Q:

normalization of a curve, simplest example

I am learning about normalization of nodal curves and I am trying to understand the simplest example: $xy=0$
As far as I understand its coordinate ring is $k[x]\oplus k[y]$ (let $k$ be an algebraically closed field) 
If I look at a smooth point, $(1,0)$ for instance, then its local ring is $k(x)\oplus k[y]$ since it is just localizing $k[x]\oplus k[y]$ on the ideal (y). and its ring of fractions is $k(x)\oplus k(y)$ I think that indeed $k(x)\oplus k[y]$ is integrally closed in $k(x)\oplus k(y)$ as desired. 
Now for the nodal point $(0,0)$, its local ring is $k[x]\oplus k[y]$  but I can not say whether it is integrally closed or not. Can someone help me? What am I doing wrong?
Thank you

A:

I've tried to read the comments, but it's not clear to me if $$K[X,Y]/(XY)\not\simeq K[X]\oplus K[Y]$$ is settled down or not. The argument is very simple: the first ring has only two idempotents, while the second has four.
In your example the smooth point $(1,0)$ corresponds to the maximal ideal $\mathfrak m=(x-1,y)$ and $R_{\mathfrak m}=K[X]_{(X-1)}$ is a DVR, so it is integrally closed.

