An exercise for a subring with finite index. Exercise
Let $A$ be a commutative ring,and let $B\subset A$ be a subring for which $A/B$ is finite.  


*

*Prove that there is an ideal I of A with $I\subset B$ for which $A/I$ is finite.  

*Prove that  the unit group of $A$ and $B$ satisfy $A^* \cap B=B^*$,and that the abelian group $A^*/B^*$ is finite.

 A: 1) Let $A/B = \{x_0+B,\ldots,x_k + B\}$. Set:
\begin{equation}
I= \{y \in B :  x_i y \in B, \quad \forall\, i=0,\ldots,k\}.
\end{equation}
This is the tentative definition of ""biggest" possible subset of $B$ that is hopefully an ideal of $A$, taking into account that $A/B$ is finite". One has to check that $I$ is actually an ideal of $A$. First, $I$ is clearly closed under addition. Then, let $a \in A$ and $x \in I$, and consider $ax \in A$. First, there exists an integer $m$ such that $a-x_m =b \in B$. Hence,
\begin{equation}
ax = (b+x_m)x = bx +x_mx
\end{equation}
where clearly $bx \in I$ and $x_mx \in B$ by definition of $I$. Next, check that for all $i=0,\cdots,k$, the element $x_i x_m x$ lies in $B$. Again, there exists an integer $h$ and $b' \in B$ such that $x_ix_m = x_h+b'$, and $x_i(x_m x) = x_h x + b'x$, which lies in $B$ because $x \in I$.
Next, we prove that $A/I$ is finite. As abelian groups, we have
\begin{equation}
A/B \cong \frac{A/I}{B/I},
\end{equation}
so it is sufficient to prove that $B/I$ is finite. We note that, as an additive subgroup of $B$, $I$ is by definition the kernel of the following homomorphism of abelian groups:
\begin{align*}
B & \to (A/B)^{\oplus k}, \\
b &\mapsto (x_1b,\ldots,x_kb).
\end{align*}
From this, we find an injection $B/I \hookrightarrow (A/B)^{\oplus k}$, which proves that $B/I$ is finite, as we wanted.
2) To check the first assertion, we need to prove that if $b \in B$ has an inverse $x \in A$ (such that $bx=1$), then actually $x \in B$. Consider the set of powers of $x$: $\{1,x,x^2,\ldots\}$. Since $A/B$ is finite, the classes $[x^i]$ can not be all different in $A/B$, namely, there are integers $n,m$ (assume $n<m$) such that $x^n - x^m \in B$. Multiply by $b^n$ and find out that $1- b^nx^m = 1-x^{m-n}\in B$, hence $x^{m-n} \in B$, from which we deduce that $x \in B$.
Next, we check that $A^*/B^*$ is finite. Consider the homomorphism of multiplicative groups
\begin{equation}
f \colon A^* \to (A/I)^*
\end{equation}
induced by the natural projection $A \to A/I$, where $I$ is the ideal we defined in part 1). Let $x \in \ker(f)$. This means that $[x]=[1]$ in $A/I$, so $x-1 \in I$, so in particular $x-1 \in B$ and hence $x \in B$. So, using what we proved so far, we get that
\begin{equation}
\ker(f) \subseteq B \cap A^* = B^* \subseteq A^*.
\end{equation}
From this, we find an injection $A^*/\ker(f) \hookrightarrow (A/I)^*$, so $A^*/\ker(f)$ is finite; hence, its subgroup $B^* / \ker(f)$ is finite and
\begin{equation}
A^*/B^* \cong \frac{A^*/\ker(f)}{B^*/\ker(f)}
\end{equation}
is finite.
