Algebraic geometry

It gives an exact dictionary between algebra and geometry: radical ideals of the polynomial ring correspond bijectively to affine varieties, so questions about shapes can be answered by pure ring computations, and vice versa. Without the hypothesis that $k$ is algebraically closed the dictionary breaks (over the reals, $x^2+1$ generates a nontrivial ideal with empty zero set).

We sketch the weak form first: if $J$ is a proper ideal of $k[x_1,\ldots,x_n]$ then $V(J)$ is nonempty. Pick a maximal ideal $\mathfrak{m}$ containing $J$; the quotient $k[x_1,\ldots,x_n]/\mathfrak{m}$ is a field, and it is a finitely generated $k$-algebra. Zariski's lemma says any such field extension of $k$ is a finite algebraic extension of $k$; since $k$ is algebraically closed, this forces $k[x_1,\ldots,x_n]/\mathfrak{m} = k$. The images of $x_1,\ldots,x_n$ under this quotient map give a point $a \in k^n$ with $g(a)=0$ for every $g \in J$, so $a \in V(J)$.

For the strong form $I(V(J)) = \sqrt{J}$, the inclusion $\sqrt{J} \subseteq I(V(J))$ is immediate: if $f^r \in J$ then $f$ vanishes wherever every element of $J$ vanishes. The reverse inclusion uses the Rabinowitsch trick: to show $f \in I(V(J))$ implies $f \in \sqrt{J}$, introduce a new variable $y$ and consider the ideal generated by $J$ together with $1 - yf$ in $k[x_1,\ldots,x_n,y]$.

This enlarged ideal has empty zero set (any common zero would need every generator of $J$ to vanish, forcing $f$ to vanish there too, which contradicts $1-yf=0$). By the weak Nullstellensatz just proved, the enlarged ideal must be the whole ring, so $1$ is a polynomial combination of the generators. Substituting $y = 1/f$ and clearing denominators produces an explicit expression showing some power $f^r$ lies in $J$, which is exactly $f \in \sqrt{J}$.

In the ordinary real plane, two curves can miss each other when roots become complex or when intersection points escape to infinity. Working over an algebraically closed field $k$ in projective space $\mathbb{P}^2(k)$ and counting tangencies with their natural algebraic multiplicity restores complete uniformity: the intersection count depends only on the degrees $\deg(C_1)$ and $\deg(C_2)$.

Let $F(x,y,z)$ and $G(x,y,z)$ be the homogeneous polynomials of degrees $d_1 = \deg(C_1)$ and $d_2 = \deg(C_2)$ defining $C_1$ and $C_2$. Choose projective coordinates so that the point $[0:1:0]$ lies on neither curve and so that no two intersection points share the same $[x:z]$-line through $[0:1:0]$.

Regard $F$ and $G$ as polynomials in the single variable $y$ with coefficients that are homogeneous polynomials in $(x,z)$. Their Sylvester resultant $R(x,z) = \operatorname{Res}_y(F, G)$ is a nonzero homogeneous polynomial in $(x,z)$ (nonzero because $F$ and $G$ share no common factor), and homogeneity calculation on the Sylvester matrix shows that $R(x,z)$ has degree exactly $d_1 d_2$.

Over the algebraically closed field $k$, any homogeneous polynomial in two variables of degree $d_1 d_2$ factors completely into $d_1 d_2$ linear forms, counted with multiplicity. Each linear factor corresponds to a line through $[0:1:0]$ containing a common zero of $F$ and $G$, and the multiplicity of the factor matches the local intersection multiplicity $\dim_k \mathcal{O}_{P}(C_1 \cap C_2)$ at that point, giving the total $\left| C_1 \cap C_2 \right| = \deg(C_1) \cdot \deg(C_2)$.