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\begin{document}

\begin{center}
\textbf{Spring 2022, Math 621:\ Week 1 Problem Set} \\
\textbf{Due:\ Thurday, February 3rd, 2022} \\
\textbf{Numerical and Affine Semigroups}
\end{center}

\begin{center}
\begin{tabular}{r@{\,\,\,}p{4in}}
Name: & 
% INSERT NAME HERE
X
\\
Collaborators: & 
% LIST COLLABORATORS HERE
X, X, X
\end{tabular}
\end{center}


\bigskip
\noindent
\textbf{Homework problems.}
You must submit \emph{all} homework problems in order to receive full credit.  

\begin{enumerate}[(H1)]

\item 
Fix a commutative semigroup $(S, +)$.  
A relation $\til$ on $S$ is a \emph{congruence} if (i)~$\til$ is an equivalence relation, and (ii) $\til$ is closed under \emph{translation} (i.e., $a \sim b$ implies $a + c \sim b + c$ for all $a, b, c \in S$).  

\begin{enumerate}[(a)]
\item 
Given a collection $\til_1, \ldots, \til_r$ of congruences on $S$, define the \emph{common refinement} $\til$ by $a \sim b$ whenever $a \sim_i b$ for all $i$ (we often write $\til = \bigcap_i \til_i$).  Prove that $\til$ is a congruence on $S$.  

\begin{proof}

\end{proof}


\item 
Prove that if $\til$ is a congruence on $S$, then the set $S/\til$ of equivalence classes of $\til$ is a semigroup under the operation $[a] + [b] = [a + b]$.  Be sure to prove well-definedness!  

\begin{proof}

\end{proof}


\item 
Given a semigroup homomorphism $\varphi:S \to T$, the \emph{kernel} of $\varphi$, denoted $\ker\varphi$, is the relation $\til$ on $S$ setting $a \sim b$ whenever $\varphi(a) = \varphi(b)$.  
Prove $\ker\varphi$ is a congruence on~$S$.  

\begin{proof}

\end{proof}


\item 
State and prove a version of the first isomorphism theorem for semigroups.  

\begin{proof}

\end{proof}

\end{enumerate}


\item 
Fix a field $\kk$, and let $R = \kk[x_1, \ldots, x_k]$.  In what follows, for $a \in \ZZ_{\ge 0}^k$, we use the shorthand 
$$x^a = x_1^{a_1} \cdots x_k^{a_k}.$$
A \emph{(unital) binomial} is a polynomial of the form $x^a - x^b \in R$ for some $a, b \in \ZZ_{\ge 0}^k$.  

\begin{enumerate}[(a)]
\item 
Fix an ideal $I \subset R$.  Define a relation $\til_I$ on $\ZZ_{\ge 0}^k$ by
$$a \sim_I b \qquad \text{whenever} \qquad x^a - x^b \in I$$
for $a, b \in \ZZ_{\ge 0}^k$.  
Prove that $\til_I$ is a congruence on $\ZZ_{\ge 0}^k$.  

\begin{proof}

\end{proof}


\item 
Consider the semigroup homomorphism $\varphi:\ZZ_{\ge 0}^3 \to \ZZ_{\ge 0}$ given by
$$(a,b,c) \longmapsto 4a + 5b + 6c.$$
Locate an ideal $I \subset \kk[x,y,z]$ such that $\til_I = \ker \varphi$.  Find a finite generating set for $I$.  

\begin{proof}

\end{proof}


\item 
Consider the semigroup homomorphism $\varphi:\ZZ_{\ge 0}^5 \to \ZZ_{\ge 0}$ given by
$$(a_1,a_2,a_3,a_4,a_5) \longmapsto 15a_1 + 97a_2 + 152a_3 + 153a_4 + 201a_5.$$
Locate an ideal $I \subset \kk[x_1,x_2,x_3,x_4,x_5]$ such that $\til_I = \ker \varphi$.  

Hint:\ do \textbf{not} attempt to write down generators for $I$!  Instead, specify $I$ as the kernel of a carefully chosen ring homomorphism.  

\begin{proof}

\end{proof}

\end{enumerate}


\item 
Determine whether each of the following statements is true or false.  Prove your assertions.  

\begin{enumerate}
\item 
For any $\alpha_1, \ldots, \alpha_k \in \ZZ_{\ge 0}^d$ with $d \ge 2$, the set $\<\alpha_1, \ldots, \alpha_k\> \setminus \<\alpha_1, \ldots, \alpha_{k-1}\>$ is infinite.  

\begin{proof}[Answer]

\end{proof}


\item 
Given any congruence $\til$ on $\ZZ_{\ge 0}^k$, the semigroup $S = \ZZ_{\ge 0}^k/\til$ is \emph{cancellative}, that is, $a + c = b + c$ implies $a = b$ for all $a, b, c \in S$.  

\begin{proof}[Answer]

\end{proof}

\end{enumerate}

\end{enumerate}



\bigskip
\noindent
\textbf{Challenge problems.}
Challenge problems are not required for submission, but bonus points will be awarded for submitting a partial attempt or a complete solution.  

\begin{enumerate}[(C1)]
\item 
Locate a commutative, cancellative semigroup $S$ such that (i)~$S$ is finitely generated, (ii)~$0 \in S$ is the only element of $S$ with an inverse, and (iii)~$S$ is not isomorphic to any affine semigroup.  

\begin{proof}

\end{proof}

\end{enumerate}

\end{document}

