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\begin{document}
\centerline{Math 361 lecture for Wednesday, Week 5}
\bigskip

\centerline{Hilbert basis theorem}
\bigskip

As a motivation question, let~$t$ be an indeterminate, and consider the ideal
\[
  I=\{f\in \R[x,y,z]: f(t,t^{2}, t^{3})=0\}.
\]
Examples of elements of~$I$ include $x^{i}y^{j}-z^{k}$ for all $(i,j,k)\in\N$
such that $i+2j=3k$, i.e., $x^{3k-2j}y^{j}-z^{k}$ where $3k\geq 2j$.  For
example, $xy-z, x-z^3$, etc. Similarly, $I$ includes all~$x^{i}z^{j}-y^{k}$ such
that~$(i,j,k)\in\N$ and $i+3k=2j$.  For example, $x-y^{2}\in I$  It also includes all $\C[x,y,z$-linear
combinations of these elements.
\medskip 

\textbf{Question.} Is this ideal finitely generated?
\bigskip

\begin{thm} (Hilbert basis theorem.)
  Let~$R$ and~$S$ be rings with~$R\subseteq S$.  Suppose that~$S$ is finitely
  generated as a ring over~$R$ and~$R$ is Noetherian.  Then~$S$ is a Noetherian
  ring.
\end{thm}
\medskip

\begin{remark}\
  \begin{itemize}
    \item For~$S$ to be \emph{finitely generated as a ring} over~$R$, we mean
      that there exist $s_{1},\ldots,s_{n}\in S$ such that
      $S=\{f(s_{1},\ldots,s_{n}): f\in R[x_{1},\ldots,x_{n}]\}$.
    \item Recall that to say a ring~$R$ is Noetherian, we consider~$R$ as
      an~$R$-module.  The submodules of~$R$ are exactly the ideals in~$R$. Thus,
      $R$ is Noetherian if all of its ideals are finitely generated.
    \item Letting~$R=K$ be a field and $S=K[x_{1},\ldots,x_{n}]$, the Hilbert
      basis theorem says that every ideal a polynomial ring with coefficients in
      a field is finitely generated.  For instance, in the example above, we
      have $I=(y^{2}-xz, xy-z, x^{2}-y)$.   [Aside: Consider the curve in $\R^{3}$ 
      parametrized by $t\mapsto (t,t^{2},t^{3})$.  This curve is the
      intersection of the two surfaces, $S=\left\{(x,y,z)\in\R: y=x^{2}\right\}$
      and math $T=\left\{(x,y,z)\in\R: z=x^{3}\right\}$.  One might think that~$I$ would generated by just the two equations $y-x^{2}$ and
      $z-x^{3}$, but it turns out that a minimum of three generators is required.
  \end{itemize}
\end{remark}
\bigskip

\begin{proof}[Proof of Hilbert basis theorem]
  Suppose that $s_{1},\ldots,s_{n}$ generate~$S$ as a ring over~$R$. Then
  there exists an $R$-module surjection
  \begin{align*}
    R[x_{1},\ldots,x_{n}]&\to S\\
    f(x_{1},\ldots,x_{n})&\mapsto f(s_{1},\ldots,s_{n}).
  \end{align*}
  Since the image of a Noetherian $R$-module is Noetherian, it suffices to prove
  that the polynomial ring~$R[x_{1},\ldots,x_{n}]$ is a Noetherian $R$-module.
  Since~$R[x_{1},\ldots,x_{n}]=R[x_{1},\ldots,x_{n-1}][x_{n}]$, by induction, it
  suffices to show that the polynomial ring $R[x]$ is Noetherian.

  So we need to show that~$R[x]$ is Noetherian.   Let~$I$ be an ideal in~$R[x]$.
  We must show that~$I$ is finitely generated, i.e,.~$I=(f_{1},\ldots,f_{k})$
  for some polynomials $f_{1},\ldots,f_{k}\in R[x]$ for some~$k$.\footnote{If
  $R$ is a field, we would be done since $R[x]$ would then be a PID by the
  division algorithm.  However, we only know that~$R$ is Noetherian.}  Given any
  polynomial~$f=\sum_{i=1}^{d} a_i x^{i}\in R[x]$ with $a_{d}\neq0$, we
  call~$a_{d}$ the \emph{leading coefficient of~$f$}. For completeness, we say
  that zero polynomial has leading coefficient~0.  Let~$A$ be the collection
  of all leading coefficients of elements of $I$.  We claim that~$A$ is an ideal
  in~$R$.  We must show that~$A$ is nonempty, closed under addition, and closed
  under ``in-out'' multiplication.
  Since~$I$ is an ideal, it contains the
  zero polynomial, whose leading coefficient is~$0\in R$.  Thus, $0\in A$.
  Next, suppose that~$a,b\in A$.  Then there exist $f,g\in R[x]$ with leading
  terms $a,b$, respectively.  Without loss of generality, say
  $\deg(f)\leq\deg(g)$, and let~$j=\deg(g)-\deg(f)$.  Define~$h:=x^{j}f$.  Then
  $h\in I$, the leading term of~$h$ is~$a$, and $\deg(h)=\deg(g)$.  Then $h+g\in
  I$ and the leading term of~$h+g$ is $a+b$.  We have shown that~$A$ is closed
  under addition.  Finally, suppose that~$f\in I$ with leading coefficient~$a$
  and~$r\in R$.  Then~$ra\in A$ since it is the leading coefficient of~$rf\in
  I$.  This proves the claim.

  Since~$R$ is Noetherian, and $A$ is an ideal in~$R$, it follows
  that~$A=(a_{1},\ldots,a_{s})$ for some~$a_{i}\in R$ and some~$s\geq1$.
  For each~$i$, take $g_{i}\in I$ with leading coefficient~$a_{i}$.  By
  multiplying by appropriate powers of~$x$, we may assume that all of
  the~$g_{i}$ had the same degree~$d$.  Let~$I_{<d}\subset R[x]$ be the set of
  all elements of $I$ that have degree strictly less than~$d$.  Then~$I_{<d}$ is
  an $R$-module (but not an ideal in~$R[x]$).  Let $M$ be the $R$-submodule of
  $R[x]$ generated by~$1,x,\ldots,x^{d-1}$.  Since~$R$ is Noetherian and~$M$ is
  a finitely generated~$R$-module,~$M$ is Noetherian.   Since~$I_{<d}$ is a
  submodule of~$M$, it follows that~$I_{<d}$ is a finitely
  generated~$R$-module.  Say $h_{1},\ldots,h_{t}$ are generators.  So every
  element of~$f\in I_{<d}$ may be written~$f=\sum_{i=1}^{t} r_{i}h_{i}$ for
  some~$r_{i}\in R$. 
  \medskip

  We finish by showing that~$I$ is generated by $s+t$ elements:
  \[
    I = (g_{1},\ldots,g_{s},h_{1},\ldots,h_{t}).
  \]
  First note that the $g_{i}$ and $h_{j}$ are elements of~$I$.  Then take $f\in I$.
  We will prove by induction on the degree of~$f$ that~$f$ is an $R[x]$-linear
  combination of the $g_{i}$ and~$h_{j}$.  If $\deg(f)<d$, then $f\in I_{<d}$,
  an~$R$-submodule of~$R[x]$ generated by the~$h_{i}$.  So in that case,
  $f=\sum_{i=1}^{t} r_{i}h_{i}$ for some~$r_{i}\in R$, and we are done.

  Next, suppose that~$\deg(f)=e\geq d$.  Say~$a\in R$ is the leading coefficient
  of~$f$.  So $a\in A$, and it follows that~$a=\sum_{i=}^{s} r_{i}a_{i}$ for
  some~$r_{i}\in R$.  Recall that all~$g_{i}$ have degree~$d$.  Therefore,
  \[
    f-\sum_{i=1}^{s} r_{i}x^{e-d}g_{i}
  \]
  is an element of~$I$ having degree strictly less than~$e$.  By  induction,
  $f-\sum_{i=1}^{s} r_{i}x^{e-d}g_{i}$ is an~$R[x]$-linear combination of
  the~$g_{i}$ and~$h_{j}$.  Hence, so is~$f$.
\end{proof}
\end{document}
