Let \(R\) be a subring of a commutative ring \(S\), and suppose the additive index \([S:R]\) is finite and equal to \(n\). Let \(m\) be an integer relatively prime to \(n\). Prove that the natural map \[ R/mR\longrightarrow S/mS \] is a ring isomorphism.
Solution
Let \[ A\coloneqq S/R \] be the finite additive quotient group. Then \(\abs A=n\).
1. Multiplication by \(m\) is an automorphism of the additive group\(A\).
Since \(\gcd(m,n)=1\), choose integers \(u,v\) with \[ um+vn=1. \] For every \(a\in A\), Lagrange’s theorem gives \(na=0\). Hence \[ u(ma)=uma=(1-vn)a=a. \] Thus multiplication by \(u\) is an inverse to multiplication by \(m\) on \(A\).
2. The natural map\[ \phi:R/mR\longrightarrow S/mS \]is surjective.
Let \(s\in S\). By step 1, multiplication by \(m\) is surjective on \(A\), so there is \(t\in S\) such that \[ m(t+R)=s+R. \] Equivalently, \[ r\coloneqq s-mt\in R. \] Then \(s-r=mt\in mS\), so \(s+mS=r+mS\). Thus every class in \(S/mS\) is the image of a class from \(R/mR\).
3. The map \(\phi\) is injective.
Suppose \(r\in R\) maps to zero in \(S/mS\). Then \[ r=ms \] for some \(s\in S\). In \(A=S/R\) this says \[ m(s+R)=r+R=0. \] Multiplication by \(m\) is injective on \(A\) by step 1, so \(s+R=0\), hence \(s\in R\). Therefore \[ r=ms\in mR, \] and the class of \(r\) in \(R/mR\) is zero.
4. The natural map is a ring isomorphism.
The map is induced by the inclusion \(R\hookrightarrow S\), so it is a ring homomorphism. Steps 2 and 3 show that it is bijective.
5. Q.E.D.
Step 4 proves the assertion.