Can a vector field $\mathbf{A}: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ always be expressed as the sum of a solenoidal (divergenceless) part $\mathbf{A}_S$, where $\nabla \cdot \mathbf{A}_S = 0$, and an irrotational part $\mathbf{A}_R$, where $\nabla\times\mathbf{A}_R = 0$?
$$ “\,\forall{\mathbf{A}}\; \exists{\mathbf{A}_S}\; \exists{\mathbf{A}_R} : \mathbf{A} = \mathbf{A}_S + \mathbf{A}_R \;\wedge\; \nabla \cdot \mathbf{A}_S = 0 \;\wedge\; \nabla \times \mathbf{A}_R = 0 \,”\,? $$
Since irrotational vector fields are gradients of some scalar field ($\mathbf{A}_R = \nabla R$), this is equivalent to asking:
$$ “\, \forall\mathbf{A}\; \exists\mathbf{A}_S\; \exists R \in \left(\mathbb{R}^3\rightarrow\mathbb{R}\right) : \mathbf{A} = \mathbf{A}_S + \nabla R \;\wedge\; \nabla\cdot\mathbf{A}_S = 0 \,”\,? $$
What’s a proof of the result? If true, are $\mathbf{A}_R$ and $\mathbf{A}_S$ unique? If it’s not always true, then what kind of fields can/cannot be decomposed in this way?
If true, this would be a very nice result: any vector field could then be thought of in terms of its unique solenoidal and irrotational components.