Rank-one cyclotomic cell modules and their forms

For $(1,\mathrm{\varnothing })$ the cyclotomic factor is $X-b$ and the row subgroup is trivial; for $(\mathrm{\varnothing },1)$ the factor is empty and the row subgroup is trivial. Thus [F2] gives the displayed Murphy elements. In multipartition dominance the first component prefix is $1$ for $(1,\mathrm{\varnothing })$ and $0$ for $(\mathrm{\varnothing },1)$, so the stated order holds.

Expanding the relation gives $X^2=(a+b)X-ab$, hence $X^{-1}=(a+b-X)\mathrm{/}(ab)$. Therefore the Laurent quotient equals the polynomial quotient $F[X]\mathrm{/}((X-a)(X-b))$. Division by this monic quadratic gives a unique remainder of degree less than $2$, so $1,X$ is a basis. Since $X=m_1+b{m}_2$, the change from $1,X$ to $m_1,m_2$ has determinant $-1$; these Murphy elements are also a basis.

Direct multiplication gives$(X-b)X=X^2-bX=a(X-b)\mathrm{.}$ Thus for any $h=A(X-b)+B$ in the basis of step 2.1, $(X-b)h=(A(a-b)+B)(X-b)$, which is the fixed-row law in the upper cell. In the lower cell quotient, $X=(X-b)+b\equiv b$ modulo the upper ideal, so $1\cdot h\equiv h(b)\cdot 1$. The involution fixes $X$ and hence both Murphy elements. These two calculations verify the cell axiom for both labels, so the displayed basis is cellular.

The upper form scalar is the coefficient in$(X-b{)}^2=(a-b)(X-b),$ so it is $a-b$. The lower basis element satisfies $1\cdot 1=1$, giving form scalar $1$.

If $a=b=c\mathrm{\ne }0$, then the defining polynomial is $(X-c{)}^2$. Translation by $c$ identifies the quotient with $F[ϵ]\mathrm{/}({ϵ}^2)$, where $ϵ=X-c$ is nonzero by the basis of step 2.1. The upper cell form is zero and the lower form is $1$. To see directly that there is one simple module, let $S$ be simple. The subspace $ϵS$ is a submodule. If it is nonzero then $ϵS=S$, which implies ${ϵ}^{2}S=S$, contradicting ${ϵ}^2=0$ and $S\mathrm{\ne }0$. Thus $ϵS=0$, and $S$ is a simple module over $H\mathrm{/}(ϵ)\cong F$, hence one-dimensional.