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I know that $\mathbb{Q}[x]/(x^2 + 1) \cong \mathbb{Q}[i]$ and $\mathbb{Q}[x]/(x^2 - 5) \cong \mathbb{Q}[\sqrt{5}]$. Is this true in general? As in, is it always the case that $\mathbb{Q}[x]/(x^2 - D) \cong \mathbb{Q}[\sqrt{D}]$? If not, under what conditions is it true?

Edit: $D$ is an integer.

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    I'm guessing it's true if and only if $D$ is not a ratio of perfect squares, since otherwise $\sqrt D \in \Bbb Q$ and then $\Bbb Q[\sqrt D] = \Bbb Q$, but I don't think that $\Bbb Q[x] / (x^2 - D) \cong \Bbb Q$ in that case. But I'm not sure enough to post an actual answer. – 2017-02-01
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    @tilper Sorry, I meant to add that $D$ is an integer. Does that make it easier? – 2017-02-01
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    That would only change "ratio of perfect squares" to "perfect square" in my first comment and I'd be just as unsure either way. – 2017-02-01

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This is true if $D$ is not a perfect square. This can be proved via the fundamental homomorphism theorem. Define a map $$\phi:\mathbb{Q}[x]\to\mathbb{Q}(\sqrt{D})$$ by $\phi(f)=f(\sqrt{D})$. It is straightforward to check that $\phi$ is a surjective homomorphism. Therefore, $\mathbb{Q}[x]/\ker\phi\cong\mathbb{Q}(\sqrt{D})$. It is left to check that $\ker\phi=(x^2-D)$. Indeed, certainly $(x^2-D)\subset\ker\phi$. On the other hand, as $x^2-D$ is irreducible over $\mathbb{Q}$ it follows that $(x^2-D)$ is maximal. As $\ker\phi\neq\mathbb{Q}[x]$, we must have $\ker\phi=(x^2-D)$.

Now, as pointed out in the comments, if $D=d^2$, then $\mathbb{Q}(\sqrt{D})=\mathbb{Q}$. However, using the Chinese Remainder Theorem, we have $$\mathbb{Q}[x]/(x^2-D)\cong\mathbb{Q}[x]/(x-d)\oplus\mathbb{Q}[x]/(x+d)\cong\mathbb{Q}\oplus\mathbb{Q}$$ so they are not isomorphic.

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    According to your answer, is it correct to say that $\mathbb{Q}(\sqrt{D_1}) \cong \mathbb{Q}(\sqrt{D_1})$ if and only if $D_1$ and $D_2$ have the "same squarefree parts", as in they are the same after dividing out any squares? – 2017-02-01
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    If you write $D_2=a^2D_1$, then $\mathbb{Q}(\sqrt{D_2})=\mathbb{Q}(a\sqrt{D_1})=\{\alpha+\beta a\sqrt{D_1}\mid \alpha,\beta\in \mathbb{Q}\}$. You should be able to write an explicit isomorphism with $\mathbb{Q}(\sqrt{D_1})=\{\alpha+\beta\sqrt{D_1}\mid \alpha,\beta\in\mathbb{Q}$. – 2017-02-01