I would mention B\'ezout's Bézout's theorem. Forgetting the complicated general definition of the intersection index one of the consequences is: whenever two curves in $\mathbf{P}^2(K), K$ an algebraically closed field have no common component, the number of the intersection points is finite and is always at most the product of the degrees. Moreover, it is the product of the degrees provided no intersection point is a singular point and all intersections are transversal. To state this over $\mathbf{C}$ we essentially only need multivariable calculus. But a proof requires a bit of algebraic geometry.
I would mention B\'ezout's theorem. Forgetting the complicated general definition of the intersection index one of the consequences is: whenever two curves in $\mathbf{P}^2(K), K$ an algebraically closed field have no common component, the number of the intersection points is finite and is always at most the product of the degrees. Moreover, it is the product of the degrees provided no intersection point is a singular point and all intersections are transversal. To state this over $\mathbf{C}$ we essentially only need multivariable calculus. But a proof requires a bit of algebraic geometry.