Is there an easy proof of the Nullstellensatz that avoids the standard Noether-normalization techniques?
One proof I know proves first the 'weak' Nullstellensatz which ensures that maximal ideals correspond to points (using normalization), and then the stronger version by introducing another variable and the Rabinovich trick.
Best Answer
There is a cheap proof of the weak Nullstellensatz in Artin's Algebra which goes like this: suppose $m$ is a maximal ideal of $F[x_1, \dots x_n]$ where $F$ is an uncountable algebraically closed field. Then $F[x_1, \dots x_n]/m$ is a field extension of $F$ which is either $F$ itself or transcendental. In the first case, $m$ is of the form $(x_i - a_i)$ for some $a_i$. In the second case, a transcendental extension of $F$ must have uncountable dimension, since if $t$ is transcendental then $\{ \frac{1}{t - a} : a \in k \}$ is linearly independent. But $F[x_1, \dots x_n]/m$ has at most countable dimension over $F$. Hence the second case does not occur.
Edit, 2/28/10: See also question #15611, in which Brian Conrad gives an argument that one can reduce the general case to the uncountable case.