<s>[INST] normalization of the nodal cubic Consider the nodal cubic $R = k[x,y]/(y^2 - x^3 + x^2)$, so $x,y$ satisfy $y^2 = x^3 - x^2$.
The idea behind normalization is to adjoin "$\frac{y}{x}$", since then in the fraction field $K$ of $R$, one has $\left(\frac{y}{x}\right)^2 = x-1$.
It's straightforward to show that the subring $S$ of $K$ generated by $R$ and $\frac{y}{x}$ is normal and integral over $R$, hence $S := R[\frac{y}{x}]$ is the normalization/integral closure of $R$.
It's also not hard to show that $S\cong R[t]/(xt - y, t^2-x+1)$.
My question is - do you need both $xt -y$ AND $t^2-x+1$?
Is $S\cong R[t]/(xt - y)$?
Is $S\cong R[t]/(t^2-x+1)$?
 [/INST] If $xt=y$ and $t^2-x+1=0$, then multiplying the  second equation by $x^2$ and using the first one shows that $y^2-x^3+x^2=0$. This means that you forget about this last equation, and $S$ is $k[x,t,y]/(xt-y,t^2-x+1)$. 
Now in this quotient, we have $x=t^2+1$ so we can remove the generator $x$, and it is isomorphic to $k[y,t]/(t^2+1)t-y)$. Of course, now we can also remove $y$, and we end up with $S=k[t]$.
Using this description of $S$, can you answer your questions?

Notice that $R[t]/(xt-y)=k[x,y,t]/(y^2-x^3+x^2,xt-y)$. As $y=xt$ here, we can remove the generator $y$, and $R[t]/(xt-y)$ is isomorphic to $k[x,t]/(x^2t^2-x^3+x^2)$, which is clearly not a domain.
On the other hand, $R[t]/(t^2-x+1)$ is $k[x,y,t]/(y^2-x^3+x^2,t^2-x+1)$, which, using $x=t^2+1$ to remove $x$, is isomorphic to $k[y,t]/(y^2-(t^2+1)^3+(t^2+1)^2)$. This has some singular points.
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