Two distinct lines in $\mathbb{P}^2$ meet in exactly one point. For example, $V(x)$ and $V(y)$ meet at $[0:0:1]$. This fails in $\mathbb{A}^2$ (parallel lines!), but in $\mathbb{P}^2$ every pair of lines intersects.

If the lines are $V(x)$ and $V(x+y)$, they meet at $[0:0:1]$ with $I = 1$ (transverse).

The line $V(y)$ and the conic $V(x^2 + y^2 - z^2)$ in $\mathbb{P}^2$:

Setting $y = 0$: $x^2 = z^2$, so $[1:0:1]$ and $[1:0:-1]$ (or equivalently $[-1:0:1]$). Two intersection points, each with multiplicity 1. Total: $1 \cdot 2 = 2$. ✓

The line $V(y - z)$ and the parabola $V(y z - x^2)$ (homogenization of $y = x^2$):

Setting $y = z$: $z^2 = x^2$, so $[1:1:1]$ and $[-1:1:1]$. Wait — that's still 2 distinct points, each with $I = 1$.

Now try $V(z)$ (the line at infinity) and $V(yz - x^2)$: setting $z = 0$, $x^2 = 0$, so only $[0:1:0]$ with multiplicity 2. The line at infinity is tangent to the parabola. Total: $1 \cdot 2 = 2$. ✓

$V(x^2 + y^2 - z^2)$ and $V(x^2 - y^2 - z^2)$ in $\mathbb{P}^2$:

Adding: $2x^2 - 2z^2 = 0$, so $x = \pm z$. Substituting $x = z$: $z^2 + y^2 - z^2 = y^2 = 0$, giving $[1:0:1]$. Substituting $x = -z$: similarly $[1:0:-1]$.

Wait — only 2 points? Each must have multiplicity 2. Indeed, at $[1:0:1]$, locally set $z = 1$: curves are $x^2 + y^2 - 1$ and $x^2 - y^2 - 1$. At $(1, 0)$: the Jacobian rows are $(2, 0)$ and $(2, 0)$ — same tangent direction! So $I = 2$ at each point. Total: $2 + 2 = 4 = 2 \cdot 2$. ✓

A generic line meets a smooth cubic curve in 3 points. For the cubic $V(y^2 z - x^3 + xz^2)$ (an elliptic curve), the line $V(y)$ gives:

$0 = -x^3 + xz^2 = x(z^2 - x^2) = x(z-x)(z+x)$

Three points: $[0:0:1]$, $[1:0:1]$, $[-1:0:1]$, each with $I = 1$. Total: $3 = 1 \cdot 3$. ✓

This is the geometric basis for the group law on elliptic curves: three collinear points sum to zero.

The inflection point of a cubic: at a flex point $P$, the tangent line meets the curve with multiplicity 3 (all of the "budget" $1 \cdot 3 = 3$ is concentrated at one point).

For $V(y^2z - x^3)$ (the cuspidal cubic), the tangent at the cusp $[0:0:1]$ is $V(y)$. Setting $y = 0$: $0 = -x^3$, so $[0:0:1]$ with multiplicity 3. This is an extreme case where the tangent line has maximal contact.

Two general cubics in $\mathbb{P}^2$ meet in $3 \cdot 3 = 9$ points. The Cayley–Bacharach theorem states:

If two cubics $C_1, C_2$ meet in 9 points, and a third cubic $C_3$ passes through 8 of them, then $C_3$ also passes through the 9th.

This is the geometric heart of the group law on elliptic curves: if $E$ is a smooth cubic (an elliptic curve) and $L_1, L_2$ are two lines, then $V(L_1 L_2) = V(L_1) \cup V(L_2)$ is a (degenerate) cubic meeting $E$ in 6 points. Cayley–Bacharach applied to these 6 points (via a suitable third cubic) yields the associativity of the group law.

In $\mathbb{A}^2$: the circle $x^2 + y^2 = 1$ and the line $x = 2$.

$4 + y^2 = 1 \Rightarrow y^2 = -3$: no real solutions! Over $\mathbb{C}$: $y = \pm i\sqrt{3}$, so 2 intersection points. Bézout gives $1 \cdot 2 = 2$. ✓ (over $\mathbb{C}$).

Now consider $x^2 + y^2 = 1$ and $x = 0$ (the $y$-axis). Two intersection points $(0, \pm 1)$. All accounted for.

But $x^2 + y^2 = 1$ and the "line at infinity": homogenize to $X^2 + Y^2 - Z^2$ and $Z = 0$. Setting $Z = 0$: $X^2 + Y^2 = 0$, giving $[1:i:0]$ and $[1:-i:0]$ over $\mathbb{C}$. These are the circular points at infinity — they lie on every circle!

For a smooth projective curve $C \subseteq \mathbb{P}^2$ of degree $d$, the genus is

$g = \frac{(d-1)(d-2)}{2}.$

This follows from Bézout + the adjunction formula. Examples:

| Degree $d$ | Genus $g$ | Type | |---|---|---| | 1 | 0 | Line ($\cong \mathbb{P}^1$) | | 2 | 0 | Conic ($\cong \mathbb{P}^1$) | | 3 | 1 | Cubic (elliptic curve) | | 4 | 3 | Quartic | | 5 | 6 | Quintic | | $d$ | $\frac{(d-1)(d-2)}{2}$ | General |

The genus measures the "complexity" of the curve. Curves of genus 0 are rational (birational to $\mathbb{P}^1$). Curves of genus 1 are elliptic. Curves of genus $\geq 2$ have finite automorphism groups (Hurwitz).

Three generic quadric surfaces in $\mathbb{P}^3$ meet in $2 \cdot 2 \cdot 2 = 8$ points. For example:

$V(x_0^2 - x_1^2), \quad V(x_0^2 - x_2^2), \quad V(x_0^2 - x_3^2)$

give $x_0 = \pm x_1 = \pm x_2 = \pm x_3$, i.e., 8 points $[\pm 1 : \pm 1 : \pm 1 : \pm 1]$ (up to overall scaling: $[1:\pm 1:\pm 1:\pm 1]$, which is 8 points). Total: 8. ✓