Simple and Multiple Points of Affine Plane Curves
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# Simple and Multiple Points of Affine Plane Curves

Recall from the Affine Plane Curves page that we said that two polynomials $F, G \in K[x, y]$ are said to be equivalent if there exists a nonzero $\lambda \in K$ such that $F = \lambda G$, and we said that an affine plane curve is an equivalence class of a nonconstant polynomial in $K[x, y]$ with this equivalence relation.

We will now begin to classify special points of affine plane curves.

 Definition: Let $K$ be a field and let $F \in K[x, y]$ be an affine plane curve. Let $\mathbf{p} = (a, b) \in F$, that if, $\mathbf{p}$ is a point which lies on $F$. 1) $\mathbf{p}$ is a Simple Point if either $F_x(\mathbf{p}) \neq 0$ or $F_y(\mathbf{p}) \neq 0$. 2) $\mathbf{p}$ is a Multiple Point or Singular Point if both $F_x(\mathbf{p}) = 0$ and $F_y(\mathbf{p}) = 0$.

Here, the notation $F_x$ and $F_y$ are used to denote the partial derivatives of $F$ with respect to $x$ and $F$ with respect to $y$.

For example, consider the affine plane curve $F(x, y) = x^2 - yx$. The partial derivatives of $F$ are:

(1)
\begin{align} \quad F_x(x, y) = 2x - y \quad , \quad F_y(x, y) = x \end{align}

Observe that $F_x = 0$ whenever $y = 2x$, and $F_y = 0$ whenever $x = 0$. Therefore, every point except $(0, 0)$ is a simple point of $F$, while the point $(0, 0)$ is a singular point of $F$.

 Definition: Let $K$ be a field and let $F \in K[x, y]$ be an affine plane curve. If $\mathbf{p} = (a, b)$ is a simple point then the Tangent Line of $F$ at $\mathbf{p}$ is the line given by the equation $F_x(\mathbf{p})(x - a) + F_y(\mathbf{p})(y - b) = 0$.

For example, consider the affine plane curve $F(x, y) = y^2 - x^3 + x$. The partial derivatives of $F$ are:

(2)
\begin{align} \quad F_x(x, y) = -3x^2 + 1 \quad , \quad F_y(x, y) = 2y \end{align}

The point $\mathbf{p} = (1, 0)$ is a simple point of $F$, with:

(3)
\begin{align} \quad F_x(\mathbf{p}) = -2 \quad , \quad F_y(\mathbf{p}) = 0 \end{align}

Therefore the tangent line of $F$ at $\mathbf{p}$ is:

(4)
\begin{align} \quad -2(x - 1) + 0(y - 0) = 0 \\ \end{align}
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