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What are the conditions for integers $D_1$ and $D_2$ so that $\mathbb{Q}[\sqrt {D_1}] \simeq \mathbb{Q}[\sqrt {D_2}]$ as fields.

Here $\mathbb{Q}[\sqrt {D}] := \{a + b \sqrt D \mid a,b \in \mathbb{Q} \}$

Really not sure where to begin with this sort of problem. I was thinking that I should split into cases where the integer is a square or not.

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    If the integer is a square, then $\mathbb{Q}[\sqrt{D}]\cong\mathbb Q$. – 2017-02-02
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    Okay, so the remaining cases are $\mathbb{Q}[\sqrt{\pm D}]$ where $D$ is a positive non-square. – 2017-02-02
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    Let $d$ be the 'squarefree' part of $D$; then analagously, $\mathbb{Q}[\sqrt{D}]=\mathbb{Q}[\sqrt{d}]$. Now note that in $\mathbb{Q}[\sqrt{d}]$ there's an element $x$ such that $x^2=d$. Can you prove that no such element exists in $\mathbb{Q}[\sqrt{d'}]$ if $d'$ is a squarefree integer $\neq d$? – 2017-02-02
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    How could I prove that? – 2017-02-02
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    Think about subextensions – 2017-02-02
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    Well, all elements of $\mathbb{Q}[\sqrt{d'}]$ are of the form $a+b\sqrt{d'}$. If the square of one of these elements is equal to $d$, then it's certainly a member of $\mathbb{Q}$. Start by figuring out what the conditions are on $a$ and $b$ that allow $(a+b\sqrt{d'})^2\in\mathbb{Q}$, and then analyze the cases. – 2017-02-02
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    @svelaz I encountered this in a ring course not sure what that means – 2017-02-02
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    @MathMajor A subextension is just a subfield containing $\mathbb{Q}$. – 2017-02-02
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    Also the brackets in my answer denote the dimension of the field as a $\mathbb{Q}$-linear space – 2017-02-02

4 Answers 4

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If $\Bbb{Q}[\sqrt{D}] = L \cong \Bbb{Q}[\sqrt{K}]= F$, where $D$ and $K$ are integers, then $D$ must have a square root in $F$. We may assume that $K$ is squarefree. If $D = (a + b \sqrt{K})^2 = a^2 + b^2K+ 2ab\sqrt{K}$, with $a, b \in \Bbb{Q}$, then since $\sqrt{K}$ is irrational either $a = 0$ or $b = 0$. In the first case $D = b^2K$, in the second case $D = a^2$, which is a contradiction. Therefore the first case holds and since $K$ is squarefree $b$ must be an integer. It follows that $L \cong F$ iff $D = b^2K$ for some nonzero integer $b$.

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There is a bijection between the set of squarefree integers and the set of quadratic extensions of $\mathbb{Q}$, given by $d \mapsto \mathbb{Q}(\sqrt{d})$. Exclude $1$ from being squarefree.

Surjectivity: a quadratic extension of $\mathbb{Q}$ takes the form $\mathbb{Q}(\sqrt{D})$ for some rational number $D = \pm p_1^{e_1} \cdots p_r^{e_r}$, where $p_i$ are prime numbers and $e_i$ are integers. If $e_i$ is even, then clearly one can obtain the same extension by removing $p_i^{e_i}$ from the factorization, so we may assume all the $e_i$ are odd. Then $\mathbb{Q}(\sqrt{D}) = \mathbb{Q}(\sqrt{\pm p_1 \cdots p_r})$, keeping the same sign as $D$.

Injectivity: Let $d_1, d_2$ be squarefree integers. To say that $\mathbb{Q}(\sqrt{d_1}) = \mathbb{Q}(\sqrt{d_2})$ is to say that $\sqrt{d_1} = a + b \sqrt{d_2}$ for rational numbers $a, b$, or $d_1 = a^2 + d_2 b^2 + 2ab \sqrt{d_2}$. Since $\sqrt{d_2} \not\in \mathbb{Q}$, we must have either $a$ or $b$ equal to $0$. Thus either $d_1 = a^2$ or $d_1 = d_2 b^2$. But $d_1$ is squarefree, so $a = 0$ and $b = \pm 1$, so $d_1 = d_2$.

Thus to determine whether $\mathbb{Q}(\sqrt{D_1}) = \mathbb{Q}(\sqrt{D_2})$ for rational numbers $D_i$, one can do the following: write $D_1$ as a product of prime numbers $\pm p_1^{e_1} \cdots p_r^{e_r}$ for integers $e_i$. Let's say that $e_1, ... , e_t$ are odd integers, and $e_{t+1}, ... , e_r$ are even. Then $\mathbb{Q}(\sqrt{d_1})$, where $d_1 = \pm p_1 \cdots p_t$, keeping the same sign as before. Do the same for $D_2$, obtaining $\mathbb{Q}(\sqrt{D_2}) = \mathbb{Q}(\sqrt{d_2})$ for some squarefree integer $d_2$. By the result above, the given quadratic extensions are equal if and only if $d_1 = d_2$.

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Same extension if and only if $D_1$ and $D_2$ are in the same squareclass. This means $D_1 D_2$ is a square.

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Hint: First note that if $a,b$ are rational, then ${\bf Q}[\sqrt a]\cong {\bf Q}[\sqrt b]$ if and only if the two fields are equal (just take the isomorphism and look where it takes $\sqrt a$; there are only two possibilities, and both lead to the conclusion that $\sqrt a\in {\bf Q}[\sqrt b]$).

Next, take any $b\in {\bf Q}$, and any $x\in {\bf Q}\left[\sqrt b\right]$ and using the fact that $\sqrt b$ is not rational unless $b$ is a square (by definition!), show that if $x^2$ is a rational number, then $x\in {\bf Q}$ or $x\in \sqrt b\cdot {\bf Q}$. Conclude that if $\sqrt a\in {\bf Q}[\sqrt b]$, then $a$ is a square or $a$ is a square times $b$. Conclude that ${\bf Q}[\sqrt b]={\bf Q}[\sqrt a]$ for $b\neq 0$ if and only if $\frac{a}{b}$ is a square.

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    The first statement you make is incorrect, unless you have some conditions on $a, b$. – 2017-02-02
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    I didn't downvote. Even for $a, b$ rational it's not true. $\mathbb{Q}(\sqrt{12}) = \mathbb{Q}(\sqrt{3})$ – 2017-02-03