7
$\begingroup$

I'm reading a paper in which they solve the equation: $a^3-2b^3=\pm 1$ in integers using algebraic number theory.

The number $a-b\alpha$, with $\alpha=\sqrt[3]{2}$, is a unit in $\mathbb{Z}[\alpha]$. The units of this ring are, up to sign, powers of the single unit $1+\alpha+\alpha^2$. With some work one finds that $|a-b\alpha|$ can only be the zero'th power, so that $a=\pm 1$ and $b=0$.

1) Why are the units of the ring only powers of $1+\alpha+\alpha^2$? I can't find anything to this effect in my textbook and web searches won't turn up with anything.

2) Can't $|a-b\alpha|$ be the $(-1)$th power as well and $a=b=\pm 1$?

The second question I've concluded is a minor error but I can't be satisfied with this solution without a proof for my first question.

The Paper in question: http://www.ams.org/journals/bull/2004-41-01/S0273-0979-03-00993-5/S0273-0979-03-00993-5.pdf

Relevant section is towards the end of second page.

  • 0
    Dirichlet's unit theorem is one of the most beautiful theorems in algebraic number theory. It lets you describe the structure of the group of units of nice rings in terms of geometry.2012-11-28

1 Answers 1

5

As noted in comments, Dirichlet's unit theorem shows that the group of units in $\mathbb Z[2^{1/3}]$ (which is the full ring of integers in $\mathbb Q(2^{1/3})$ --- see this question) consists of elements of the form $\pm \eta^n$ for some fundamental unit $\eta$. The proof of Dirichlet's theorem should be effective in principal (perhaps this book adopts a perspective which helps with making things effective), and I guess in practice for sufficiently simple fields (such as $\mathbb Q(2^{1/3})$). In any case, you can certainly use a computer algebra package (such as sage) to determine that $1+\alpha + \alpha^2$ is a fundamental unit (assuming that it is; I didn't check).

You are correct that $(1+\alpha + \alpha^2)^{-1} = -1 + \alpha,$ and so the authors of the paper you are reading mistated their claim. [In the context of the paper you are reading, note that $a = 1, b = 0$ actually leads to the solution $x = 0, y = 1$, and it is the omitted solution $a = b = 1$ which leads to $x = 2, y = 3$.]

Also, while it is easy to see that any power of $1+\alpha+\alpha^2$ has a non-zero coefficient of $\alpha^2$ (if we write $(1+\alpha+ \alpha^2)^n = a_n + b_n\alpha + c_n \alpha^2$ then there is a simple recursion for $a_n,b_n,c_n$ in terms of $a_{n-1}, b_{n-1},c_{n-1}$, and one sees that $a_n,b_n,c_n$ are always positive, because this recursion only involves addition, no subtraction, and $a_1 = b_1 = c_1 = 1$ is positive), this is less obvious (at least to me) for the negative powers, because while there is also a simple recursion for the coefficients of $(1-\alpha)^n$, in this case the recursion has a mixture of signs, and the coefficients do vary in sign, so you will have to work harder to verify that the coefficient of $\alpha^2$ is never zero when $n > 1$ (again, assuming that it's in fact true, which I didn't try to check).

  • 1
    @Gene: Dear Gene, That's great that you're trying to read the paper! Don't be deterred by the apparent difficulty of the math. Firstly, if you take a second semester in Algebra, you will learn more background material; secondly, if you would really like to understand the paper, that will give you a lot of motivation to learn some of the algebraic number theory, and learning mathematics is *much* easier when you have a real application in mind for it. Best wishes,2012-11-28