Prove that \(\ZZ\) is initial in the category of unital rings and unital ring homomorphisms. Deduce that every abelian group has a unique \(\ZZ\)-module structure.
Solution
1. \(\ZZ\) is initial among unital rings.
Let \(R\) be a unital ring. Define \[ \phi:\ZZ\to R, \qquad \phi(n)=n\cdot1_R, \] where for \(n>0\) this is a sum of \(n\) copies of \(1_R\), for \(n<0\) it is the negative of \((-n)\cdot1_R\), and \(\phi(0)=0\).
Then \[ \phi(m+n)=\phi(m)+\phi(n),\qquad \phi(mn)=\phi(m)\phi(n),\qquad \phi(1)=1_R, \] so \(\phi\) is a unital ring homomorphism.
If \(\psi:\ZZ\to R\) is any unital ring homomorphism, then \(\psi(1)=1_R\), hence additivity forces \[ \psi(n)=n\cdot1_R=\phi(n) \] for every \(n\in\ZZ\). Thus \(\phi\) is unique.
2. Every abelian group has a unique \(\ZZ\)-module structure.
For an abelian group \(A\), the endomorphism ring \(\Endo_{\mathbf{Ab}}(A)\) is unital. A \(\ZZ\)-module structure on \(A\) is equivalent to a unital ring homomorphism \[ \ZZ\to\Endo_{\mathbf{Ab}}(A). \] By 1 there is exactly one such homomorphism. Explicitly, \[ n\cdot a= \begin{cases} a+\cdots+a,&n>0,\\ 0,&n=0,\\ -((-n)\cdot a),&n<0. \end{cases} \]