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.)
