Let \(R=(R,+,\cdot)\) be an integral domain. Let \(C\) be the additive group of \(R\) generated by the identity, that is, let \(C\) be the smallest subgroup of \(R\) such that \(C\) contains \(0\) and \(1\) and \(C\) is closed under \(+\).
Show that \(C=\{n\cdot 1 : n\in\mathbb{Z}\}\) and hence is a subring of \(R\). Here if \(n\ge 0\), then we can define \(n\cdot 1\) by induction as \(0\cdot 1=0\), \(1\cdot 1=1\) and \((n+1)\cdot 1=1+(n\cdot 1)\), and if \(n<0\), we define \(n\cdot 1=-(|n|\cdot 1)\).
Show that \(\phi\colon\mathbb{Z}\to C\) defined by \(\phi(n)=n\cdot 1\) is a surjective ring homomorphism.
Prove that either \(C\) is isomorphic to \(\mathbb{Z}\) or \(C\) is isomorphic to \(\mathbb{Z}_p\) for some prime \(p\).
Solution
1. One has\[ C=\{n\cdot1:n\in\mathbb Z\}, \]and this is a subring of \(R\).
The set \[ D:=\{n\cdot1:n\in\mathbb Z\} \] contains \(0\) and \(1\), and is closed under addition and additive inverses because \[ (m\cdot1)+(n\cdot1)=(m+n)\cdot1, \qquad -(n\cdot1)=(-n)\cdot1. \] Thus \(D\) is an additive subgroup of \(R\) containing \(1\). By minimality of \(C\), we have \(C\subseteq D\). Conversely, every additive subgroup containing \(1\) contains every integer multiple \(n\cdot1\), so \(D\subseteq C\). Hence \(C=D\).
Moreover \[ (m\cdot1)(n\cdot1)=(mn)\cdot1, \] so \(C\) is closed under multiplication and is therefore a subring of \(R\).
2. The map\[ \phi:\mathbb Z\to C, \qquad \phi(n)=n\cdot1, \]is a surjective ring homomorphism.
For \(m,n\in\mathbb Z\), \[ \phi(m+n)=(m+n)\cdot1=m\cdot1+n\cdot1=\phi(m)+\phi(n), \] and \[ \phi(mn)=(mn)\cdot1=(m\cdot1)(n\cdot1)=\phi(m)\phi(n). \] Also \(\phi(1)=1\). By 1, every element of \(C\) is \(n\cdot1\) for some \(n\in\mathbb Z\), so \(\phi\) is surjective.
3. Either\[ C\cong\mathbb Z \]or\[ C\cong\mathbb Z/p\mathbb Z \]for some prime \(p\).
Because \(C\) is a subring of the integral domain \(R\), it is itself an integral domain. Hence the kernel of the surjective homomorphism \(\phi:\mathbb Z\to C\) is a prime ideal of \(\mathbb Z\).
Every ideal of \(\mathbb Z\) is of the form \(d\mathbb Z\) for a unique \(d\ge0\). If \(d>0\) and \(d\mathbb Z\) is prime, then \(d\) must be prime: if \(d=ab\) with \(1<a,d\) and \(1<b<d\), then \[ ab\in d\mathbb Z \] but neither \(a\) nor \(b\) lies in \(d\mathbb Z\), contradicting primality. Thus \[ \ker\phi=0 \quad\text{or}\quad \ker\phi=p\mathbb Z \] for a prime \(p\).
By the first isomorphism theorem for rings, \[ C\cong\mathbb Z/\ker\phi. \] Therefore either \(C\cong\mathbb Z\), or \(C\cong\mathbb Z/p\mathbb Z=\mathbb Z_p\) for some prime \(p\).