Why doesn't this argument prove every nonsingular curve embeds into the projective plane?

Question

It is a statement that some curves can never be embedded in the plane, but only in projective 3-space. Why does the following argument not prove that every nonsingular curve can be embedded in the projective plane, though?

Let $X$ be a nonsingular complete curve over an algebraically closed field $k$, with function field $F.$ We prove that $X$ can be embedded in $\mathbb{P}^2_k.$

1. $F$ has transcendence degree 1 over $k,$ and so we may find some $x \in F$ so that $F/k(x)$ is an algebraic extension. In fact, since $F$ is finite type over $k,$ $F/k(x)$ is even a finite extension.
2. As $F/k(x)$ is a finite extension, and as $k(x)$ is separable, we deduce that $F/k(x)$ is a finite separable extension. The primitive element theorem then tells us that $F = k(x, y)$ for some $y \in F.$
3. Let $P \in k(x)[T]$ denote the monic minimal polynomial of $y$ over $k(x).$ Then $P(y) = 0$ looks like some polynomial equation $$y^n + f_{n-1}(x)y^{n-1} + \cdots + f_0(x) = 0,$$ where each $f_i\in k(x).$ Clearing denominators, we find that there is some polynomial $Q \in k[T_1, T_2]$ so that $Q(x,y) = 0.$
4. Let $X'$ denote the subscheme of $\mathbb{P}^2$ cut out by $Q$. In the standard affine chart, $X'$ looks like $\operatorname{Spec} k[T_1, T_2]/(Q).$ The fraction field of the ring $k[T_1, T_2]/(Q)$ is just $F.$
5. Thus $X'$ is a projective curve with fraction field $F,$ implying it is isomorphic to $X.$

Where does this argument go wrong? I can think of two possible problems.

1. In step 4, is it possible that $X'$ is a singular curve?
2. Also in step 4, is it possible that $F$ is not isomorphic to the field of fractions of $k[T_1, T_2]/(Q)?$

I don't have any indication of why either 1 or 2 would fail; maybe someone could clarify if these are the problem, or if there was something else I missed?

Answer 1

As in Cranium Clamp's comment, $X'$ may be singular. Here is an example:

Let $X=\mathbb P^1$, so $F=k(t)$. Following your steps:

1. Let $x=t^2\in F$, so $F/k(t^2)$ is a degree $2$
2. Let $y=t^3$
3. $P=T^2-t^3\in F[T]$. Thus you are looking at the curve $P(y)=y^2-x^3=0$. Moreover, $P=Q$, and $X'=\{[\alpha:\beta:\gamma]\in\mathbb P^2:\beta^2\gamma-\alpha^3=0\}$.
4. On the standard chart, $X'\backslash\{\gamma=0\}=\mathrm{Spec}\ k[x,y]/(y^2-x^3)$.

However, since both partial derivatives of $y^2-x^3$ disappears at $(x,y)=(0,0)$, the curve $X'$ is singular.