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\begin{document}
\title{Hilbert's Nullstellensatz}
\author{Daniel Allcock}
\address{Department of Mathematics, University of Texas, Austin}
\email{allcock@math.utexas.edu}
\urladdr{http://www.math.utexas.edu/\textasciitilde allcock}
\date{January 9, 2005}
\thanks{Partly supported by NSF grants DMS-0245120 and DMS-0231585.}
\maketitle

\noindent
This is the simplest proof of  the Nullstellensatz that I have been
able to come up with.   It is meant for students learning commutative
algebra for the first time---students perhaps lost in the sea of new
vocabulary, with no clear guidance about which concepts are
all-important (e.g., Noetherianness, and integrality and finiteness of
ring extensions) and which are less so.  Accordingly, we use nothing
beyond unique factorization in one-variable polynomial rings and the
basics of field extensions.  

Dan Bernstein led me to some references, and it turns out that my
proof is the same in its essentials as one by Zariski \cite{zariski}.
Zariski's proof led to the definition of a class of rings called
either Jacobson rings or Hilbert rings, which are defined
as ``the class of rings to which this argument applies''; see
\cite{eisenbud} for a discussion.  Also, our arguments about
denominators motivate the definition of a finite extension of rings,
although we avoid using this language explicitly.

I am also grateful to Keith Conrad for his helpful comments.

\medskip\noindent
{\bf Theorem.}\kern1em
{\it Let $k$ be a field and $K$ a field extension which is finitely
  generated as a $k$-algebra.  Then $K$ is algebraic over $k$.}

\medskip\noindent
{\it Example of Proof.} Suppose $k$ is infinite and $K$ is the simple
transcendental extension $k(x)$.  We claim that if $f_1,\dots,f_m\in
K$, then the $k$-algebra $A$ they generate is smaller than $K$.  To see
this, choose $c\in k$ away from the poles of the rational
functions $f_i$.  Then no element of $A$ can have a pole at $c$, so
$1/(x-c)$ is not in $A$, and $A$ is smaller than $K$.  Embellishing 
this argument yields the full proof:

\begin{proof}
  We will assume throughout that $K$ is transcendental over $k$ and
  finitely generated as a $k$-algebra, and deduce that $K$ is not
  finitely generated as a $k$-algebra, a contradiction.
  
  Suppose first that $K$ has transcendence degree one; this means that
  it contains a subfield $k(x)$ which is a copy of the one-variable
  rational function field, and that $K$ is algebraic over $k(x)$.
  This, together with the finite generation of $K$, shows that $K$ has
  finite dimension as a $k(x)$-vector space. Choose a basis
  $e_1,\dots,e_\ell$ and write down the multiplication table for
  $K$:
$$e_ie_j=\sum_k\frac{a_{ijk}(x)}{b_{ijk}(x)}e_k\;,$$
with  the $a$'s and $b$'s in $k[x]$.  
We will show that
for any $f_1,\dots,f_m\in K$, the $k$-algebra $A$ they generate is
smaller than $K$.   It is convenient to adjoin $f_0=1$ as a 
generator.  Express  $f_0,\dots,f_m$ in terms of our basis:
$$f_i=\sum_j\frac{c_{ij}(x)}{d_{ij}(x)}e_j\;,$$ with the $c$'s and
$d$'s in $k[x]$.  Now, an element $a$ of $A$ is a $k$-linear
combination of $f_0=1$ and products of $f_1,\dots,f_m$.  Expanding in
terms of our basis, we see that $a$ is a $k(x)$-linear combination of
products of the $e_i$, with the special property that the denominators
of the coefficients involve only the $d$'s.  Using the multiplication
table repeatedly, we see that $a$ is a $k(x)$-linear combination of
the $e_i$, whose coefficients' denominators involve only the $b$'s and
$d$'s.  A precise way to state the result of this argument is: when
$a$ is expressed as a $k(x)$-linear combination of the $e_i$, with
every coefficient in lowest terms, then all its coefficients'
denominators' irreducible factors are among the irreducible factors of
the $b$'s and $d$'s.  Therefore
$$
\frac{1}{\hbox{some other irreducible polynomial}}\kern1em\hbox{cannot lie
  in $A$},$$
and $A$ is smaller than $K$.  

This argument requires the existence of infinitely many irreducible
polynomials
in $k[x]$; to prove this one can mimic Euclid's proof of the
infinitude of primes in $\Z$.  
(If $k$ is infinite then one can just take the infinitely many linear
polynomials $x-c$, $c\in k$.)

Now suppose $K$ has transcendence degree~$>1$ over $k$, and
choose a subextension $k'$ over which $K$ has transcendence degree~$1$.  By the above, $K$ is not finitely generated as a $k'$-algebra,
so it isn't as a $k$-algebra either.  (To build $k'$ explicitly,
choose $k$-algebra generators $x_1,\dots,x_n$ for $K$ over $k$ and set
$k'=k(x_1,\dots,x_{\ell-1})$, where $x_\ell$ is the last of the $x$'s which is
transcendental over the field generated by its predecessors.)
\end{proof}

\medskip\noindent
{\bf `Weak' Nullstellensatz.}\kern1em
{\it Let $k$ be an algebraically closed field.  Then every maximal
  ideal in the polynomial ring $R=k[x_1,\dots,x_n]$ has the form
  $(x_1-a_1,\dots,x_n-a_n)$ for some $a_1,\dots,a_n\in k$.  As a
  consequence, a family of polynomial functions on $k^n$ with no
  common zeros generates the unit ideal of $R$.}

\begin{proof}
If $\m$ is a maximal ideal of $R$ then $R/\m$ is a field which is
finitely generated as a $k$-algebra.  By the previous theorem it is an
algebraic extension of $k$, hence equal to $k$.  Therefore each $x_i$
maps to some $a_i\in k$ under the natural map $R\to R/\m=k$, so
$\m$ contains the ideal $(x_1-a_1,\dots,x_n-a_n)$.  This is a maximal
ideal, so it equals $\m$.  To see the second statement, consider the
ideal generated by some given polynomial functions with no common
zeros.  If it lay in some maximal ideal, say
$(x_1-a_1,\dots,x_n-a_n)$, then all the functions would vanish at
$(a_1,\dots,a_n)\in k^n$, contrary to hypothesis.  Since it doesn't
lie in any maximal ideal, it must be all of $R$.
\end{proof}

\noindent
{\bf Nullstellensatz.}\kern1em
{\it Suppose $k$ is an algebraically closed field and $g$ and $f_1,\dots,f_m$
  are members of $R=k[x_1,\dots,x_n]$, regarded as polynomial
  functions on $k^n$.  If $g$ vanishes on the common zero-locus of the
  $f_i$, then some power of $g$ lies in the ideal  they
  generate.} 

\begin{proof}
Probably no one will ever improve on the trick of Rabinowitsch: the
polynomials $f_1,\dots,f_m$ and $x_{n+1}g-1$ have no common zeros in
$k^{n+1}$, so by the weak Nullstellensatz we can write
$$1=p_1f_1+\dots+p_mf_m+p_{m+1}\cdot(x_{n+1}g-1)\;,$$
where the $p$'s are polynomials in $x_1,\dots,x_{n+1}$.  Taking the
image of this equation under the homomorphism $k[x_1,\dots,x_{n+1}]\to
k(x_1,\dots,x_n)$ given by $x_{n+1}\mapsto 1/g$, we find
$$1=p_1\!\left(x_1,\dots,x_n,\frac{1}{g}\right)f_1+\dots+p_m\!\left(x_1,\dots,x_n,\frac{1}{g}\right)f_m\;.$$
After multiplying through by a power of $g$ to clear denominators, we
have  Hilbert's theorem.
\end{proof}

\begin{thebibliography}{9}

\bibitem{eisenbud}
D. Eisenbud, {\it Commutative Algebra with a View Toward Algebraic
  Geometry}, Springer-Verlag 1995.

\bibitem{zariski} 
O.~Zariski, A new proof of Hilbert's Nullstellensatz, {\it B.A.M.S.}
{\bf 53} (1947) 362--368.
       
\end{thebibliography}

\end{document}