My question arises from following former thread of mine:Category Equivalence between Curves and Fields of Transcendence degree $1$
and refers to a comment of @reuns:
Let $K/k(t)$ a finite extension and $R=k[x_1,…,x_n]/I$ the integral closure of $k[t]$. $V(I)$ is a smooth affine curve (and its function field is $K$) but its projective closure doesn't need to be smooth...
My question is how to see that the resulting scheme $V(I) = Spec(k[x_1,…,x_n]/I$ is a smooth curve?
My considerations:
By definition of smoothness this is equivalent to show that $R=k[x_1,…,x_n]/I $localized at every prime $p⊂R$ is regular. Since $R$ is integral over $k[t]$ so $dim(R)=1$ and it suffice to show that the max ideal mp of local ring $R_p $ is principal or (by Nakayama) equivalently $m_p/m^2_p$ is one dimensional vector space. But from here I'm stuck in finding an argument.