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\title{Algebra Homework Set 5}
\author{James Cummings (jcumming@andrew.cmu.edu)}


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   Due by class time on Wednesday 12 October. Homework must be typeset and submitted by email as a PDF file. 

\begin{enumerate}


\item Let $R$ be a ring. A set $S \subseteq R$ is {\em multiplicatively closed (MC)} if $1 \in S$ and
  $S$ is closed under $\times$. Prove that if $S$ is an MC set with $0 \notin S$, and $I$ is a maximal
  element of the poset of ideals which are disjoint from $S$ ordered by inclusion, then $I$ is prime.  

\item Let $R$ be a ring. A ring element $r$ is {\em nilpotent} if $r^n = 0$ for some $n > 0$.
  Prove that if $r$ is nilpotent then $1 + r$ is a unit. Prove that the nilpotent elements
  of $R$ form an ideal.

\item Let $\phi: R \rightarrow S$ be a ring HM and let $J$ be an ideal of $S$. Prove that
  $R/\phi^{-1}[J]$ is isomorphic to a subring of $S/J$, and use this to conclude that if $J$ is a prime ideal of $S$
   then $\phi^{-1}[J]$
  is a prime ideal of $R$. Give an example to show that in general the preimage of a maximal ideal
   under a HM is not maximal. 

 \item Let $R_n = \mathbb Z[x_1, \ldots x_n]$ (the ring of polynomials in $n$ variables) and
   let $R = \bigcup_n R_n$. It is easy to see that $R$ is a ring (you need not prove this).
   Let $I$ be the ideal generated by $\{ x_n : n \in {\mathbb N} \}$, equivalently the set of
   polynomials with zero constant term. Show that $I$ is not an fg ideal.

   Hint: suppose that $I$ is generated by elements $f_1, \ldots f_k$ and consider a variable
   $x_k$ where the index $k$ is greater than the index of any variable appearing in the $f_i$. 
   
 \item Prove that the product $R \times S$ of two Noetherian rings $R$ and $S$ is Noetherian.
   

   
\item  Recall that we may view the free group $F$ generated by a set $X$ as follows: the underlying
  set of $F$ is the set of reduced words on $X$, and the group operation is ``concatenate and reduce''
  Let $H \le F$ be an arbitrary subgroup of $F$. Prove that there is a set $T \subseteq F$ such that
  \begin{enumerate}
  \item $T$ meets each left coset of $H$ in exactly one element.
  \item $T$ is ``closed under suffixes'', that is to say if $w \in T$ then every final segment
    of $w$ is in $T$.
  \end{enumerate}

  Hint: Show that we can apply ZL to the poset $\mathbb P$ of sets which are closed under suffixes and meet each left
  coset at most once. If $T$ is maximal and fails to meet each left coset, let $w$ be of minimal length with
  $w H \cap T = \emptyset$ and consider the word $u$ obtained by deleting the first entry in $w$. 
    
\end{enumerate}
  
  
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