\documentclass{article}
\usepackage{amssymb}
\newtheorem{problem}{Problem}
\newtheorem{lemma}{Lemma}
\newtheorem{sublemma}{Sublemma}
\title{CHM: A.23}
\author{Po-Shen Loh}
\date{29 March 2001}
\begin{document}
\maketitle

\begin{problem}[A.23]
  Let $R$ be an infinite ring such that every subring of $R$ different
  from $\{0\}$ has a finite index in $R$.  (By the index of a subring,
  we mean the index of its additive group in the additive group of
  $R$.)  Prove that the additive group of $R$ is cyclic.
\end{problem}

\noindent \textbf{Solution:}

Our goal will be to use the Fundamental Theorem of Finitely Generated
Abelian Groups, from which we conclude that the additive group of
$R$ (let us call it $A$) is isomorphic to $\mathbb{Z}^n$.  Then we
will use an idea similar to ``logarithms'' to prove that it is really
just $\mathbb{Z}$.

First, we will show that $A$ is torsion free as an abelian group.

\begin{lemma}
  \label{lem:tor-free}
  $A$ is torsion-free.
\end{lemma}

\noindent \textbf{Proof of Lemma \ref{lem:tor-free}:}

We proceed by indirect proof; suppose, for the sake of contradiction,
that there exist $n \in \mathbb{Z}$ and $r \in R$ such that $nr = 0$.
We can immediately proceed to the case of $n$ prime, because if $n =
pq$ for prime $p$, then we can just consider $qr$ instead of $r$.  Let
$H = \langle r \rangle$ and $K = \langle r^2 \rangle$.  By the given
information, it is obvious that all subrings of the infinite $R$ must
likewise be infinite.  We will demonstrate that if $nr$ is indeed
zero, then $H$ is finite.

Again we have $|H:K| < \infty$, so let us try to mine something out of
the coset structure.  Concentrate on the cosets $r+K$, $r^3+K$,
$r^5+K$, \ldots (the cosets corresponding to the odd powers of $r$).
By finiteness, eventually they will repeat, so we will have the form
$r^a - r^b \in K$.  Since the elements of $K$ can be written as finite
polynomials in $r^2$, by the oddity of $a$ and $b$, we have a nonzero
polynomial $f(x) \in \mathbb{Z}[x]$ that evaluates to zero at $r$.
But since $n$ is a prime and $nr = 0$, we can work with coefficients
in the finite field $\mathbb{Z}_n$; we find a corresponding polynomial
$g(x) \in \mathbb{Z}_n[x]$ with the same property.  Let $cx^d$ be its
leading term.  Since $\mathbb{Z}_n$ is a field, we can normalize
$g(x)$ by multiplying it by $c^{-1}$.  This translates into a relation
$x^d = p(x)$ for some integral polynomial $p(x)$ with degree less than
$d$.  Then all elements of $H$ corresponding to polynomials of degree
at least $d$ can be reduced to polynomials of degree under $d$
(replace $x^d$ terms by polynomials of degree less than $d$); so the
elements of $H$ can be completely described by the polynomials of
finite degree less than $d$ with coefficients in the finite field
$\mathbb{Z}_n$; this is a finite set, so $|H| < \infty$ and we have
the contradiction predicted at the beginning.  Our lemma is proven.


Next we show that $A$ is finitely generated.  

\begin{lemma}
  \label{lem:fin-gen}
  $A$ is finitely generated.
\end{lemma}

\noindent \textbf{Proof of Lemma \ref{lem:fin-gen}:}

Suppose that there exists a subring $T \subset R$ such that its
additive group $B$ is finitely generated by some set $\mathcal{G}$.
Now, from the given information, $|A:B| < \infty$, so we have only
finitely many cosets.  Let $\mathcal{C} = \{c_1, c_2, c_3, \ldots,
c_k\}$ be representatives from the cosets, and consider the set
$\mathcal{G} \cup \mathcal{C}$.  This is a finite set, and it
generates $R$, so we only need to find a subring of $R$ with finitely
generated additive group.  We digress to prove an auxiliary lemma that
will provide us with such a subring.

\begin{sublemma}
  \label{sublem:r-poly}
  For any $r \in R$, there exist $n \in \mathbb{Z}$ and $p(x) \in
  \mathbb{Z}[x]$, with the constant term of $p(x)$ equal to zero, such
  that $nr = p(r)$.  Here, we write $nr$ with $n \in \mathbb{Z}$ and
  $r \in R$ to denote the summation of $n$ copies of $r$.  (If $n$ is
  negative, we add the appropriate number of copies of $-r$.
\end{sublemma}

\noindent \textbf{Proof of Sublemma \ref{sublem:r-poly}:}

Let subrings $D$ and $E$ be defined as follows: $D = \langle r
\rangle$ and $E = 0 \oplus r\mathbb{Z}[r]$.  In this definition,
$r\mathbb{Z}[r]$ refers to the evaluation at $r$ of all polynomials
with zero constant term.  Clearly, $E \subset D$, and since $D \subset
R$, it follows that $|D:E| < \infty$.  Let us look at the cosets
$0+E$, $r+E$, $2r+E$, $3r+E$, \ldots; since we have a finite index,
the cosets must eventuall repeat, so there exist $n \in \mathbb{Z}$
such that $nr \in E$, from which our claim is immediate.

Returning to the problem at hand, let us choose an arbitrary nonzero
$r \in R$.  Applying the lemma, we find some corresponding polynomial
$p(x)$.  Let $d$ and $a$ be the respective degree and coefficient of
the polynomial's leading term; we will be able to prove that $\langle
ar \rangle$ works for $B$.  This is because $(ar)^d = a^dr^d =
a^{d-1}(ar^d) = a^{d-1}q(r)$ for some $q \in \mathbb{Z}[x]$, and since
the degree of $q(x)$ is less than $d$, the final expression can be
written as $s(ar)$ for some $s \in \mathbb{Z}[x]$.  Since all elements
in $\langle ar \rangle$ can be expressed as finite integral
polynomials in $ar$, any such polynomial with degree $d$ or higher can
be re-expressed as a polynomial of degree less than $d$ by applying
the relation $(ar)^d = s(ar)$.  Therefore the finite set $\{ar,
(ar)^2, (ar)^3, \ldots, (ar)^{n-1}\}$ generates $\langle B \rangle$,
as desired.  Our lemma is proven.



At this point we can invoke the Fundamental Theorem of Finitely
Generated Abelian Groups: $A \cong \mathbb{Z}^n$.  The remainder of
this proof will show that $n = 1$.

\begin{lemma}
  \label{lem:n-1}
  $n = 1$.
\end{lemma}

\noindent \textbf{Proof of Lemma \ref{lem:n-1}:}

Suppose that the $n$ generators of the free group $A$ are $\{x_1, x_2,
\ldots, x_n\}$.  Define the family of free subgroups $A_k =
\mathbb{Z}x_k$; as we go through the all the powers of $x_i$, we must
enter all of the subgroups.  That is, for each $A_k$, there exists
some $a_k$ for which $x_i^{a_k} \in A_k$.  This is because if we did
not enter some subgroup $A_m$, then for every $z \in \mathbb{Z}$ we
have a distinct coset $za_m + \langle x_i \rangle$.  But $|R : \langle
x_i \rangle| < \infty$, so we have a contradiction.  Therefore, we
must enter every coset as we go through the powers of $x_i$.  Let us
go through the powers of $x_1$ and see what happens.

Let $\mathcal{S} = \{k \in \mathbb{Z}^+ | x_1^k \in A_1\}$.  Clearly,
if $a, b \in \mathcal{S}$ then $a+(b-1) \in \mathcal{S}$ because
$x_1^{a+b-1} = x_1^{a-1}x_1^b = x_1^a \in A_1$.  Similarly, as long as
$a-(b-1)$ is positive, it must also be in $\mathcal{S}$.  Therefore,
since $1 \in \mathcal{S}$, if $2 \in \mathcal{S}$ we will have
$\mathcal{S} = \mathbb{Z}^+ \rightarrow n=1$.  This is our final
sublemma.

\begin{sublemma}
  \label{sublem:2-in}
  $2 \in \mathcal{S}$.
\end{sublemma}

\noindent \textbf{Proof of Sublemma \ref{sublem:2-in}:}

Again proceed by indirect proof; assume for the sake of contradiction
that $2 \not \in \mathcal{S}$.  Let $y$ be the smallest member of
$\mathcal{S}$ other than $1$.  By the above result, $\mathcal{S}$ must
contain all positive integers congruent to $1$ modulo $y$, and nothing
more.  Define $M = \langle x_1 \rangle$ and $N = \langle x_1^y
\rangle$.  Again $|M:N| < \infty$ so we can extract information by
looking at the family of cosets $0+N$, $x_1+N$, $2x_1+N$, $3x_1+N$,
\ldots (this is infinite since $R$ is torsion-free and these are
distinct cosets because $\mathcal{S} = \{k \in \mathbb{Z}^+ | k \equiv
1 \pmod{y}\}$).  We discover that for some $z \in \mathbb{Z}$, $zx_1
\in N$; the elements of $N$ can be expressed as polynomials in
$x_1^y$, so we find $p(x_1) \in A_1$ for some integral polynomial with
all exponents divisible by $y$.  

Write $p(x_1)$ as $\sum_k c_k x_1^{yk}$.  By assumption $y \neq 2$, so
none of the terms in $p(x_1)$ are in $A_1$.  Therefore when we write
$p(x_1) = zx_1$, we have a nontrivial linear dependence relation
between the basis elements of our free abelian group.  This is a
contradiction, so $y = 1$ as required.  Our sublemma is proven.


And we are done.

\end{document}
