The branching filtration need not split in modular characteristic

By [F2] the tabloids $v_1,v_2,v_3$ form a basis of $V$ and the transposition $\tau =(12)\in S_2$ acts by $\tau \cdot v_1=v_2$, $\tau \cdot v_2=v_1$ and $\tau \cdot v_3=v_3$: it permutes the labels $1,2$ of the singleton second row and fixes $3$.

Put $t=\begin{array}{cc}1& 2\\ 3& \end{array}$ and $u=\begin{array}{cc}1& 3\\ 2& \end{array}$, the two standard $(2,1)$-tableaux. Their column stabilizers are $C_t=\{1,(13)\}$ and $C_u=\{1,(12)\}$ by [F3], and the associated tabloids are $\{t\}=v_3$, $(13)\cdot \{t\}=v_1$, $\{u\}=v_2$, $(12)\cdot \{u\}=v_1$. Since all signs equal $1$ in $K$ by [F1], [F3] gives $e_t=v_3+v_1$ and $e_u=v_2+v_1$, and these two vectors are linearly independent by their coefficients at the basis vectors $v_3$ and $v_2$. For any $(2,1)$-tableau with first row $(a,b)$ and second row $(c)$, the column stabilizer is $\{1,(ac)\}$, so its polytabloid is $v_c+v_a$. Each such pair sum lies in the span of the two displayed vectors: the only other pair sum is $v_2+v_3=(v_2+v_1)+(v_3+v_1)$ in characteristic two. Thus all polytabloids lie in this span, and $S_K^{(2,1)}={\mathrm{span}}_K\{v_3+v_1,v_2+v_1\}$ is two-dimensional.

The fixed space of $\tau$ on $S_K^{(2,1)}$ is one-dimensional: writing $x=a(v_3+v_1)+b(v_2+v_1)=(a+b)v_1+b{v}_2+a{v}_3$ with $a,b\in K$, step 1.1 gives $\tau \cdot x=(a+b)v_2+b{v}_1+a{v}_3=b{v}_1+(a+b)v_2+a{v}_3$, and $\tau \cdot x=x$ forces $a+b=b$ and $b=a+b$, that is $a=0$; conversely every $x=b(v_2+v_1)$ is fixed. So the fixed space is ${\mathrm{span}}_K\{v_2+v_1\}$, of dimension one.

The removable rows of $(2,1)$ are $r_1=1$ and $r_2=2$, with ${\lambda }^{(1)}=(1,1)$ and ${\lambda }^{(2)}=(2)$. By [F4] the restriction of $S_K^{(2,1)}$ to $S_2$ has the filtration $0⊊V_1⊊V_2=S_K^{(2,1)}$ with $V_1\mathrm{/}{V}_0\cong S_K^{(1,1)}$ and $V_2\mathrm{/}{V}_1\cong S_K^{(2)}$, where $V_1$ is spanned by the standard polytabloids whose tableaux have largest label $3$ in row $r_1=1$, that is $V_1={\mathrm{span}}_K\{e_u\}={\mathrm{span}}_K\{v_2+v_1\}$. Both quotient modules are one-dimensional over $K$ and trivial for $S_2$: $S_K^{(2)}$ is spanned by the unique $(2)$-tabloid, on which $S_2$ acts trivially, and $S_K^{(1,1)}$ is spanned by $e_w$ for a column tableau $w$, which is invariant under the transposition because the two $(1,1)$-tabloids are exchanged; in particular the submodule $V_1$ is exactly the fixed space computed in step 2.1.

Suppose the restriction were the direct sum of the two removable-corner factors, that is $S_K^{(2,1)}\cong S_K^{(1,1)}\oplus S_K^{(2)}$ as $S_2$-modules. Since both summands are trivial by step 3.1, the right-hand side would be a two-dimensional trivial $S_2$-module, on which $\tau$ acts as the identity by [F5], so every vector of $S_K^{(2,1)}$ would be fixed by $\tau$. This contradicts the fixed space computed in step 2.1, which is one-dimensional. Equivalently, $V_1$ equals the full fixed space, so an $S_2$-complement to $V_1$ would be a submodule contained in the fixed space $V_1$ and hence would be zero, showing that the extension $0\to V_1\to S_K^{(2,1)}\to S_K^{(2,1)}\mathrm{/}{V}_1\to 0$ of two trivial one-dimensional $S_2$-modules does not split. Thus the removable-corner filtration of $S_K^{(2,1)}$ over $K$ of characteristic two is a nonsplit extension of the two one-dimensional Specht factors $S_K^{(1,1)}$ and $S_K^{(2)}$, both trivial for $S_2$, and the field-independent branching rule is refuted.