Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?

In this post we saw isomorphism of vector spaces over $\mathbb{Q}$. Just came across this question:

• Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?

I know these as $\mathbb{Q}$-Vector spaces, are isomorphic from the linked post. But as fields are they isomorphic? I neither know how to prove it nor how to disprove it.

More generally: suppose $d$ and $d'$ are both squarefree integers, both different from $1$, and consider $F_1 = \mathbb{Q}(\sqrt{d})$ and $F_2 = \mathbb{Q}(\sqrt{d'})$.

They are both isomorphic as $\mathbb{Q}$-vector spaces, since they are both of dimension $2$; or more explicitly, every element of $F_1$ can be written uniquely as $a+b\sqrt{d}$ with $a,b\in\mathbb{Q}$ (unique because $\sqrt{d}\notin\mathbb{Q}$), and every element of $F_2$ can be written uniquely as $x+y\sqrt{d'}$ with $x,y\in\mathbb{Q}$. The map $f\colon F_1\to F_2$ given by $f(a+b\sqrt{d}) = a + b\sqrt{d'}$ is additive and $\mathbb{Q}$-homogeneous, clearly bijective, so $F_1$ and $F_2$ are isomorphic as vector spaces over $\mathbb{Q}$.

However, they are never isomorphic as fields; clearly, $d$ is a square in $F_1$. I claim $d$ can only be a square in $F_2$ if $d=d'$. Indeed, if$(x+y\sqrt{d'})^2 = d$. That means that $x^2 + d'y^2 + 2xy\sqrt{d'} = d$, hence $2xy = 0$ and $x^2+d'y^2=d$. If $x=0$, then $d=d'y^2$, so clearing denominators you get $da^2 = d'b^2$ for some $a,b\in\mathbb{Z}$, $\gcd(a,b)=1$; since both $d$ and $d'$ are squarefree, it follows that $|a|=|b|=1$, so $d=d'$. If $y=0$, then $d=x^2$, so $d$ is the square of a rational, contradicting the fact that it is a squarefree integer different from $1$. Thus, of $d$ is a square in $F_2$, then $d=d'$. Hence, if $F_1\cong F_2$, then $d=d'$ (converse is immediate).

Now, since every quadratic extension of $\mathbb{Q}$ is equal to $\mathbb{Q}(\sqrt{d})$ for some squarefree integer $d$ different from $1$, you conclude that any two quadratic extensions are either identical or not isomorphic as fields.

To prove this: Suspect the fields are not isomorphic, then we can attempt to find a property which holds inside one and does not the other - but whose truth is preserved by isomorphism.

In the field $\mathbb{Q}(\sqrt{2})$ there is an element which satisfies the field property $x^2=2$. There is no element in $\mathbb{Q}(\sqrt{3})$ which satisfies this, but suppose for a contradiction that there was an isomorphism $\psi : \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{3})$ we would have $\psi(x^2) = \psi(2)$ which is equivalent to $\psi(x)^2 = \psi(1)+\psi(1)$ and since $\psi(1) = 1$ we have an element of $\mathbb{Q}(\sqrt{3})$ which, squared, is 2.

Hint $\ $ Compare discriminants: $\{\:\!(\alpha-\alpha'\big)^2:\ \alpha \in \mathbb Q(\sqrt 2)\:\!\} =\,2\:\!\mathbb Q^{2}\, $ vs. $\, 3\:\!\Bbb Q^{2}\, $ for $\rm\ \mathbb Q(\sqrt 3)\ $

Note that if $\rm\ \alpha,\: \alpha'\ \not\in\mathbb Q\ $ are conjugate then they remain so under any field isomorphism since their minimal polynomial $\rm\ (x-\alpha)\ (x-\alpha')\ $ is in $\rm\:\mathbb Q[x]\:$ so it is fixed by any isomorphism.

In fact quadratic fields are characterized uniquely by their discriminant.

Short answer: the splitting fields of $x^2-2$ and $x^2-3$ over $\mathbb{Q}$ cannot be the same field or isomorphic fields. For instance, there are infinite odd primes $p$ such that $\left(\frac{2}{p}\right)=1$ and $\left(\frac{3}{p}\right)=-1$ (any prime $\equiv 17\pmod{24}$, for instance) and infinite odd primes $p$ such that $\left(\frac{2}{p}\right)=-1$ and $\left(\frac{3}{p}\right)=1$ (any prime $\equiv13\pmod{24}$, for instance). In particular $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a quadratic extension of both $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ and these fields cannot be isomorphic.