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Let $m$ be an integer such that $m \equiv 2 \pmod 3$. Show that the number $$\dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$$ is an algebraic integer.

The usual technique, doing $x = \dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$ and trying to find an algebraic expression in terms of $x$ seems not to work in this case (at least I couldn't do it). Can anyone help me? This is a question from an old exam.

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  • $\begingroup$ You can find a polynomial with your number a root directly and show that it is monic. $\endgroup$
    – Asvin
    May 4, 2015 at 21:49

5 Answers 5

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Here is a complete answer.

Let $x=\dfrac{m-\sqrt[3]{2}}{\sqrt[3]{3}}$

$\implies x\sqrt[3]{3}=m-\sqrt[3]{2}$

$\implies x\sqrt[3]{3}+ \sqrt[3]{2}=m$

$\implies 3x^3+2+3x^2.(3)^{2/3}.2^{1/3}+3x.(3)^{1/3}.2^{2/3}=m^3$

$\implies 3x^3+2+3x.(3)^{1/3}.2^{1/3}[x\sqrt[3]{3}+ \sqrt[3]{2}]=m^3$

$\implies 3x^3+2+3x.(3)^{1/3}.2^{1/3}.(m)=m^3$

$\implies 3x.(3)^{1/3}.2^{1/3}.(m)=m^3-3x^3-2$

Now cubing both sides, we get

$-162x^3m^3=(3x^3-(m-2))^3=27x^9-27x^6m+x^3[9m^2-36m+90]-[m^3-6m^2+12m-8]$

$\implies 27x^9-27x^6m+x^3[162m^3+9m^2-36m+90]-[m^3-6m^2+12m-8]=0$ $\hspace{0.4cm}$ $(\star)$

Now we have $m \equiv\ 2\ \text{mod}\ {3} \implies m=3k+2$ for some integer $k$. Now this implies that $$m^2=9k^2+12k+4$$ & $$m^3=27k^3+54k^2+36k+8$$.

Now putting these values in the coefficient of $x^3$ i.e. $162m^3+9m^2-36m+90$, becomes $27[6m^3+3k^2+2]$ and the constant $[m^3-6m^2+12m-8]$ becomes $27k^3$, thus equation in $\star$ becomes $$27x^9-27x^6m+27[6m^3+3k^2+2]x^3-27k^3=0$$

$\implies $ $$x^9-x^6m+[6m^3+3k^2+2]x^3-k^3=0$$ which is monic and coefficients are integers and satisfies $\dfrac{m-\sqrt[3]{2}}{\sqrt[3]{3}}$, thus $\dfrac{m-\sqrt[3]{2}}{\sqrt[3]{3}}$ is an algebraic integer. $\hspace{15cm} \blacksquare$

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  • $\begingroup$ Excellent!. I knew there was something straightforward to do but couldn't see that substitution step. $\endgroup$
    – tomi
    May 5, 2015 at 7:32
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$x\sqrt[3]{2} = m - \sqrt[3]{2} \to \sqrt[3]{2} = \dfrac{m}{x+1} \to 2 = \dfrac{m^3}{x^3+3x^2+3x+1} \to 2x^3+6x^2+6x+2-m^3=0$. Thus $x$ is algebraic number.

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  • $\begingroup$ This is not a complete answer but can be turned into one easily. It is not clear to me if this is by intent or due to not reading the question completely. The question asks for "algebraic integer" not "algebraic number." $\endgroup$
    – quid
    May 4, 2015 at 21:57
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    $\begingroup$ You have to show that it is a algebraic integer, that is the polynomial is monic. This only happens when $2|m$. The question seems wrong as is. $\endgroup$
    – Asvin
    May 4, 2015 at 21:58
  • $\begingroup$ Sorry guys, I made a mistake, actually is $\sqrt[3]{3}$ downstairs. The result polynomial is still not monic and I cannot find any way to turn it into a monic one. The hypotesis $m\equiv2 (mod3)$ doesn`t fit $\endgroup$
    – Lucas
    May 4, 2015 at 22:04
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Another proof: it is enough to prove $N(m-\sqrt[3]{2})$ is an integer. Now the norm is the product of the conjugates: if $j=\mathrm e^{\tfrac{2\mathrm i\pi}3}$, $$N\biggl(\frac{m-\sqrt[3]{2}}{\sqrt[3]{3}}\biggr)=\frac{(m-\sqrt[3]{2})(m-j\sqrt[3]{2})(m-j^2\sqrt[3]{2})}{\sqrt[3]3\cdot j\sqrt[3]{3}\cdot j^2\sqrt[3]{3} }\equiv \frac{m^3-2}3, $$ As $m\equiv 2\mod3$, $\,m^3\equiv 8\equiv 2\mod 3$, so $m^3-2\equiv 0\mod 3$, which proves the assertion.

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  • $\begingroup$ That's nice. You just want to change the $2$ in the denominator for a $3$. And put the actual number in the $N()$ I think. $\endgroup$
    – quid
    May 4, 2015 at 23:55
  • $\begingroup$ Oh! Yes I concentrated all my attention on the numerator and forgot there was a denominator. I'll change that in a moment. Thanks. $\endgroup$
    – Bernard
    May 5, 2015 at 0:11
  • $\begingroup$ I don't think it is enough to prove that norm is integer, as there are numbers which are not algebraic integers but have integer norm. $\endgroup$ May 5, 2015 at 1:18
  • $\begingroup$ 3 + 2√(-3) is not an integer in the ring of of Q(√(-3)) but its norme is a rational integer. I am something confused, The integers of this field are of the form [a + b√(-3)] / 2 and must have the same parity however that element is root of a monic polynomial. Can you clarify me this, please. $\endgroup$
    – Piquito
    May 5, 2015 at 1:21
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Let $x=\frac {m-\sqrt [3] 2}{\sqrt [3] 3}$

Let $m=3k+2$

Then $x=\frac {3k+2-\sqrt [3] 2}{\sqrt [3] 3}$

$x\sqrt [3] 3=3k+2-\sqrt [3] 2$

$x\sqrt [3] 3+\sqrt [3] 2 =3k+2$

$\left( x\sqrt [3] 3+\sqrt [3] 2 \right)^3 =\left(3k+2 \right)^3$

$3x^3+3\sqrt [3] {18}x^2+3\sqrt [3] {12}x+2=27k^3+54k^2+36k+8$

... no not getting anywhere helpful!

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Let $x=\frac{m-\sqrt[3]2}{\sqrt[3]3}$, and let $y=\sqrt[3]3x$.
Then $p(y)=(y-m)^3+2=0$.
Let $\omega$ be a complex cube-root of $1$.
$p(y)p(\omega y)p(\omega^2y)=q(y^3)=q(3x^3)$ is a cubic in $y$ with integer coefficients. It is an integer polynomial in $x$ with lead coefficient $27$, and hopefully the other coefficients are multiples of $27$.

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