\documentclass[11pt]{article}
\usepackage{amssymb,amsmath, amsthm, amsfonts}
%\usepackage{setspace}\setstretch{1.1}
\setlength{\textwidth}{6.5in}
\setlength{\textheight}{9in}
\setlength{\oddsidemargin}{0in}
\setlength{\evensidemargin}{0in}
\setlength{\topmargin}{0in}
\pagestyle{empty}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\C}{\mathbb{C}}
\begin{document}
\noindent Math 463/563 - \textbf{Homework 1 Solutions}
\bigskip
\begin{enumerate}
\item Decide which of the following (a) -- (f) are subrings of $\Q$.
\begin{enumerate}
\item the set of all rational numbers with odd denominators (when written in lowest terms)\medskip
\noindent \textbf{Solution:} Yes, this is a subring.\smallskip
\item the set of all rational numbers with even denominators (when written in lowest terms)\medskip
\noindent \textbf{Solution:} No, this is not a subring because it does not contain $1_{\Q}$.\smallskip
\item the set of nonnegative rational numbers\medskip
\noindent \textbf{Solution:} No, this is not a subring because, for example, 1 does not have an additive identity.\smallskip
\item the set of squares of rational numbers\medskip
\noindent \textbf{Solution:} No, this is not a subring because, for example, 1 does not have an additive identity.\smallskip
\item the set of all rational numbers with odd numerators (when written in lowest terms)\medskip
\noindent \textbf{Solution:} No, this is not a subring because it is not closed under addition: for example, $1 + 1 = 2$ is not in the ring.\smallskip
\item the set of all rational numbers with even numerators (when written in lowest terms)\medskip
\noindent \textbf{Solution:} No, this not is a subring because it does not contain $1_{\Q}$.\smallskip
\end{enumerate}
\item Let $R$ be a ring and let $S$ be a subring of $R$. Prove that if $u$ is a unit in $S$ then $u$ is a unit in $R$. Show by example that the converse is false.\medskip
\noindent \textbf{Solution:} Let $u \in S$ be a unit. Then there exists $v \in S$ such that $uv = vu = 1_S$. Since $S$ is a subring of $R$, we have that $1_S = 1_R$ and $u, v \in R$, hence $u$ is a unit in $R$.
\noindent Converse: ``Let $u \in S$. If $u$ is a unit in $R$ then $u$ is a unit in $S$.''
\noindent Counter-example: $R = \Q$, $S = \Z$, $u = 2$. Notice that 2 is a unit in $\Q$ but $2$ is not a unit in $\Z$.
\smallskip
\item Let $R$ be a set with two laws of composition satisfying all the ring axioms except the commutative law for addition. Use the distributive law to prove that the commutative law for addition holds, so that $R$ is a ring.\medskip
\noindent Let $a, b \in R$. We want to show that $a + b = b + a$. Consider the element $a + b \in R$. Notice that its additive inverse is $- b - a$, which by the distributive property is equal to $-(b+a)$. Hence
$$a + b - (b+a) = 0.$$
By adding $(b + a)$ on the right on both sides we obtain
$$a + b = b + a,$$
as desired.
\noindent \textbf{Aside:} Notice that what we are really saying is that $a + b$ and $b + a$ have the same additive inverse, hence must be the same element, since inverses are unique.
\smallskip
\item \textbf{Required only for 563.} Let $\Q[\alpha, \beta]$ denote the smallest subring of $\C$ containing the rational numbers $\Q$ and the elements $\alpha = \sqrt{2}$ and $\beta = \sqrt{3}$. Let $\gamma = \alpha + \beta$. Is $\Q[\alpha, \beta] = \Q[\gamma]$? Is $\Z[\alpha, \beta] = \Z[\gamma]$?\medskip
\noindent \textbf{Solution:} We begin by showing that $\Q[\gamma] \subseteq \Q[\alpha, \beta]$ and $\Z[\gamma] \subseteq \Z[\alpha, \beta]$.
Note that $\Q[\gamma]$ (resp. $\Z[\gamma]$) is the smallest ring containing $\gamma$ and $\Q$ (resp. $\Z$), hence we need only show that $\gamma \in \Q[\alpha, \beta]$ and $\gamma \in \Z[\alpha, \beta]$, but this is immediate since both rings contain $\alpha$ and $\beta$ and therefore contain $\alpha + \beta = \gamma$.
Now we will show that $\Q[\alpha, \beta] \subseteq \Q[\gamma]$, hence the rings are equal. Again, by the fact that $\Q[\alpha, \beta]$ is the smallest ring containing $\Q$, $\alpha$, and $\beta$, it suffices to show that $\alpha, \beta \in \Q[\gamma]$. In fact, since $\beta = \gamma - \alpha$, we only need to show that $\alpha \in \Q[\gamma]$.
To see this, notice that
$$\gamma^3 = (\sqrt{2} + \sqrt{3})^3 = 11\sqrt{2} + 9\sqrt{3},$$
so $\gamma^3 - 9\gamma = 2\sqrt{2}$, hence $\alpha = \frac{1}{2}(\gamma^3 - 9\gamma) \in \Q[\gamma]$. Therefore $\Q[\gamma] = \Q[\alpha, \beta]$.\bigskip
Now we need to show that $\Z[\alpha, \beta] \nsubseteq \Z[\gamma]$. An element of $\Z[\gamma]$ is of the form
$$a_0 + a_1 \gamma + \cdots + a_n \gamma^n,$$
where $a_i \in \Z$. We claim that
$$\gamma^k =
\begin{cases}
(2a + 1)\alpha + (2b + 1)\beta, & \text{if $k$ is odd} \\
(2c + 1) + 2d \alpha\beta, & \text{if $k$ is even}.
\end{cases}$$
We proceed by induction on $k$. Note that we have to treat the two cases differently, but the statement is that the equality above holds for all $k$. That being said, in order to use induction, we need to first show that it holds for $k = 0$ and $k = 1$, then we may assume that it holds for general $k$ (even or odd) and show that it holds for $k + 1$.
It is clear that the equality holds for $k = 0$ and $k = 1$, so assume that it holds for $k$.
If $k$ is odd, then the induction hypothesis tells us that
$$\gamma^{k+1} = \gamma^k \gamma = ((2a + 1)\sqrt{2} + (2b + 1)\sqrt{3})(\sqrt{2} + \sqrt{3}) = (2(2a + 3b + 2) + 1) + 2(a + b + 1)\sqrt{6},$$
as desired.
If $k$ is even, then the induction hypothesis tells us that
$$\gamma^{k+1} = \gamma^k \gamma = ((2c + 1) + 2d\sqrt{6})(\sqrt{2} + \sqrt{3}) = (2(c + 3d) + 1)\sqrt{2} + (2(c + 2d) + 1)\sqrt{3},$$
as desired.
Since $\alpha\beta \in \Z[\alpha, \beta]$, it suffices to show that it is not in $\Z[\gamma]$. If $\alpha\beta \in \Z[\gamma]$, then we should be able to write $\alpha \beta$ as a linear combination of powers of $\gamma$, but every such linear combination will have an even coefficient of $\alpha \beta$, therefore $\Z[\alpha, \beta] \ne \Z[\gamma]$.
\end{enumerate}
\end{document}