Normalization of $k[t^2,t^3]$ 
I am trying to prove that, for some field $k$, $k[t]$ is the normalization of the ring $k[t^2,t^3] \cong k[x,y]/(x^2 - y^3)$. 

I trying to do it as follows: Consider $R = k[t^2,t^3]$ and its normalization $\bar{R}$. Then, I want to show that $\bar{R} = k[t]$. I have shown that $k[t] \subseteq \bar{R}$, but I am struggling a lot showing that $\bar{R} \subseteq k[t]$. Could anyone help me out with this or give me hints on how to do it? Thanks a lot!
 A: So you've already shown that $k[t] \subseteq \bar R$. You also know that $R \subseteq k[t]$, which implies that $\bar R \subseteq \overline{k[t]}$. Therefore, as Rene pointed out, it only remains to show that $\overline{k[t]} = k[t]$, i.e. it only remains to show that $k[t]$ is integrally closed in its field of fractions.
In fact, any unique factorization domain is integrally closed in its field of fractions. And since $k[t]$ is a UFD, this solves your problem.
To prove my claim, suppose $R$ is a UFD and ${\rm Frac}(R)$ is its field of fractions. Suppose $x \in {\rm Frac}(R)$ obeys an equation of the form
$$ x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 =0,$$
where each $a_i$ is in $R$. Since $R$ is a UFD, $x$ can be written as a fraction of the form $x = u /v$, where $u, v \in R$ have no common irreducible factors. By clearing denominators, we get an equation of the form
$$ u^n = - v \left( a_{n-1}u^{n-1} + \dots  +a_1 u v^{n-2} + a_0 v^{n-1} \right) = 0,$$
which implies that $v$ divides $u^n$. But $u$ and $v$ have no common irreducible factors, so the only possibility is that $v$ is a unit, which implies that $x \in R$.
