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\begin{document}

\noindent Math 746, Spring 2016\\
Homework 4\\
Due: April 11 (Monday)

\section{Exercises}

\begin{enumerate}[(1)]

\item Let $m,n$ be positive integers and pick $1 \le r \le \min(m,n)$. Let $R$ be a commutative ring. Set $S = R[x_{ij} \mid 1 \le i \le m,\ 1 \le j \le n]$. Let $I_r$ be the ideal generated by the $r \times r$ minors of the generic $m \times n$ matrix $X = (x_{ij})$.

  \begin{enumerate}
  \item If $R$ is a field, call a monomial ordering {\bf anti-diagonal} if the initial monomial of a minor is the product of the entries along the anti-diagonal of the submatrix. For example, the anti-diagonal monomial of $\det \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix}$ is $x_{12}x_{21}$. Show that anti-diagonal monomial orders exist.

  \item Again, assuming $R$ is a field, it is true that the $r \times r$ minors form a Gr\"obner basis for any anti-diagonal monomial ordering. Prove this in the case $r=m=2$.\footnote{If you're feeling ambitious, do the general case, though I hope this special case already illustrates some interesting points.} Conclude that ${\rm in}(I_r)$ is radical.

  \item Use the fact in (b) to show that $I_r$ is a prime ideal when $R$ is a field.

  \item Use the fact in (b) to show that $S/I_r$ is a free $R$-module (first do the case $R = \bZ$), and hence flat over $R$.
  \end{enumerate}

\bigskip

\item Let $R$ be a noetherian ring and let $M$ be an $R$-module. Let $I,J \subset R$ be ideals.
  \begin{enumerate}
  \item Show that there is a natural exact sequence
\[
0 \to \rH^0_{I+J}(M) \to \rH^0_I(M) \oplus \rH^0_J(M) \to \rH^0_{I \cap J}(M),
\]
and that the last map is surjective if $M$ is an injective module.
\item Use injective resolutions to construct the Mayer--Vietoris sequence
\[
\xymatrix{
0 \ar[r] &  \rH^0_{I+J}(M) \ar[r] &  \rH^0_I(M) \oplus \rH^0_J(M) \ar[r] &  \rH^0_{I \cap J}(M) \ar[lld] & \\
& \rH^1_{I+J}(M) \ar[r] &  \rH^1_I(M) \oplus \rH^0_J(M) \ar[r] &  \rH^1_{I \cap J}(M) \ar[r] & \cdots 
}
\]
  \end{enumerate}

\bigskip

\item Let $k$ be a field and $S = k[x_1,\dots,x_n]$. Let $I \subset S$ be an ideal. The {\bf arithmetic rank} of $I$ is the smallest $r$ such that there exist $f_1,\dots,f_r$ with $\sqrt{(f_1,\dots,f_r)} = \sqrt{I}$.

  \begin{enumerate}
  \item If $\rH_I^i(S) \ne 0$, show that the arithmetic rank of $I$ is at least $i$.
  \item Set $n=4$ and $I = (x_1,x_2) \cap (x_3,x_4) = (x_1x_3, x_1x_4, x_2x_3, x_2x_4)$. Use the Mayer--Vietoris sequence to show that $\rH_I^3(S) \ne 0$.
  \item Continue (b). Show that $I = \sqrt{(x_1x_3, x_2x_4, x_1x_4 + x_2x_3)}$ and conclude that $I$ has arithmetic rank $3$.\footnote{Give an argument for the radical equality that avoids the use of software.}
  \end{enumerate}

\end{enumerate}

\section{Further reading}

\begin{itemize}
\item The determinantal ring $S/I_r$ is Cohen--Macaulay when $R$ is Cohen--Macaulay; this essentially reduces to the case of a field, and in that case, one can prove this by showing that $S/{\rm in}(I_r)$ is Cohen--Macaulay. There are special techniques for showing that quotients by squarefree monomial ideals are Cohen--Macaulay. For example, see Theorem 5.53 of 

Ezra Miller, Bernd Sturmfels, {\it Combinatorial Commutative Algebra}, Graduate Texts in Math. {\bf 227}, Springer-Verlag, 2005.

See Theorem 16.43 for a reference specific to this example.

\item The arithmetic rank of any ideal in $k[x_1,\dots,x_n]$ with $k$ a field is always at most $n$. For a proof and more general results, see:

David Eisenbud, E. Graham Evans, Every algebraic set in $n$-space is the intersection of $n$ hypersurfaces, {\it Invent. Math.} {\bf 19} (1973), 107--112.
\end{itemize}

\end{document}
