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\lhead{MATH 522--01}
\chead{Problem Set \#1\answer{ solutions}}
\rhead{}% Your name here!
\lfoot{}
\cfoot{\answer{Page \thepage\ of \pageref{LastPage}}}
\rfoot{due Thursday, January 22, 2015}

\begin{document}
\begin{enumerate}
\item \question{Let $U$ be a nonempty set and let $R$ be the set of
    all subsets of $U$ (i.e. the power set of $U$). For the two given
    proposed definitions of ``addition'' and ``multiplication'',
    determine whether $R$ is a ring or not; if it is not a ring,
    explain why, and if it is a ring, identify its identity elements.}
  \begin{enumerate}
  \item \question{$A+B=A\cup B$ and $A\cdot B=A\cap B$.}

    \answer{This is not a ring, because there is no inverse under
      addition. To determine an identity, we note that $A+0=A\cup 0=A$
      is only universally true if $0=\emptyset$; however, then since
      for nonempty $A$, there is no set $(-A)$ such that
      $A+(-A)=A\cup(-A)=\emptyset$.}

  \item \question{$A+B=(A\cup B)-(A\cap B)$ (also known as the
      ``symmetric difference'') and $A\cdot B=A\cap B$.}

    \answer{This is in fact a ring: closure is easily demonstrated,
      and associativity is a bit tedious but straightforward ($A+B+C$
      consists of those elements lying in exactly 1 or 3 of the sets
      $A$, $B$, and $C$). The distributeive law may be derived from
      noting that $[(A\cup B)-(A\cap B)]\cap C=[(A\cap C)\cup(B\cap
      C)]-(A\cap B\cap C)$ The additive identity is $\emptyset$, and
      every element of $R$ is its own additive inverse. There is also
      a multiplicative identity, $U$ itself.}

  \end{enumerate}

\item \question{Prove that for a commutative ring $R$ and $x\in R$,
    $x$ is a unit of $R$ if and only if $x$ divides every element of
    $R$.}
  
  \answer{First, let us note that the statement is vacuously true if
    $R$ lacks a multiplicative identity, as \emph{neither} of the
    conditions can be met. Clearly, $x$ can be a unit only if $xy=1$
    for some $x,y\in R$, which would require that $1\in R$. Similarly,
    if $x$ divides every element of $R$, then in particular $x$
    divides $x$, so $x=kx$ for some $k\in R$; now, for any $r\in R$,
    there is an $\ell\in R$ such that $r=\ell x$, and so $r=\ell
    x=\ell kx=k(\ell x)=kr$, making $k$ a multiplicative
    identity. Henceforth, we may thus assume that $R$ has a
    multiplicative identity.

    Let us suppose that $x$ is a unit, so that there is a $y\in R$
    such that $xy=1$. Then for any $r\in R$, $(xy)r=1r=r$, so
    $r=x(yr)$. Thus $x$ divides $r$.

    Conversely, if we suppose that $x$ divides every element of $R$,
    then, in particular, $x$ divides $1$, so there is a $k\in R$ such
    that $xk=1$; since $k$ is definitionally the inverse of $x$, $x$
    is a unit.}

\item \question{Prove the following two statements:}
  \begin{enumerate}
  \item \question{For any ring $R$ and a (not necessarily finite)
      collection $\mathcal R$ of subrings of $R$, their intersection
      $$\bigcap_{S\in\mathcal R} S$$
      is also a subring of $R$. (Hint: you may find it easier to warm
      up by addressing just an intersection of two subrings of $R$)}

    \answer{Let $T=\bigcap_{S\in\mathcal R} S$ for brevity. Then if
      $a,b\in T$, it follows that $a,b\in S$ for every single ring
      $S\in\mathcal R$. Since each $S$ is a ring, $ab$ and $a+b$ are
      in $S$ for every $S\in\mathcal R$, and thus $ab$ and $a+b$ are
      in $T$, demonstrating closure of $T$ under addition and
      multiplication.

      Likewise, since each $S$ is a ring, $-a\in S$ for every
      $S\in\mathcal R$, and so $-a\in T$, demonstrating closure under
      additive inversion.

      Finally, since every $S$ is a ring, $0\in S$ and thus $0\in T$.}

  \item \question{For a ring $R$ and a subset $S$ of $R$, there is a
      unique \emph{smallest} subring $\langle S\rangle$ of $R$ which
      contains all the elements of $S$ (hint: choose an appropriate
      collection $\mathcal R$ for the above statement).}

    \answer{Let $\mathcal R$ contain every ring which is a superset of
      $S$ and a subring of $R$. By the previous part,
      $T=\bigcap_{X\in\mathcal R}X$ is a subring of $R$, and for every
      ring $X$ which is a superset of $S$ and a subring of $R$,
      $X\subseteq T$, we may assert that $T$, as it contains $S$ and
      is contained in every ring containing $S$, is the smallest such
      ring.}
  \end{enumerate}

\item \question{Address the two following questions about subrings:}
  \begin{enumerate}
  \item \question{Let $\mathbb Q[\sqrt 2,\sqrt 3]$ denote the
      smallest subring of $\mathbb R$ which contains every rational
      number, the irrational number $\sqrt 2$, and the irrational
      number $\sqrt 3$. Let $\mathbb Q[\sqrt 2+\sqrt 3]$ denote the
      smallest subring of $\mathbb R$ which contains every rational
      number and the irrational number $\sqrt 2+\sqrt 3$ (note that
      the previous question guarantees that these structures are
      well-defined). Prove that $\mathbb Q[\sqrt 2,\sqrt 3]=\mathbb
      Q[\sqrt 2+\sqrt 3]$.}

    \answer{It suffices to show that $\sqrt 2$ and $\sqrt 3$ are
      elements of $\mathbb Q[\sqrt 2+\sqrt 3]$ to prove that $\mathbb
      Q[\sqrt 2,\sqrt 3]\subseteq\mathbb Q[\sqrt 2+\sqrt 3]$;
      likewise, if we show that $\sqrt 2+\sqrt 3\in\mathbb Q[\sqrt
      2,\sqrt 3]$, then $\mathbb Q[\sqrt 2+\sqrt 3]\subseteq\mathbb
      Q[\sqrt 2,\sqrt 3]$, collectively showing the desired equality.

      The second of these is trivial: $\sqrt 2\in\mathbb Q[\sqrt
      2,\sqrt 3]$ and $\sqrt 3\in \mathbb Q[\sqrt 2,\sqrt 3]$, so by
      closure of addition, $\sqrt 2+\sqrt 3\in\mathbb Q[\sqrt 2,\sqrt
      3]$.

      The first membership, however, is more difficult. Since $(\sqrt
      2+\sqrt 3)\in\mathbb Q[\sqrt 2+\sqrt 3]$, we may use closure of
      multiplication to find that its square $5+2\sqrt 6$ is also in
      $\mathbb Q[\sqrt 2+\sqrt 3]$. $-5$ and $\frac12$ are rational,
      and can be added and multiplied in turn to the above to find
      that $\sqrt 6\in\mathbb Q[\sqrt 2+\sqrt 3]$. Finally, we
      multiply $\sqrt 6$ by $(\sqrt 2+\sqrt 3)$ once more to find that
      $2\sqrt 3+3\sqrt 2$ is in $\mathbb Q[\sqrt 2+\sqrt 3]$, and
      subtracting twice and thrice $(\sqrt 2+\sqrt 3)$ demonstrates
      that $\sqrt 2$ and $-\sqrt 3$ respectively are in $\mathbb
      Q[\sqrt 2+\sqrt 3]$.}
    
  \item \question{Let $\mathbb Z[\sqrt 2,\sqrt 3]$ and $\mathbb
      Z[\sqrt 2+\sqrt 3]$ be defined as in the previous part except
      with ``rational number'' replaced by ``integer'' in the
      definition. Demonstrate that $\mathbb Z[\sqrt 2,\sqrt
      3]\neq\mathbb Z[\sqrt 2+\sqrt 3]$.}

    \answer{The same argument as above can be used to show that
      $\mathbb Z[\sqrt 2+\sqrt 3]\subseteq\mathbb Z[\sqrt 2,\sqrt 3]$;
      however, our argument for the reverse containment will not work
      as designed because we cannot multiply by $\frac12$. It is,
      however, easy to follow the above argument without the division
      to guarantee that $2\sqrt 2$, $2\sqrt 3$, and $2\sqrt 6$ are
      all elements of $\mathbb Z[\sqrt 2+\sqrt 3]$. However, half of
      each of these values is \emph{not} in the ring: note that the set
      \[\{a+2b\sqrt2+2c\sqrt 6+d(\sqrt 2+\sqrt 3):a,b,c,d\in\mathbb
      Z\}\] is a ring containing $\mathbb Z$ and $\sqrt 2+\sqrt 3$
      but not $\sqrt 2$ or $\sqrt 3$.}
  \end{enumerate}

\item \question{Prove that if the additive group of a ring $R$ is
    cyclic, then $R$ is commutative.}

  \answer{Let $g$ be a generator of the additive group. Thus, for any
    $a,b\in R$, either $a$ or $-a$ may be written as
    $\underbrace{g+g+\cdots+g}_{k\text{ times}}$ for some nonnegative
    integer $k$; likewise $b$ or $-b$ may be written as
    $\underbrace{g+g+\cdots+g}_{\ell\text{ times}}$ for some
    non-negative integer $\ell$. Using the distributive law, the
    product $ab$ can be written as
    $\pm\underbrace{g+g+\cdots+g}_{k\ell\text{ times}}$; however,
    likewise $ba$ could be written as
    $\pm\underbrace{g+g+\cdots+g}_{\ell k\text{ times}}$; since $k$
    and $\ell$ are ordinary integers, the number of addends is the
    same in both expressions, so $ab=ba$.}

\end{enumerate}
\end{document}