Let $A = F[x,y] / \langle y^2 - x^3 \rangle$. In a past exam question (from last year), students were asked to show that $A$ is an integral domain, and also that it is not integrally closed in its field of fractions.
A solution was provided by the university, which I will put in my own words below. The problem is that the ideas involved are quite unfamiliar, and I've never seen a similar strategy employed before. Could anybody give me some examples of similar constructions, susceptible to similar methods?
To be clear, I do understand the solution below; I am happy that each step follows from the last. However, I am not confident that I would be able to come up with such a solution myself in a similar context, and I am looking for more examples, so that I can become more familiar with the ideas.
Solution
Let $\varphi:F[x, y] \to F[t]$ be the unique homomorphism with $\varphi(x) = t^2$ and $\varphi(y) = t^3$. Then $\varphi(y^2 - x^3) = 0$, so $\varphi$ descends to a ring homomorphism $\bar{\varphi}:A \to F[t]$.
Let $\bar{x}, \bar{y}$ be the images of $x$ and $y$ respectively in $A$. Then $A = F[\bar{x}] \oplus \bar{y}F[\bar{x}]$, where directness of the sum comes from considering degrees. If $f(\bar{x}) + \bar{y}g(\bar{x}) = 0$ then (taking the image under $\bar{\varphi}$) also $f(t^2) + t^3g(t^2) = 0$, which means that $f = g = 0$ since all terms of $f(t^2)$ have even degree and all terms of $t^3g(t^2)$ have odd degree. Therefore $A$ is a domain.
Now, the image of $\bar{\varphi}$ is $F[t^2, t^3, t^4, \ldots]$, and in particular $t \not \in \bar \varphi(A)$. However, taking the natural extension of $\bar \varphi$ to the field of fractions of $A$, we have $\bar \varphi(\bar y/\bar x) = t^3/t^2 = t$, so $\bar y / \bar x \not \in A$. However, $(\bar{y} / \bar x)^2 = \bar y^2 / \bar x^2 = \bar x^3 / \bar x^2 = \bar x$, so $\bar y / \bar x$ is a root of the monic polynomial $X^2 - \bar x \in A[X]$, hence is integral over $A$.