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I'm trying to prove the following:

Find the necessary and sufficient conditions for the integers $D_1$ and $D_2$ for the fields $$\mathbb Q(\sqrt{D_1}) \simeq \mathbb Q(\sqrt{D_2})$$

to be true.

I have no clue how to approach this.. Any help would be great!

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    Wasn't this questin answered very recently? Did the other one get deleted? – 2017-02-02
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    @MarianoSuárez-Álvarez Which post is that?? – 2017-02-02

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If $D_1$ is a square in $\mathbb{Q}$ (i.e, there exists $q \in \mathbb{Q}$ such that $q^2 = D_1$), then $\mathbb{Q}(\sqrt{D_1})= \mathbb{Q}$ and hence $\mathbb{Q}(\sqrt{D_1})\cong \mathbb{Q}(\sqrt{D_2})$ if and only if $D_2$ is also a square in $\mathbb{Q}.$ So, suppose $D_1$ is not a square in $\mathbb{Q}.$

Suppose first that $D_1$ is a positive integer with $\sqrt{D_1} \notin \mathbb{Q}$. Write $\sqrt{D_1}=p\sqrt{d_1}$ where $d_1$ is the squarefree part of $D_1$. Then $\mathbb{Q}(\sqrt{D_1})=\mathbb{Q}(\sqrt{d_1}).$

Now, if $\varphi:\mathbb{Q}(\sqrt{d_1})\to \mathbb{Q}(\sqrt{D_2})$ is an isomorphism, then $\varphi$ must fix $\mathbb{Q}$, that is, for every $p/q \in \mathbb{Q}$ we must have $\varphi(p/q)=p/q$. Indeed, if $m \in \mathbb{Z}^+$, then $\varphi(m)=\varphi(1+\cdots+1)=m\varphi(1)=m.$ Similarly $\varphi(-m)=\varphi(-1)\varphi(m)=-m$. Since $\varphi(0)=\varphi(0\cdot 0)=\varphi(0)\varphi(0),$ then $\varphi(0)=0$ and thus $\varphi$ fixes $\mathbb{Z}.$ Furthermore, $$\frac{m}{m}=1=\varphi(1)=\varphi(\frac{m}{m})=\varphi(\frac{1}{m})\varphi(m)=\varphi(\frac{1}{m})\cdot m \hspace{0.4cm}\Rightarrow \hspace{0.4cm} \varphi(\frac{1}{m}) = \frac{1}{m}.$$ Therefore, if $m/n \in \mathbb{Q},$ then $\varphi(m/n)=\varphi(1/n)\varphi(m)=m/n$, so $\varphi$ fixes $\mathbb{Q},$ as claimed.

Write $\sqrt{D_2}=q\sqrt{d_2}$ where $d_2$ is the square free part of $D_2.$ Then $\mathbb{Q}(\sqrt{D_2})=\mathbb{Q}(\sqrt{d_2}).$ Thus $\varphi:\mathbb{Q}(\sqrt{d_1})\to \mathbb{Q}(\sqrt{d_2})$ and we must have $\varphi(\sqrt{d_1})=\sqrt{d_2}.$

Note that $\sqrt{d_1}$ is a root of the polynomial $x^2-d_1$ over $\mathbb{Q}.$ Furthermore, $$\varphi(\sqrt{d_1})^2=\varphi(\sqrt{d_1}^2)=\varphi(d_1)=d_1,$$ so $\varphi(\sqrt{d_1})^2$ is a root of $x^2-d_1$ as well. Thus, $\varphi(\sqrt{d_1})=\pm \sqrt{d_1}$ and since $\sqrt{d_2}$ is positive, then $d_1=d_2.$ Hence, $D_1$ and $D_2$ have the same square-free part.

If $D_1$ is a negative integer, we write $D_1=pd_1$ $(p>0, d_1<0)$ where $-d_1$ is the least integer $0<-d_1\leq -D_1$ such that $\sqrt{-d_1} \notin \mathbb{Q}$ and note that $\sqrt{d_1}=i\sqrt{-d_1}$ is a root of $x^2+d_1$ and the procedure is analogous.

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    What do you mean $\psi$ must fix $\mathbb Q$? – 2017-02-01
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    @user282639 See my edit. – 2017-02-01
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    I see that makes sense... but why should $\psi(\sqrt{D_1}) = \sqrt{D_2}$? Did you mean that $\psi(\sqrt{D_1})^2$ is a root of $x^2-D_2$ as well? – 2017-02-01
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    Since $\varphi$ fixes all elements in $\mathbb{Q},$ if $\sqrt{D_1} \notin \mathbb{Q}$ then the only possibility to $\varphi(\sqrt{D_1})$ is $\sqrt{D_2}$ since $\varphi$ is bijective. – 2017-02-01
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    Since $\varphi(\sqrt{D_1})^2-D_1=0$, then $\varphi(\sqrt{D_1})$ is a root of $x^2-D_1.$ – 2017-02-01
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    How does $D_1$ being a square guarantee $\mathbb Q[\sqrt{D_1}]$ being isomorphic to $\mathbb Q$? – 2017-02-02
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    I don't think it is right that $D_1=D_2.$ Is it not true that $\mathbb{Q}(\sqrt{8}) = \mathbb{Q}(\sqrt{2})$? I think the correct condition is that $D_1$ and $D_2$ have the same square-free part. – 2017-02-02
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    @spaceisdarkgreen You're right. Thanks. I fixed it. – 2017-02-02
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    @user282639 If $D_1$ is a square in $\mathbb{Q}$ then $\sqrt{D_1} \in \mathbb{Q}$, so $\mathbb{Q}(\sqrt{D_1})$ equals $\mathbb{Q}.$ – 2017-02-02
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    Is it necessary to prove that given that $D_1$ and $D_2$ have the same square free part, that the map $\psi: \mathbb Q[\sqrt{D_1}] \to \mathbb Q[\sqrt{D_2}]$ is an isomorphism? – 2017-02-02
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    And just not fully sure why $\sqrt{d_1}$ needs to map to $\sqrt{d_2}$ in $\psi$ – 2017-02-02
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    @user282639 if $D_1$ and $D_2$ have the same square free part, say $d$, then $\mathbb{Q}(\sqrt{D_1})=\mathbb{Q}(\sqrt{d})=\mathbb{Q}(\sqrt{D_2}).$ Now, every element of $\mathbb{Q}(\sqrt{d_1})$ has the form $p+q\sqrt{d_1}$ and every element of $\mathbb{Q}(\sqrt{d_2})$ has the form $p+q\sqrt{d_2},$ since $d_1$ and $d_2$ are square free and $\varphi$ is an isomorphism, is must be $\varphi(d_1)=\pm\sqrt{d_2}$ and thus $d_1=d_2.$ – 2017-02-02