A partition with too many rows vanishes at fixed rank

$S_{(1,1,1)}({\mathbb{C}}^2)=0$, because $\mathrm{\ell }((1,1,1))=3>2=\dim V$ [F1]. In rank $3$ the same module is nonzero, since $S_{(1,1,1)}({\mathbb{C}}^3)={\mathrm{\Lambda }}^3{\mathbb{C}}^3\cong \mathbb{C}$ [F2]; so the vanishing is a fixed-rank phenomenon, not a vanishing of the coefficient.

The coefficient is nonzero: $c_{(1,1),(1)}^{(1,1,1)}=1$, because the skew diagram $(1,1,1)\mathrm{/}(1,1)$ is the single box $(3,1)$, the unique semistandard tableau of that shape and content $(1)$ carries the letter $1$, and its reading word $1$ is a lattice word. Equivalently, vertical Pieri [F3] assigns coefficient one to this shape; at rank $2$ its Schur module is zero, while at rank $3$ it is a nonzero summand.

At rank $r=2$ the honest decomposition is $S_{(1,1)}({\mathbb{C}}^2)\otimes {\mathbb{C}}^2\cong S_{(2,1)}({\mathbb{C}}^2)$ without the summand $S_{(1,1,1)}({\mathbb{C}}^2)=0$ of step 1.1; its dimensions follow directly from tableaux: shape $(1,1)$ on two letters has its unique column $1,2$, while shape $(2,1)$ has that forced first column and a top-right entry $1$ or $2$ (Semistandard tableaux expand Schur characters). Hence $\dim S_{(1,1)}({\mathbb{C}}^2)\cdot \dim {\mathbb{C}}^2=1\cdot 2=2=\dim S_{(2,1)}({\mathbb{C}}^2)$, so no room remains for a second nonzero summand. Hence the term with $\mathrm{\ell }(\nu )>r$ is a zero module: suppressing the bound $\mathrm{\ell }(\nu )\le r$ and claiming that every $\nu$ with $c_{\lambda \mu }^{\nu }\mathrm{\ne }0$ contributes a nonzero summand of the tensor product is false, although the coefficient itself is $1$ and the corresponding stable statement at large rank is true.

Consistently, the rank-$2$ Schur polynomial vanishes, $s_{(1,1,1)}(x_1,x_2)=0$, whereas $s_{(1,1)}{s}_{(1)}=s_{(2,1)}+s_{(1,1,1)}$ holds as an identity of symmetric functions; the specialization to two variables drops the second term by definition, and at rank $3$ it is a genuine summand.