TITLE: Equivalence of weak forms of Hilbert's Nullstellensatz
QUESTION [7 upvotes]: The version of the Nullstellensatz with which I am familiar states that if $K$ is an algebraically closed field, and $f_1,\dots,f_n\in K[X_1,\dots,X_m]$, then the family $\{f_i\}$ has a common zero iff $\langle f_1,\dots,f_n\rangle\neq K[X_1,\dots,X_m]$.
However, another form I have heard of states that if $K$ is algebraically closed, then the maximal ideals of $K[X_1,\dots,X_m]$ are precisely those of form $(X_1-a_1,\dots,X_m-a_m)$ for some $a_i\in K$. 
Can anybody explain how the first form implies the second?

REPLY [6 votes]: Let $\mathfrak m$ be a maximal ideal.  By the Hilbert Basis Theorem,
it is generated by some $f_1,\ldots,f_n$.  By the first form of the Nullstellensatz that you cite,
they have a common zero, say $(a_1,\ldots,a_m)$.
Then the polynomials $f_1,\ldots,f_n,x_1-a_1,\ldots,x_m - a_m$ have a common zero (namely the same point $(a_1,\ldots,a_n)$), and so they generate a proper ideal,
which contains $\mathfrak m = (f_1,\ldots,f_n)$.  Thus it must equal $\mathfrak m$ (as $\mathfrak m$ is maximal), and
so $\mathfrak m  \supseteq (x_1-a_1,\ldots,x_m-a_m)$.  Since $(x_1-a_1,\ldots,x_m-a_m)$
is clearly maximal, this inclusion is an equality.  QED