Showing that the field of fractions of $\mathbb{Z}[\sqrt{d}]$ is $\mathbb{Q}[\sqrt{d}].$ 
Let $d\in\mathbb{Z}$ be an integer that it not a square. Let $\sqrt{d}\in \mathbb{C}$ be a square root of $d$. Let  $\mathbb{Z}[\sqrt{d}]:=\left\{a+b\sqrt{d}:a,b\in\mathbb{Z}\right\}$. Let $F$ be its field of fractions. Show that $F$ can be identified with $\mathbb{Q}[\sqrt{d}]:=\left\{a+b\sqrt{d}:a,b\in\mathbb{Q}\right\}$.

My solution. With $f:F\to \mathbb{Q}[\sqrt{d}]$ by $f((a+b\sqrt{d},a'+b'\sqrt{d}))=\frac{a+b\sqrt{d}}{a'+b'\sqrt{d}}=\frac{aa'-bb'd}{a'^2-b'^2d}+\frac{a'b-ab'}{a'^2-b'^2d}\sqrt{d}$. It must be shown that $f$ is a ring isomorphism and bijective, but it is too tedious.
Is there a simpler way to prove that $F$ and $\mathbb{Q}[\sqrt{d}]$ are isomorphic?
 A: Nice question. Your approach is fine, and you're right that it's tedious. Here's another one. Let $R$ be an integral domain and $F$ a field together with a map $R \to F$. We can show that $F$ is the field of fractions of $R$ by showing that it is the smallest field containing $R$.
Now it is obvious that $\mathbb{Q}[\sqrt{d}]$ contains $\mathbb{Z}[\sqrt{d}]$, and it is obvious that any field containing $\mathbb{Z}[\sqrt{d}]$ has to contain $\mathbb{Q}[\sqrt{d}]$, since elements of the form $m$ and $n \sqrt{d}$ have to be invertible. So the only thing to check is that $\mathbb{Q}[\sqrt{d}]$ is a field, i.e. you have to prove that non-zero elements have inverses.
A: If $\alpha\in\Bbb{C}$ is algebraic, meaning that there exists a polynomial $p\in\Bbb{Q}[x]$ such that $p(\alpha) = 0,$ we in fact have
$$
\operatorname{Frac}(\Bbb{Z}[\alpha]) = \Bbb{Q}[\alpha],
$$
where
\begin{align*}
\Bbb{Z}[\alpha] &:= \{p(\alpha)\in\Bbb{C}\mid p\in\Bbb{Z}[x]\}\\
\Bbb{Q}[\alpha] &:= \{p(\alpha)\in\Bbb{C}\mid p\in\Bbb{Q}[x]\}.
\end{align*}
That is, elements $\beta\in\Bbb{Z}[\alpha]$ (respectively, $\beta\in\Bbb{Q}[\alpha]$) are complex numbers of the form
$$
\beta = \sum_{i = 0}^n a_i \alpha^i,
$$
where $a_i\in\Bbb{Z}$ (respectively, $a_i\in\Bbb{Q}$).
As in hunter's answer, we see that any field containing $\Bbb{Z}[\alpha]$ must contain $\Bbb{Q}[\alpha],$ so all that remains is to prove that $\Bbb{Q}[\alpha]$ is a field.
You can prove this by using the polynomial Euclidean algorithm. Let $g\in\Bbb{Q}$ be a polynomial such that $g(\alpha)\neq 0.$ We want to prove that $\frac{1}{g(\alpha)}\in\Bbb{Q}[\alpha].$ So, first observe that $g$ is relatively prime to the minimal polynomial $f$ of $\alpha.$ The Euclidean algorithm then implies that you can find two new polynomials $h,k\in\Bbb{Q}[x]$ such that $fh + gk = 1.$ Evaluating at $\alpha$, we find
\begin{align*}
1 &= f(\alpha)h(\alpha) + g(\alpha)k(\alpha)\\
&= 0 + g(\alpha)k(\alpha),
\end{align*}
so that $k(\alpha) = \frac{1}{g(\alpha)}.$ Thus, $\Bbb{Q}[\alpha]$ is a field.
