Let \(R\) be a commutative algebra over \(\mathbb C\). A derivation of \(R\) is a \(\mathbb C\)-linear map \(D:R\rightarrow R\) such that (i) \(D(1)=0\) and (ii) \(D(ab)=D(a)b+aD(b)\) for all \(a,b\in R\).
Describe all derivations of the polynomial ring \(\mathbb C[x]\).
Let \(A\) be the subring (or \(\mathbb C\)-subalgebra) of \(\mathrm{End}_{\mathbb C}(\mathbb C[x])\) generated by all derivations of \(\mathbb C[x]\) and the left multiplications by \(x\). Prove that \(\mathbb C[x]\) is a simple left \(A\)-module. > Note that the inclusion \(A\rightarrow\mathrm{End}_{\mathbb C}(\mathbb C[x])\) defines a natural left \(A\)-module structure on \(\mathbb C[x]\).
Solution
Goal. Describe all derivations of \(\CC[x]\), and show \(\CC[x]\) is a simple left \(A\)-module.
1. All derivations of \(\CC[x]\).
1.1. A derivation \(D\) is determined by \(D(x)\).
\(D(1) = 0\) and the Leibniz rule give \(D(x^n) = n x^{n-1} D(x)\) by induction, so \(D\) is determined by its value on the generator \(x\).
1.2. Conversely, for any \(p(x) \in \CC[x]\), the map \(D_p(f) = p(x) f'(x)\) is a derivation.
\(D_p(1) = 0\) and \(D_p(fg) = p(f'g + fg') = D_p(f)g + fD_p(g)\).
1.3. Hence the derivations of \(\CC[x]\) are exactly \(\theset{p(x)\frac{d}{dx} : p \in \CC[x]}\).
1.1 and 1.2.
2. \(\CC[x]\) is a simple left \(A\)-module.
2.1. \(A\) contains the derivation \(\frac{d}{dx}\) and the multiplication operator \(x\).
\(\frac{d}{dx}\) is a derivation (with \(p = 1\)), and \(x\) is a left multiplication.
2.2. The operators \(x\) and \(\frac{d}{dx}\) satisfy \([\frac{d}{dx}, x] = 1\).
\(\frac{d}{dx}(x f) - x \frac{d}{dx}(f) = f + x f' - x f' = f\), so \([\frac{d}{dx}, x] = \operatorname{id}\).
2.3. \(A\) is the Weyl algebra \(A_1(\CC)\).
\(A\) is generated by \(x\) and \(\frac{d}{dx}\) subject to the relation \([\frac{d}{dx}, x] = 1\), which is the defining presentation of the Weyl algebra.
2.4. \(\CC[x]\) is a simple \(A_1(\CC)\)-module.
the Weyl algebra acts on \(\CC[x]\) with \(x\) acting by multiplication and \(\frac{d}{dx}\) by differentiation; this module is simple (any nonzero submodule contains a nonzero polynomial, and applying \(\frac{d}{dx}\) repeatedly gives a nonzero constant, hence \(1\), hence the whole module).
2.5. Hence \(\CC[x]\) is a simple left \(A\)-module.
2.3 and 2.4.
3. Q.E.D.
1 describes the derivations; 2 proves simplicity.