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.
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.