Published online by Cambridge University Press: 20 November 2018
In this note we prove the following surprising characterization: if   $X\,\subset \,{{\mathbb{A}}^{n}}$  is an (embedded, non-empty, proper) algebraic variety deûned over a field
 $X\,\subset \,{{\mathbb{A}}^{n}}$  is an (embedded, non-empty, proper) algebraic variety deûned over a field   $k$  of characteristic zero, then
 $k$  of characteristic zero, then   $X$  is a hypersurface if and only if the module
 $X$  is a hypersurface if and only if the module   ${{T}_{{{O}_{{{\mathbb{A}}^{n\,/k}}}}}}(X)$  of logarithmic vector fields of
 ${{T}_{{{O}_{{{\mathbb{A}}^{n\,/k}}}}}}(X)$  of logarithmic vector fields of   $X$  is a reflexive
 $X$  is a reflexive   ${{O}_{{{\mathbb{A}}^{n}}}}$ -module. As a consequence of this result, we derive that if
 ${{O}_{{{\mathbb{A}}^{n}}}}$ -module. As a consequence of this result, we derive that if   ${{T}_{{{O}_{{{\mathbb{A}}^{n\,/k}}}}}}(X)$  is a free
 ${{T}_{{{O}_{{{\mathbb{A}}^{n\,/k}}}}}}(X)$  is a free   ${{O}_{{{\mathbb{A}}^{n}}}}$ -module, which is shown to be equivalent to the freeness of the
 ${{O}_{{{\mathbb{A}}^{n}}}}$ -module, which is shown to be equivalent to the freeness of the   $t$ -th exterior power of
 $t$ -th exterior power of   ${{T}_{{{O}_{{{\mathbb{A}}^{n\,/k}}}}}}(X)$  for some (in fact, any)
 ${{T}_{{{O}_{{{\mathbb{A}}^{n\,/k}}}}}}(X)$  for some (in fact, any)   $t\,\le \,n$ , then necessarily
 $t\,\le \,n$ , then necessarily   $X$  is a Saito free divisor.
 $X$  is a Saito free divisor.