Direct proof of classification of $\textrm{Spec}(k[X])$, $k$ an algebraically closed field Let $k$ be an algebraically closed field.  Combining the Nullstellensatz, the fact that $\textrm{Dim } k[X,Y] = 2$, and Krull's height theorem, one can show that the prime ideals of $k[X,Y]$ consist of $(0), (X-a,Y-b)$, or $(f)$ for $f \in k[X,Y]$ irreducible.  How can this be proved without assuming these results?
 A: Let $(f)$ be a nonzero principal prime ideal of $R = k[X,Y]$.  Since $R$ is a unique factorization domain, $f$ is irreducible if and only if it is prime, and in any integral domain, a principal ideal is prime if and only if some (equivalently each) generator is prime.  
Let $P$ be a prime ideal of $R$ which is not principal.   It is easy to see that any set of generators for $P$ can be replaced by a set of irreducible generators.  In particular, there exist $f, g \in P$ which are irreducible and nonassociate.  

(Gauss's Lemma): Let $S$ be a unique factorization domain with quotient field $K$.  Let $a \in S[X]$ be nonconstant.  Then $a$ is irreducible in $S[X]$ if and only if it is irreducible in $K[X]$.

By Gauss's Lemma, $f$ and $g$, which are irreducible in $k[X,Y] = k[X][Y]$, remain irreducible in $k(X)[Y]$, and they are obviously still non associate here: otherwise, $f(X,Y) = \frac{h_1(X)}{h_2(X)} g(X,Y)$ for some nonzero $h_i \in k[X]$, or $h_2(X)f(X,Y) = h_1(X)g(X,Y)$.  The irreducible factors of $h_i$ remain irreducible in $k[X,Y]$.  Writing $h_1$ and $h_2$ as products of irreducibles, we contradict the unique factorization of $k[X,Y]$.
So there exist $m_1, m_2 \in k(X)[Y]$ such that $m_1(X)f(X,Y) + m_2(X)g(X,Y) = 1$.  Clearing denominators, we obtain an $h \in k[X]$ which lies in $P$.  There then exists an irreducible factor $(X-a)$ of $h$ which lies in $P$.
By the same argument, there exists a $b \in k$ such that $(Y-b) \subseteq P$.  Then $(X-a,Y-b) \subseteq P$, hence $(X-a,Y-b) = P$.
