Integral closure in field of fractions.

Question

Let $I$ be the ideal generated by $2xy+x^2+y^3$ in $\mathbb{R}[x,y]$. Define $A:=\mathbb{R}[x,y]/I$, I want to find the normalisation of $A$, that is, the set $B= \{ a \in \operatorname{Frac} A : a\text{ integral over }A \}$. What standard methods are used to do this? Any hints?

Answer 1

$\mathbb R[\bar x,\bar y]=\mathbb{R}[x,y]/(2xy+x^2+y^3)$ is an integral domain. (Here $\bar x,\bar y$ denote the residue classes of $x,y$ modulo the ideal $(2xy+x^2+y^3)$.)
Then $\bar x/\bar y$ belongs to the field of fractions, and moreover $(\bar x/\bar y)^2+2(\bar x/\bar y)+\bar y=0$, that is, $\bar x/\bar y$ is integral over $\mathbb R[\bar x,\bar y]$.

Now let's take a look at the ring $\mathbb R[\bar x,\bar y,\bar x/\bar y]$: it is in fact isomorphic to $$\mathbb R[x,y,t]/(2xy+x^2+y^3,ty-x,t^2+2t+y)$$ which is, at its turn, isomorphic to $\mathbb R[y,t]/(t^2+2t+y)\simeq\mathbb R[t]$, and therefore integrally closed.

We thus proved that the integral closure of $\mathbb R[\bar x,\bar y]$ is $\mathbb R[\bar x,\bar y,\bar x/\bar y]$.