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I just realized that one special case of Bézout theorem is very easy.

Bézout theorem: two homogeneous polynomials $P, Q \in \mathbb{C}[X, Y, Z]$ of degree $p, q$ with no common factor define two projective algebraic curves in $P_2(\mathbb{C})$ that intersect at $pq$ points counted with multiplicity.

Special case: Suppose that $P, Q$ do not depend on $Z$ (they can be considered homogeneous in $\mathbb{C}[X, Y]$). Then (as explained here) they factor as products of $p$ (resp. $q$) terms in the form $aX + bY$ (with $a,b \in \mathbb{C}$) so that the two curves are actually one union of $p$ lines, and one union of $q$ other lines that intersect $pq$ times with multiplicity (well, in this case they will all intersect at the point $[0:0:1]$, but we can extend to any homogeneous polynomial that can be expressed as a factor of linear terms).

Question: Can we go on and prove Bézout theorem starting from this special case? (something in the spirit of that proof of the genus-degree formula that starts from a singular case for instance)

Weier
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    The step where $P$ is arbitrary, but $Q$ is still a product of linear polynomials $Q_i$ is still easy: Then for each line $L = V(Q_i) \subset V(Q)$, $P$ becomes a polynomial on $L$, and hence has exactly $\deg P$ zeroes counted with multiplicity by the fundamental theorem of algebra. – red_trumpet Jun 21 '23 at 19:19

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