Let \(R=\ZZ[x]\) and let \(J=(2,x)\trianglelefteq R\). Prove that the \(R\)-module \(J\) is not a direct sum of cyclic \(R\)-modules.
Solution
1. The ideal \(J=(2,x)\) is not principal.
Suppose \(J=(f)\) for some \(f\in\ZZ[x]\). Since \(2\in(f)\), the polynomial \(f\) divides the constant polynomial \(2\). Hence \(f\) is a nonzero constant integer. Since also \(x\in(f)\), this integer divides the polynomial \(x\), so it must be a unit \(\pm1\). Thus \((f)=R\).
But \(J\) is proper: the quotient map \[ \ZZ[x]\longrightarrow\FF_2, \qquad h(x)\longmapsto h(0)\bmod2 \] has kernel \((2,x)\). Hence \(J\ne R\), a contradiction. Therefore \(J\) is not principal.
2. Any two nonzero cyclic \(R\)-submodules of \(J\) have nonzero intersection.
Let \[ M=R\alpha, \qquad N=R\beta \] with \(0\ne\alpha,\beta\in J\). Since \(R=\ZZ[x]\) is an integral domain, \[ \alpha\beta\ne0. \] Moreover, \[ \alpha\beta\in R\alpha=M \qquad\text{and}\qquad \alpha\beta\in R\beta=N. \] Thus \[ 0\ne\alpha\beta\in M\cap N. \]
3. In any internal direct-sum decomposition of \(J\) into cyclic submodules, at most one summand can be nonzero.
Suppose \[ J=\bigoplus_{i\in I}M_i \] with each \(M_i\) cyclic. Distinct summands in an internal direct sum have zero intersection. By 2, two distinct nonzero cyclic summands would have nonzero intersection. Hence at most one \(M_i\) is nonzero.
4. No such direct-sum decomposition exists.
The module \(J\) is nonzero, so in a decomposition as in 3 exactly one summand would be nonzero. Then \(J\) itself would be cyclic as an \(R\)-module. Since \(J\) is an ideal of \(R\), a cyclic \(R\)-module structure would mean \[ J=R\alpha=(\alpha) \] for some \(\alpha\in R\), making \(J\) principal. This contradicts 1.