Writing a rational number $D$ as $f^2D'$

Question

I need help proving the following fact.

Let $D$ be a rational number that is not a perfect square in $\mathbf Q$, i.e., $\sqrt D$ is not rational. The rational number $D$ may be written $f^2D'$ for some rational number $f$ and a unique integer $D'$ where $D'$ is not divisible by the square of any integer greater than $1$, i.e., $D'$ is either $-1$ or $\pm 1$ times the product of distinct primes in $\mathbf Z$ (for example, $8/5 = (2/5)^2\cdot 10$).

This is a part of the examples of rings in Abstract Algebra by Dummit and Foote, and they state this without proof. I am not sure where to begin proving this.

Answer 1

I'm gonna assume all variables in this post take values in the positive integers, just because it simplifies things. It's easy to generalize.

Every (nonzero, nonunit) rational number has a unique factorization into powers of a finite number of distinct prime integers, if we allow negative exponents. Suppose $D = p_1^{a_1}...p_n^{a_n}$, where the $p_i$ are prime and the $a_i$ are nonzero. By assumption that $\sqrt{D}$ is irrational, not all $a_i$ are even. Let $D' = \prod_{i: a_i = 2k + 1} p_i$. Then you can let $f = \sqrt{D/D'}$, and $D = f^2D'$ as desired.

Now, if you can find another rational number $g$ and a squarefree integer $Q$ such that $D = g^2Q$, then $\frac{g^2}{f^2} = \frac{D'}{Q}$, but the left hand side is a rational square and the right side cannot be unless $D' = Q$, since both are squarefree integers. We conclude that $D'$ is uniquely specified.