 Solution

The equality ${\sum }_{\lambda }(\dim S^{\lambda }{)}^2=n!$ exhausts the dimension of the regular representation. There is no room for another irreducible constituent, since it would contribute its positive squared dimension. ThusEach has the standard-polytabloid basis just proved. The usual split characteristic-zero statement follows over any characteristic-zero field because these integral Specht models remain irreducible after extension to an algebraic closure and already exhaust all dimensions there. In the literal wording of the source, polytabloids themselves are vectors; it is the Specht modules with those bases that form the family of representations.