The quadratic fields are of the form $\Bbb Q(\sqrt d) $

Question

We want to prove that the quadratic fields have form $\Bbb Q(\sqrt d) $, where $d\neq 0,1$ is a square free integer.

Proof. Let $F$ be a quadratic field. So we have $F=\Bbb Q(\alpha)$, for some $\alpha \in \Bbb C$. So, the degree of the minimal polynomial of $\alpha \in \Bbb C$ is $\deg f_\alpha = 2$. So, this polynomial has the form $$f_\alpha(X) =X^2+pX+q,$$ where $p, q\in \Bbb Q$. So, since $\alpha $ is a root, it must be $$\alpha=-p/2\pm \sqrt D, $$ where $D = p^2 /4-q \in \Bbb Q$. Clearly $D $ is not $0$. Then, $$F= \Bbb Q(\alpha) =\Bbb Q(-p/2 \pm \sqrt D) =\Bbb Q(\sqrt D). $$

Since $D\in \Bbb Q$, $D=a/b$, for some integers $a,b $ with $b≠0$. So, $b^2D=ab$.

In this point, in my notes it is written *"collecting all squares in $ab$ together, we can write $ab = c^ 2d$, with $c, d \in \Bbb Z \backslash \{0\}$ and d squarefree." *

I can not understand this step. It might be an easy one, but frankly I can't see it. Could you please elaborate?

Answer 1

Every integer can be reduced to prime factorization. So, collect all maximum even powers of primes into some number $c^2$.

Example: $3^5 5^8 7^4 11^1 13^5 = (3^2 5^4 7^2 13^2)^2 (3 \cdot 11 \cdot 13)$

Now $D=\left(\dfrac c b\right)^2 d$

So $\sqrt{D}= \dfrac{c}{b}\sqrt{d}$ with $d$ a square free integer.

Answer 2

As an alternative to InterstellarProbe's prime-factorization approach:

For any integer $n\not=0$ (e.g., $n=ab$), let $c$ be the largest positive integer such that $c^2\mid n$. (There must be such a largest $c$, since $1^2\mid n$ and $c^2\not\mid n$ if $c\gt|n|$.) Let $d=n/c^2$. If $d$ were not squarefree -- i.e., if $r^2\mid d$ with $r\gt1$ -- then we would have $(rc)^2\mid n$ with $rc\gt c$, which contradicts the defining property of $c$. We thus have $n=c^2d$ with $c,d\in\mathbb{Z}\backslash\{0\}$ and $d$ squarefree.