Let \(R\) be a commutative integral domain. Show that the following are equivalent:
\(R\) is a field;
\(R\) is a semi-simple ring;
Any \(R\)-module is projective.
Solution
1. If \(R\) is a field, then \(R\) is semisimple.
Viewed as a left module over itself, a field \(R\) has only the submodules \(0\) and \(R\), because its left ideals are precisely its ideals. Thus \({}_RR\) is simple. A simple module is semisimple, so \(R\) is a semisimple ring.
2. If \(R\) is semisimple, then every \(R\)-module is projective.
Let \(M\) be an \(R\)-module. Choose a surjection \[ \pi:F\twoheadrightarrow M \] from a free module \(F\). Since \(R\) is semisimple, the regular module \({}_RR\) is semisimple. Therefore every direct sum of copies of \(R\), in particular the free module \(F\), is semisimple. Hence every submodule of \(F\) is a direct summand; in particular, \[ F=\ker\pi\oplus C \] for some submodule \(C\).
The restriction \[ \pi|_C:C\longrightarrow M \] is an isomorphism: it is injective because \(C\cap\ker\pi=0\), and it is surjective because every \(f\in F\) can be written \(f=k+c\) with \(k\in\ker\pi\), so \(\pi(f)=\pi(c)\). Thus the surjection \(F\twoheadrightarrow M\) splits. Since every surjection from a free module onto \(M\) splits, \(M\) is projective.
3. If every \(R\)-module is projective, then \(R\) is a field.
Let \(0\ne a\in R\). The module \(R/(a)\) is projective by hypothesis, so the short exact sequence \[ 0\longrightarrow (a)\longrightarrow R\longrightarrow R/(a)\longrightarrow0 \] splits. Hence \[ R=(a)\oplus I \] for some ideal \(I\).
Let \(e\in(a)\) be the component of \(1\) under this direct-sum decomposition, so \[ 1=e+f, \qquad e\in(a),\ f\in I. \] Because \((a)\) and \(I\) are ideals, \[ ef\in(a)\cap I=0. \] Therefore \[ e=e(e+f)=e^2. \] Thus \(e\) is an idempotent. Since \((a)\ne0\), its direct-summand projection is nonzero, so \(e\ne0\). In an integral domain the only idempotents are \(0\) and \(1\), because \[ e^2=e\implies e(e-1)=0. \] Hence \(e=1\). Since \(e\in(a)\), we have \(1\in(a)\), so \((a)=R\) and \(a\) is a unit.
Every nonzero element of \(R\) is therefore invertible, so \(R\) is a field.
4. Consequently the three conditions are equivalent.
Steps 1, 2, and 3 give the cycle \[ R\text{ field}\Longrightarrow R\text{ semisimple} \Longrightarrow\text{every }R\text{-module projective} \Longrightarrow R\text{ field}. \]