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As the title says, I'm trying to show that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer.

I suppose there's probably some heavy duty classification theorems that give one line proofs to this but I don't have any of that at my disposable so basically I'm trying to construct a polynomial over $\mathbb{Z}$ which has this complex number as a root.

My general strategy is to raise both sides of the equation

$$x = \frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$$

to the $n^{th}$ power and then break up the resulting sum in such a way as to resubstitute back in smaller powers of $x$. Also since this root is complex I know it must come in a conjugate pair for the coefficients of my polynomial to be real, thus I know that

$$x = -\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$$

must also be a root of my polynomial. Hence from this I obtain:

$$3x^2 = (10^{\frac{2}{3}} -1 )^2$$

However since my root is pure imaginary I don't really get any more information from this, so I'm a bit stumped, I tried raising both sides of $3x^2 -1 = 10^{\frac{4}{3}} - (2)10^{\frac{2}{3}}$ to the third power but it doesn't look like it's going to break up correctly, can anyone help me with this? Thanks.

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3  
The element lies in the splitting field of $x^3-10$, which is generated by $10^{1/3}$ and $\zeta=(-1+\sqrt{-3})/2$. You could try looking the product of linear terms $x-\alpha$, where $\alpha$ is the image of $x$ under the six automorphisms; the automorphisms are easy enough. –  Arturo Magidin Jul 25 '12 at 19:57
    
I see that it does lie in said splitting field, but could you explain more what you mean by the rest of your comment? –  Ron Jeremy Jul 25 '12 at 20:01
    
If the six automorphisms are $\sigma_1,\ldots,\sigma_6$, then $x$ is a root of $(t-\sigma_1x)\cdots(t-\sigma_6x)$, which has rational coefficients. –  Arturo Magidin Jul 25 '12 at 20:02

6 Answers 6

up vote 9 down vote accepted

Your general strategy is workable, but you'll be a lot happier if you start by eliminating the cube root. Start with $$x \sqrt{-3}+1=10^{2/3} \, .$$ Cube this equation, expand the left-hand side, collect all the factors of $\sqrt{-3}$ on one side, and square the result, and you'll be left with a sextic (hopefully the same one Robert came up with, up to a constant multiple).

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The minimal polynomial, found with help from Maple, is $z^6+z^4+67 z^2+363$.

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A systematic (but possibly not particularly clever) approach would be to work in $\mathbb Q[\alpha,\beta]$ where $\alpha=\sqrt[3]{10}$ and $\beta=i\sqrt{3}$.

First note that $1/\beta = \frac{-1}{3}\beta$. Then your $x$ is some polynomial in $\alpha$ and $\beta$. Write out the powers of $x$ up to $x^6$ as similar polynomials, simplifying along the way using $\alpha^3=10$ and $\beta^2=-3$.

You now have $1, x, x^2,\ldots x^6$ expressed as polynomials of degree at most $3$ in $\alpha$ and at most $2$ in $\beta$. The space of such polynomials is a $6$-dimensional rational vector space, so you can find a nontrivial linear combination of your $7$ polynomials -- with is then a polynomial in $x$ that has your mystery number as a root.

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If we start with $\sqrt{-3}x + 1 = 10^{2/3}$, we get $$(\sqrt{-3}x+1)^3 = 100$$ hence $$100 = -3\sqrt{-3}x^3 - 9x^2 + 3\sqrt{-3}x + 1.$$ Therefore, $$3\sqrt{-3}x^3 + 9x^2 - 3\sqrt{-3}x + 99 = 0.$$ Dividing through by $3\sqrt{-3}$ we obtain $$x^3 + \frac{3}{\sqrt{-3}}x^2 - x + \frac{33}{\sqrt{-3}}=0$$ and rationalizing we get $$x^3 + \frac{3\sqrt{-3}}{-3}x^2 - x + \frac{33\sqrt{-3}}{-3} = 0$$ or $$x^3 - \sqrt{-3}x^2 - x -11\sqrt{-3}=0.$$ This is a monic polynomial with coefficients in $\mathbb{Z}[\sqrt{-3}]$; hence $x$ is integral over $\mathbb{Z}[\sqrt{-3}]$, which in turn is integral over $\mathbb{Z}$, so $x$ is integral over $\mathbb{Z}$, as desired.

To go from this to the polynomial in Robert Israel's answer, we multiply by the conjugate: $$\begin{align*} &\Bigl((x^3-x) - \sqrt{-3}(x^2+11)\Bigr)\Bigl((x^3-x)+\sqrt{-3}(x^2+11)\Bigr)\\ &\qquad = (x^3-x)^2 + 3(x^2+11)^2\\ &\qquad = x^6 - 2x^4 + x^2 + 3x^4 + 66x^2 + 363\\ &\qquad = x^6 + x^4 + 67x^2 + 363, \end{align*}$$ and we are done.

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$$\begin{eqnarray} & &\rm x &=&\,\rm \frac{\sqrt[3]{a}-1}{\sqrt{b}} \\ &\Rightarrow& 0 &=&\,\rm (\sqrt{b}\,x + 1)^3\!-a\, =\, 3b\,x^2\!+1\!-\!a + (bx^3\! + 3x)\, \color{#C00}{\sqrt{b}}\ =:\ f(\color{#C00}{\sqrt{b}}) \\ &\Rightarrow& 0 & = &\,\rm f(\color{#C00}{\sqrt{b}})\,f(\color{#C00}{-\sqrt{b}})\, =\, -b^3\,x^6 + 3b^2\,x^4 -3(2a\!+\!1)b\,x^2+(a\!-\!1)^2\\ \rm a=100,\ b = -3 &\Rightarrow& 0 &=&\,\rm 27\,(x^6 + x^4 + 67\, x^3 + 363) \\ \rm a=10,\ \ \ b = -3 &\Rightarrow\,& 0 &=&\,\rm 27\, (x^6+x^4 +\,\ 7\, x^2 + 3)\\ \end{eqnarray} $$

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A somewhat less computational approach can be based on the idea that an algebraic number is an algebraic integer if and only if it is everywhere locally integral. Certainly $\sqrt{-3}$ is a $p$-adic unit (possibly in a quadratic extension) everywhere except $3$, so the only issue is at $3$.

Second, observe/speculate that there is a cube root $\alpha$ of $10$ in $\mathbb Q_3$, using Hensel's Lemma: $10$ is pretty close to $1$ modulo powers of $3$, that is, $3$-adically. A useful form of Hensel's lemma here is that a monic polynomial $f$ with coefficients in $\mathbb Z_3$ and $x_o\in \mathbb Z_3$ such that $|f(x_o)/f'(x_o)^2|_3<1$ produces a root of $f(x)=0$ in $\mathbb Z_3$.

From this, we see that if we can find a cube root of $10$ mod $3^3=27$, we have a $3$-adic cube root. Indeed, $4^3=64=10+2\cdot 3^3$.

So there is $\alpha\in \mathbb Z_3$ with $\alpha-4=0 \mod 3$, so ${\alpha-1\over 3}$ is a $3$-adic integer, and certainly ${\alpha-1\over \sqrt{-3}}$ is integral over $\mathbb Z_3$, although in a quadratic extension.

A similar pattern occurs for all odd primes: $(1+p)^p=1+p^2 \mod p^3$, so there is a $p$-th root of $1+p^2$ in $\mathbb Z_p$.

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My first encounter with "everywhere locally implies globally (sometimes)" was when I first got to grad school, and asked Bernard Dwork (a very nice guy, in addition to his mathematical virtues) some question about algebraic number theory, which I'd been thinking about globally, and gotten tangled-up, despite working hard to be clever. Unsurprisingly, Dwork saw that the issue was local-everywhere, and dispatched it immediately, truly. I was stunned, not only that he could ascertain the facts, but that better technique was so devastatingly much better. It's not just packaging, folks. –  paul garrett Jul 27 '12 at 23:38

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