To show that curves in $P^2(k)$ always intersect for algebraically closed field $k$ Let $k$ be an algebraically closed field. How do I show, using Hilbert Nullstellensatz, that any two curves in $P^2(k)$ intersect ? That is, let $f(X,Y,Z),\ g(X,Y,Z)\in k[X,Y,Z]$ be irreducible homogeneous polynomials. Then how to show that the ideal $\langle X,Y,Z \rangle$ is not contained in the ideal $\sqrt{\langle f , g \rangle}$ ?

Let's work with Hilbert series. The ring $R=k[X,Y,Z]$ is graded
in the usual way. If $M=\bigoplus_{n=0}^\infty M_n$ is a finitely
generated $R$-module, its Hilbert series is
$$h_M(q)=\sum_{n=0}^\infty \dim_k(M_n) q^n.$$
In particular
$$h_R(q)=\frac{1}{(1-q)^3}.$$
If $I$ is a homogeneous ideal of $R$, the projective Nullstellensatz
states that $I$ has common zeroes unless it contains $R_n$ for
all large enough $n$. In other words $I$ has common zeroes unless
$h_{R/I}(q)$ is a polynomial in $q$.
Suppose that $f$ and $g$ have no common factor, and let $I=\left<f,g\right>$. Then $I$ fits into an exact sequence
$$0\to R\to R\oplus R\to I\to 0.\tag{*}$$
In (*) the surjection is $(u,v)\mapsto uf+vg$ and the injection
is $w\mapsto (wg,-wf)$. It follows that
$$h_I(q)=\frac{q^m+q^n-q^{m+n}}{(1-q)^3}$$
where $m$ and $n$ are the degrees of $f$ and $g$.
Therefore
$$h_{R/I}(q)=\frac{1-q^m-q^n+q^{m+n}}{(1-q)^3}
=\frac{(1-q^m)(1-q^n)}{(1-q)^3}.$$
This is not a polynomial, so by the Nullstellensatz, $f$ and $g$
have a common solution.
Of course, this goes part-way to proving Bezout's theorem.
