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Let $k$ be a field and $C/k$ be an affine plane curve over $k$, namely $C = \mathrm{Spec}(A)$ for some $A = k[x,y]/(f(x,y))$, here $f(x,y) \in k[x,y]$ is an irreducible polynomial. Let $B$ be the integral closure of $A$ in the quotient field $\mathrm{Frac}(A)$ of $A$. Then $C' := \mathrm{Spec}(B)$ is the normalization of $C$, which is an non-singular affine curve over $k$. Consider the base change $C_{\overline{k}} := C \times_{k} \overline{k}$. Suppose $f(x,y)$ is geometrically irreducible, then $ C_{\overline{k}} $ is an affine plane curve over $\overline{k}$ defined by an equation $f(x,y) \in \overline{k}[x,y]$. Suppose $C_{\overline{k}}$ has only ordinary singular points, (namely, the tangent lines at each singular point of $C_{\overline{k}}$ are all distinct.) One can blow up the curve $C_{\overline{k}}$ and get a non-singular affine plane curve $C'_{\overline{k}}$. (Each time, we take a linear change such that the singular point at which we want to blow up becomes origin, then use the well-known coordinate change $(x, y) \rightarrow (x, xy)$. On the other hand, if $k$ is perfect, then $ C' \times_k \overline{k}$ is also a non-singular affine curve over $\overline{k}$ and we have a natural dominant morphism $ C' \times_k \overline{k} \rightarrow C \times_k \overline{k} = C_{\overline{k}}$. Hence we have an $\overline{k}$-isomorphism $C'_{\overline{k}} \rightarrow C' \times_k \overline{k} $. This means that $C'$ is $\overline{k}$-isomorphic to a plane curve which is defined by a single equation if we work with $\overline{k}$. I heard from other people that the non-singular model of a plane curve is not always able to be defined by a single equation. (Although some of them said this with thinking the projective case.)

My question is that: With the above assumptions, does $C'$ is $k$-isomorphism to a plane curve? If yes, how to get the equation? If not, the counter-example is appreciated. (If all the singular points are $k$-rational, then the answer is positive, and the result of the blow up process for $C_{\overline{k}}$ gives the require equation.)

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This question is considered when I considered another question mathoverflow.net/questions/66548/… –  user565739 Jun 3 '11 at 23:25
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Even if the singular points are $k$ rational your claim is not correct. The map $(x,y) \to (x,xy)$ is not finite, in fact not even surjective, so the map from the inverse image of $C$ to $C$ need not be finite either. So one only gets an open covering of $C'$ by non-singular affine plane curves. –  ulrich Jun 4 '11 at 5:15
    
Yes, you are right. Before I have done the computations for some curves, like $ y^2 - x^3$ or $ y^2 - x^2(x+1)$ and what I got is a finite morphism. For example, $z := y/x$ in the quotient field of $k[x,y]/( y^2 - x^2(x+1) )$ or $k[x,y]/(y^2-x^3)$ is an integral element. But in general, we don't get an integral element. Thank you. –  user565739 Jun 4 '11 at 9:10
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