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\parbox{1.95in}{\bf Math 3110-101}
\hfill {\Large \bf Homework \#9} \hfill
\parbox{1.95in}{\bf \hfill Due: Fri., Apr. 19${}^\text{\bf th}$, 2024}

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%\noindent {\bf\large Name:} \quad \underline{\large\qquad This is my Name \qquad}

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Please remember when submitting any work via email or in person to\dots\\
\force \hfill \fbox{\fbox{\bf\large\color{red}PUT YOUR NAME ON YOUR WORK!}} \hfill \force


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\begin{description}
\item[\#1] In each of the following rings, $R$, state the characteristic of the ring. If $R$ has unity, 
     give an example of a unit and its inverse (other than $1$ itself). If no unit exists, explain why not. 
     If $R$ has zero divisors, give an example of a zero divisor. If no zero divisors exist, explain why not.
     \begin{enumerate}[(a)]
     \item $R = \mathbb{Z}_{14} \times \mathbb{Z}_{10} $
     \item $R = (2\mathbb{Z})[x]$ (polynomials with even integer coefficients)
     \item $R = (\mathbb{Z}_5)^{2 \times 2} $ ($2 \times 2$ matrices with entries in $ \mathbb{Z}_5$)
     \end{enumerate}     
          
\item[\#2] Let $R$ be a ring with $1$. Show that $r \in R$ cannot be both a zero divisor and a unit. 

If $u,v \in R$ are units, is $uv$ always a unit? [Explain why or give a concrete counterexample.]

If $a,b \in R$ are zero divisors, is $ab$ always a zero divisor? [Explain why or give a concrete counterexample.]

{\it Note:} Be careful. Do not assume $R$ is commutative.\\ \force\hfill Also, pay attention to the fact that zero divisors are by definition \underline{\it nonzero} elements.

\item[\#3] Let $S = \left\{ \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \;\middle|\; a,b \in \mathbb{Z}\right\} \subseteq \mathbb{Z}^{2\times 2}$.

   \begin{enumerate}[(a)]
   \item Prove $S$ is a subring of $\mathbb{Z}^{2\times 2}$.
   \item Show $S$ is a commutative ring with $1$.
   \item Recall that $U(S)$ are the units of $S$. Find all of them. What kind of group is the group of units of $S$?
   
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   {\it Note:} $U(\mathbb{Z}^{2\times 2}) = \mathrm{GL}_2(\mathbb{Z}) = \{ A \in \mathbb{Z}^{2\times 2} \;|\; \det(A)=\pm 1\}$.\\ \force\hfill The determinant should be helpful in figuring out what the units of $S$ are.
   
   \item $\mathbb{Z}^{2 \times 2}$ has zero divisors but $S$ does not.  Give an example of a zero divisor in $\mathbb{Z}^{2\times 2}$.
   
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   Next, we can show $S$ has no zero divisors. Fill in the details:\\
   Suppose $A$  is a zero divisor in $S$ ($\subseteq \mathbb{Z}^{2\times 2} \subseteq \mathbb{R}^{2 \times 2}$). Explain why $A^{-1}$ cannot exist in $\mathbb{R}^{2 \times 2}$.\\ What does this say about $\det(A)$? What matrices in $S$ have such a determinant?
   
   {\it Note:} Since $S$ is a commutative ring with $1\not=0$ and has no zero divisors, it is called an {\it integral domain}.
   \end{enumerate}
   
\item[\#4] Prove that $\mathbb{Q}[\sqrt{-3}] = \{ a+b\sqrt{-3} \;|\; a,b \in \mathbb{Q} \}$ is a field. 

{\it Note:} Use a subring test to show it is a ring (it is a subring of $\mathbb{C}$).\\ \force \hfill Then use the ``conjugate trick'' to show it has multiplicative inverses.

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