On prime ideals of $K[X, Y, Z]$ Let $R=K[X,Y,Z]$ be the polynomial ring in variables $X, Y, Z$ over a field $K$. Why does $R$ have infinite prime ideals?
 A: Note that it is enough to prove that $K[X,Y,Z]$ has infinitely many maximal ideals, because maximal ideals are prime.
Now, A theorem:

Let $K[x]$ be a polynomial ring over $K$, and let $J(K[x])$ be it's Jacobson radical. Then $K[x]/J(K[x])$ is never an Artinian ring.

Proof: If we let $I=\langle x + J(K[x])\rangle$, then I claim $I^k \neq I^{k+1}$ for all $k$. If this is true for some $k$, then $x^k + J(K[x]) = f(x)x^{k+1} + J(K[x])$ for some $f \in K[x]$. Hence $ x^k -f(x)x^{k+1} \in J(K[x])$ and so $g(x) = 1- x^k + f(x)x^{k+1}$ is nilpotent. But the coefficient of $x^k$ in $g(x)$ is $-1$, which is not a nilpotent element. Hence, this is a contradiction, hence there is a strictly descending chain of ideals which is never stationary : $I \supset I^2 \supset I^3 \supset ...$. Hence the ring  $K[x]/J(K[x])$ is not an Artinian ring. 
Finally, we finish it with:

Theorem: $K[x]/J(K[x])$ has infinitely many maximal ideals.

Proof: If not, then let $m_1, m_2, ... , m_i$ be the maximal ideals of $K[x]/J(K[x])$. Then by the Chinese remainder theorem, $K[x]/J(K[x]) = \bigoplus_{i} \dfrac{(K[x]/J(K[x]))}{m_i}$. Since each $\dfrac{(K[x]/J(K[x]))}{m_i}$ is a field and there are only two ideals in a field, it follows that there are atmost $2^n$ ideals in this ring. But then the ring would be Artinian, which gives a contradiction.
Hence, it follows that $K[x]$ has infinitely many maximal, hence prime ideals. Furthermore, this proof extends to $K[x,y,z]$, which is easy to see.
A: Let $R$ be a domain. The kernel $p_r$ of the ring homomorphism $f_r:R[X]\rightarrow R$, $f(X):=r$, is a prime ideal containing the polynomial $X-r$. If $r,s\in R$ such that $p_r=p_s$ one gets $X-s\in p_r$, hence $r-s=0$. Consequently, if $R$ is infinite then $R[X]$ possesses infinitely many prime ideals. The domain $K[X,Y]$ is infinite.
