The result I'm trying to prove is:

There is a bijection between square-free integers $d\neq 0,1$ and quadratic fields

One direction is:

If $d$ is a square-free integer $\neq 0,1$, then $\sqrt{d}\notin \mathbb{Q}$, so that, $[\mathbb{Q}(d):\mathbb{Q}]>1$.

But since $x^2-d$ is irreducible in $\mathbb{Q}[x]$, $[\mathbb{Q}(d):\mathbb{Q}]\leq2$.

Therefore $\mathbb{Q}(d)$ is a quadratic field for every such $d$.

Can someone help out with the rest of the proof?

• re "There is a bijection" Actually you are likely trying to prove more. If it were just there is, it'd suffices to show both are countable.
– quid
Aug 22, 2016 at 20:17

If $X$ is a quadratic field, there exists $x\in K$, $x$ is not in $Q$, since $[K:Q]=2$, there exists $a,b,c\in Q, a+bx+cx^2=0$, $c\neq 0$ otherwise $x\in Q$, we can assume $c=1$ by dividing by $c$. We have $(x+{b\over 2})^2-b^2/4+a=0$, this implies that $(x+{b\over 2})^2={{b^2-4a}\over 4}$, this implies that $x=\sqrt{e\over f}, e,f \in Z$, remark that $f\sqrt{e\over f}=\sqrt f\sqrt f\sqrt{e\over f}=\sqrt{ef}$ and $\sqrt{e\over f}={1\over f}\sqrt{ef}$, this implies that $Q(\sqrt{e\over f})=Q(\sqrt{ef})$. Remark that without restricting the generalitiy, we can replace $ef$ by a square free integer $d$, $Q(\sqrt d)\subset K$ and $[K:Q]=[Q(\sqrt d):Q]=2$ we deduce that $K=Q(\sqrt d)$. This shows that every quadratic field is of the form $Q(\sqrt d)$.

Suppose that $Q(\sqrt d)=Q(\sqrt d')$ where $d$ and $d'$ are square free integers, this implies $\sqrt d'=a+b\sqrt d$, suppose $a=0$, we deduce that $d'=b^2d$. Contradiction since $d'$ is square free. $b\neq 0$ since $\sqrt d'$ is not in $Q$, we deduce that $a,b\neq 0$. we have: $d'=a^2+db^2+2ab\sqrt d$, this implies that $2ab\sqrt d\in Q$ and $\sqrt d\in Q$ contradiction since $d$ is square free, this establishes the requested bijection.

One has to show the following: suppose $d$ and $d'$ are both squarefree integers, both different from $1$, and consider $F_1 = \mathbb{Q}(\sqrt{d})$ and $F_2 = \mathbb{Q}(\sqrt{d'})$. Then $F_1$ is not isomorphic to $F_2$, because they have different discriminants. For other proofs, and further details, see this MSE question.

• One also has to show that any quadratic field is generated by a square-free $d$. Aug 22, 2016 at 19:51
• One also needs to show that every quadratic extension of $\mathbb Q$ is $\mathbb Q[\sqrt d]$ for some square-free $d$. Aug 22, 2016 at 19:51
• You are right, I missed that detail! But I was more worried by the part mentioned by Arthur and Henning. Aug 22, 2016 at 19:53
• @HenningMakholm actually for the question as asked this is not necessary.
– quid
Aug 22, 2016 at 20:19
• @quid: The question asks for a bijection between square-free integers and quadratic fields. If there were some quadratic field it didn't it wouldn't be a bijection. Aug 22, 2016 at 20:24

For "every quadratic field is $\mathbb Q[\sqrt d]$":

If $F$ is a quadratic extension of $\mathbb Q$, then let $(1,g)$ be a basis for it as a rational vector space, and consider the minimal polynomial of $g$. This polynomial is necessarily of degree $2$, so by means of the quadratic formula you can write an expression for $g$ that uses only rational constants and a single square root. A bit of moving factors out from under the square root will make the number it applies to a square-free integer, and then you know a $d$ such that $g\in\mathbb Q[\sqrt d]$. Conclude that actually $F=\mathbb Q[\sqrt d]$.

Here’s a much more abstract and advanced viewpoint:

It’s the simplest consequence of Kummer Theory that for any field $K$ of characteristic different from $2$, the quadratic extensions of $K$ are in natural correspondence to the one-dimensional subspaces of the infinite dimensional $\Bbb F_2$-vector space $K^\times/{K^\times}^2$, in other words, in natural correspondence to the nontrivial elements of this vector space. You see that for $K=\Bbb Q$, these nontrivial elements are just the square-free integers, including $-1$.

To give a generalization for cubic extensions, you have to assume that the characteristic is not $3$, and and you have to restrict to extensions that are Galois over $K$ (and thus cyclic).