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Continued fractions and class groups

MathOverflow Asked by Stanley Yao Xiao on November 3, 2021

Let $d$ be a positive integer. It is well-known (due to Lagrange) that the continued fraction of $sqrt{d}$ is eventually periodic. Moreover, it is known that the equation

$$displaystyle x^2 – dy^2 = -1$$

is soluble in integers $x,y$ if and only if the continued fraction of $sqrt{d}$ has odd period.

An alternative criterion is given in terms of the class group $text{CL}(K_d)$ of $K_d = mathbb{Q}(sqrt{d})$ and the narrow class group $text{CL}^sharp(K_d)$. The difference is that the former mods out all principal ideals, while the latter only mods out principal ideals of positive norm. Indeed, we see that the above equation, which exactly identifies when $mathcal{O}_{K_d}$ has a unit of norm $-1$, implies that the two groups are equal. The converse is also true.

More precisely, we don’t need to look at the entire class group/narrow class group, but only the $2^k$-torsion subgroups for $k = 1, 2, cdots$. Indeed, we have the following alternative criterion: the equation $x^2 – dy^2 = -1$ is soluble in integers $x,y$ if and only if $text{CL}(K_d)[2^k] cong text{CL}^sharp(K_d)[2^k]$ for all $k geq 1$ (note by the finiteness of the class group, both sides eventually stabize).

Is there deeper connection between continued fractions of $sqrt{d}$ and the $2^k$-torsion criterion above?

For instance, we know that for all primes $p equiv 1 pmod{4}$ the equation $x^2 – py^2 = -1$ is soluble, which means that for any such prime $sqrt{p}$ always has odd period. Is there any reason why one should believe this looking only from the continued fraction side?

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