place of $x$. So your definition of $\Bbb Q[2^{1/3}]$ does not correspond to the standard definition of that notation.

That being said, it is true that $2^{2/3}$ is linearly independent of the set $\{1,2^{1/3}\}$, so in the two-dimensional vector space over $\Bbb Q$ spanned by $1$ and $2^{1/3}$ does not contain $2^{2/3}$.

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We can use $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz),$$ which gives $$a^3+2b^3-4+6ab=0.$$

Now, let $a=\frac{m}{n}$ and $b=\frac{k}{n}$, where $m$, $n$ be integers and $n$ be natural numbers.

Thus, we get $$m^3+2k^3-4n^3+6mnk=0,$$ which is contradiction because we can make here the infinite descent.

My explanation.

We see that $m$ divided by $2$. Thus, there is $m_1\in\mathbb Z$, for which $m=2m_1$ and we obtain $$4m_1^3+k^3-2n^3+6m_1nk=0.$$ Now, we see that $k$ divided by $2$, which says that there is $k_1\in\mathbb Z$, for which $k=2k_1$, which gives $$2m_1^3+4k^3-n^3+6m_1nk_1=0.$$

Now, we see again that $n$ divided by $2$, which says that there is $n_1\in\mathbb N$, for which $n=2n_1$ and we obtain $$m_1^3+2k_1^3-4n_1^3+6m_1n_1k_1=0.$$ Hence, if there are $(m,n,k)$, where $n$ is a natural numbers, for which $$m^3+2k^3-4n^3+6mnk=0$$ then there are $(m_1,n_1,k_1)$ for which this equation holds and $n_1<n$.

Since we can repeat this procedure more and more, we get an infinite sequence of natural numbers $n>n_1>...$, which is impossible.

Done!

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    $\begingroup$ How does this answer OP's question? $\endgroup$ – Anurag A Aug 11 '17 at 7:35
  • $\begingroup$ @Anurag A Yes of course, it's contradiction by infinite descent. $\endgroup$ – Michael Rozenberg Aug 11 '17 at 7:36
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    $\begingroup$ First of all why should $a$ and $b$ have the same denominator $n$ and with that being said can you explain what is the contradiction you are arriving at. $\endgroup$ – Anurag A Aug 11 '17 at 7:38
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    $\begingroup$ @Anurag If $a=\frac{2}{3}$ and $b=\frac{5}{7}$ then $a=\frac{14}{21}$ and $b=\frac{15}{21}$. $\endgroup$ – Michael Rozenberg Aug 11 '17 at 7:40
  • $\begingroup$ If $\alpha^2 $ is in the $\mathbb{Q}$-linear span of $\{1,\alpha\}$ then it is the root of $x^2+bx+c \in \mathbb{Q}[x]$. Thus $\alpha = \frac{-b\pm \sqrt{b^2-4c}}{2}$. This is a contradiction when $\alpha = 2^{1/3}$. $\endgroup$ – reuns Aug 11 '17 at 7:41

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