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\begin{document}
\title{Worksheet \#1-- Math 5320\\Fall 2019}
\author{Due Wednesday, January 15th, in Gradescope}

\maketitle
You may work in groups of up to 3 on this assignment.  Only one assignment needs to be turned in per group.  It still needs to be turned in on gradescope.
\vskip 9pt
Recall that a \emph{ring} is a set $R$ with two binary operations, $+, \cdot$ satisfying the following axioms for all $x,y,z \in R$.
\begin{multicols}{3}
    \begin{enumerate}
    \item $(x+y) + z = x+ (y+z)$.
    \item $\exists 0 \in R$ such that $x + 0 = 0 + x = x$.    
    \item $\forall x \in R, \exists -x \in R$ such that $x + (-x) = 0$.
    \item $x+y = y+x$.
    \item $x(yz) = (xy)z$.
    \item $\exists 1 \in R$ such that \\$1 x = x1 = x.$
    \item $xy = yx$ (our book assumes this, most do not, ie in most texts not all rings are \emph{commutative})
    \item $x(y+z) = xy + xz$.
    \end{enumerate}
    \end{multicols}
\noindent
{\bf 1.}  Prove that the element $1$ in a ring is unique.
\vskip 3pt
\emph{Hint:}  Suppose you have two 1s, say $1$ and $1'$.
\vskip 130pt
\noindent
{\bf 2.}  Show that if $x \in R$ then $0x = 0$.
\vskip 3pt
\emph{Hint:} We already showed this in class. 
\newpage
\noindent
{\bf 3.}  Show that $(-1) \cdot x = -x$.
\vskip 3pt
\emph{Hint:}  $-x$ is characterized uniquely by the fact that $x + (-x) = 0$ (because inverses are unique in groups, and rings are groups under $+$).
\vskip 130pt
\noindent
{\bf 4.}  Show that $(-1) \cdot (-1) = 1$ in any ring.
\vskip 130pt
\noindent
{\bf 5.}  Consider the ring $R = \bZ/10\bZ$, the integers under addition and multiplication modulo 10.  Find all solutions to the equation $5x = 0$.  
\vskip 3pt
\emph{Hint: }  There should be 5 distinct solutions.
\vskip 130pt
A ring is called an \emph{integral domain} if $ab = 0$ implies that either $a = 0$ or $b = 0$.
\vskip 3pt
\noindent
{\bf 6.}  Prove that if $R$ is an integral domain and $ax = ay$ for some $a \neq 0$, then $x = y$.  
\newpage
\noindent
{\bf 7.}  Suppose $R$ is a ring such that if $a \neq 0$ and $ax = ay$, then $x = y$.  Prove that $R$ is an integral domain.
\vskip 150pt
An element of a ring $x \in R$ is called a \emph{zero divisor} if there exists a nonzero $y \in R$ such that $xy = 0$.   If $x \neq 0$ is not a zero divisor it is called either a \emph{non-zero divisor} or a \emph{regular element}.
\vskip 3pt
\noindent
{\bf 8.}  Suppose $R$ is a ring and $x \in R$.  Show that $x$ cannot simultaneously be both a zero divisor and a unit.  On the other hand, give an example of an element in a ring that is neither a zero divisor or a unit.
\newpage
If $f(x) = \sum_{i = 0}^n a_i x^i \in R[x]$, then we use $\deg(f(x))$ to denote the largest $i$ such that $a_i \neq 0$.  It is called the \emph{degree} of $f(x)$.
\vskip 3pt
\noindent
{\bf 9.}  Suppose that $R$ is a ring, $x$ is an indeterminate, and $R[x]$ is the associated polynomial ring.  
\vskip 3pt
Show that if $R$ is an integral domain and $f(x), g(x) \in R[x]$, then 
\[
    \deg (f(x) \cdot g(x)) = \deg f(x) + \deg g(x).
\]
Give an example to show that this does not hold if $R$ is not an integral domain.
\end{document} 