Let \(\mathbb{C}[x]/\langle x^n \rangle\) denote the evident \(\mathbb{C}[x]\)-module, and let \(m, n \in \mathbb{N}\).
Show that there exist \(a_1, \ldots, a_k\) such that \[\mathbb{C}[x]/\langle x^n \rangle \otimes_{\mathbb{C}[x]} \mathbb{C}[x]/\langle x^m \rangle \cong \bigoplus_{i=1}^{k} \mathbb{C}[x]/(x^{a_i}).\]
Determine the \(a_i\) in terms of \(m,n\).
Hint: Figure out the action of \(x\) on the obvious \(\mathbb{C}\)-basis.
Solution
1. For every commutative ring \(R\) and ideals \(I,J\subseteq R\), there is a canonical isomorphism\[ (R/I)\otimes_R(R/J)\cong R/(I+J). \]
Define \[ \Phi:(R/I)\otimes_R(R/J)\longrightarrow R/(I+J), \qquad (r+I)\otimes(s+J)\longmapsto rs+(I+J). \] The map is well-defined and \(R\)-balanced: changing \(r\) by an element of \(I\) or \(s\) by an element of \(J\) changes \(rs\) by an element of \(I+J\).
Define \[ \Psi:R/(I+J)\longrightarrow (R/I)\otimes_R(R/J), \qquad r+(I+J)\longmapsto (r+I)\otimes(1+J). \] If \(r\in I+J\), write \(r=i+j\) with \(i\in I\) and \(j\in J\). Then \[ (i+I)\otimes(1+J)=0, \qquad (j+I)\otimes(1+J)=(1+I)\otimes(j+J)=0, \] so \(\Psi\) is well-defined.
Now \[ \Phi\Psi(r+(I+J))=r+(I+J), \] and, using the balancing relation, \[ \Psi\Phi\big((r+I)\otimes(s+J)\big) =(rs+I)\otimes(1+J) =(r+I)\otimes(s+J). \] Hence \(\Phi\) and \(\Psi\) are inverse isomorphisms.
2. Applying 1 with \(R=\mathbb{C}[x]\), \(I=(x^n)\), and \(J=(x^m)\) gives\[ \mathbb{C}[x]/(x^n)\otimes_{\mathbb{C}[x]}\mathbb{C}[x]/(x^m) \cong \mathbb{C}[x]/(x^{\min(m,n)}). \]
In \(\mathbb{C}[x]\), \[ (x^n)+(x^m)=(x^{\min(m,n)}). \] Substitute this into 1.
3. Thus the decomposition in part (a) exists with\[ k=1, \qquad a_1=\min(m,n), \]and these values answer part (b).
The module on the right side of 2 is already one summand of the required form \(\mathbb{C}[x]/(x^{a_1})\), so taking \(a_1=\min(m,n)\) gives the desired decomposition.