James submodule theorem over every field

Let $U\le M_F^{\lambda }$ be an $F[S_n]$-submodule. If ${\kappa }_{t}U\mathrm{\ne }0$ for some $\lambda$-tableau $t$, then ${\kappa }_{t}U$ is a nonzero subspace of ${\kappa }_t{M}_F^{\lambda }=F{e}_t$ by [F3], hence ${\kappa }_{t}U=F{e}_t$ and $e_t={\kappa }_{t}u\in U$ for some $u\in U$; since $S_F^{\lambda }=F[S_n]e_t$ by [F4] and $U$ is a submodule, $S_F^{\lambda }\le U$. If instead ${\kappa }_{t}U=0$ for every $\lambda$-tableau $t$, then for every $u\in U$ and every $t$, using ${\kappa }_t\{t\}=e_t$ and the self-adjointness of ${\kappa }_t$ from [F2], ${\beta }_F(u,e_t)={\beta }_F(u,{\kappa }_t\{t\})={\beta }_F({\kappa }_{t}u,\{t\})=0,$ so $u$ is orthogonal to every polytabloid and $U\le (S_F^{\lambda }{)}^{\perp }$. In either case one of the two alternatives of the James submodule theorem holds.

Over the arbitrary field $F$ put $R_F^{\lambda }:=S_F^{\lambda }\cap (S_F^{\lambda }{)}^{\perp }$ and $D_F^{\lambda }:=S_F^{\lambda }\mathrm{/}{R}_F^{\lambda }$, the same formulas as in the modular definition [F5]. Let $\phi :S_F^{\lambda }\to (S_F^{\lambda }{)}^{\ast }$ be $\phi (v)={\beta }_F(v,\cdot ){\mathrm{\mid }}_{S_F^{\lambda }}$. Its kernel is $S_F^{\lambda }\cap (S_F^{\lambda }{)}^{\perp }=R_F^{\lambda }$, because ${\beta }_F(v,s)=0$ for all $s\in S_F^{\lambda }$ says exactly that $v\in (S_F^{\lambda }{)}^{\perp }$ when $v\in S_F^{\lambda }$; in the standard basis of $S_F^{\lambda }$ paired with its dual basis the matrix of $\phi$ is the Gram matrix $G_F$ of ${\beta }_F$, by [F2] and [F1]. Hence by [F7] the rank of $\phi$ equals ${\mathrm{rank}}_F{G}_F$ and rank-nullity gives ${\dim }_F{S}_F^{\lambda }={\mathrm{rank}}_F{G}_F+{\dim }_F{R}_F^{\lambda },{\dim }_F{D}_F^{\lambda }={\mathrm{rank}}_F{G}_F\mathrm{.}$ In particular $D_F^{\lambda }=0$ exactly when every entry of the Gram matrix vanishes in $F$.

Let $E\mathrm{/}F$ be a field extension. By [F1] applied over the commutative rings $F$ and $E$, the identifications $M_E^{\lambda }\cong E{\otimes }_{\mathbb{Z}}{M}_{\mathbb{Z}}^{\lambda }$ and $M_F^{\lambda }\cong F{\otimes }_{\mathbb{Z}}{M}_{\mathbb{Z}}^{\lambda }$ give $E{\otimes }_F{M}_F^{\lambda }\cong M_E^{\lambda }$, and likewise $E{\otimes }_F{S}_F^{\lambda }\cong S_E^{\lambda }$ with standard bases matched; by [F2] the form ${\beta }_E$ is the scalar extension of ${\beta }_F$. For $r\in R_F^{\lambda }$ and $s\in S_F^{\lambda }$ one has ${\beta }_F(r,s)=0$, hence ${\beta }_E(1\otimes r,1\otimes s)=1\otimes {\beta }_F(r,s)=0$, and since $E{\otimes }_F{R}_F^{\lambda }\subseteq E{\otimes }_F{S}_F^{\lambda }=S_E^{\lambda }$ this gives $E{\otimes }_F{R}_F^{\lambda }\subseteq (S_E^{\lambda }{)}^{\perp }\cap S_E^{\lambda }=R_E^{\lambda }$. Therefore the natural map $E{\otimes }_F{D}_F^{\lambda }=E{\otimes }_F\left(S_F^{\lambda }\mathrm{/}{R}_F^{\lambda }\right)⟶S_E^{\lambda }\mathrm{/}\left(E{\otimes }_F{R}_F^{\lambda }\right)⟶D_E^{\lambda }$ is a surjection of finite-dimensional $E$-vector spaces.

Let $N\le S_F^{\lambda }$ be an $F[S_n]$-submodule. Applying step 1.1 to $N$ viewed as a submodule of $M_F^{\lambda }$ gives $S_F^{\lambda }\le N$ or $N\le (S_F^{\lambda }{)}^{\perp }$, and in the second case $N\le (S_F^{\lambda }{)}^{\perp }\cap S_F^{\lambda }=R_F^{\lambda }$. Hence every proper submodule of $S_F^{\lambda }$ is contained in $R_F^{\lambda }$. Moreover $R_F^{\lambda }$ is itself an $F[S_n]$-submodule: if $x\in (S_F^{\lambda }{)}^{\perp }$, $v\in S_F^{\lambda }$ and $\sigma \in S_n$, then invariance in [F2] gives ${\beta }_F(\sigma x,v)={\beta }_F(x,{\sigma }^{-1}v)=0$ because ${\sigma }^{-1}v\in S_F^{\lambda }$, so $(S_F^{\lambda }{)}^{\perp }$ is a submodule, and $R_F^{\lambda }$ is the intersection of two submodules.

By step 1.2 applied over $F$ and over $E$, ${\dim }_F{D}_F^{\lambda }={\mathrm{rank}}_F{G}_F$ and ${\dim }_E{D}_E^{\lambda }={\mathrm{rank}}_E{G}_E$. The rank of the integer matrix $G_{\lambda }$ over a field is the largest $m$ for which some $m\times m$ minor has nonzero image in that field; a minor is an integer, its image vanishes over $F$ if and only if it vanishes over $E$, and $F$ and $E$ have the same characteristic, so ${\mathrm{rank}}_F{G}_F={\mathrm{rank}}_E{G}_E$. Since ${\dim }_E(E{\otimes }_F{D}_F^{\lambda })={\dim }_F{D}_F^{\lambda }$, the surjection of step 1.3 is an isomorphism $D_E^{\lambda }\cong E{\otimes }_F{D}_F^{\lambda }\mathrm{.}$

Define $\bar{\beta }:D_F^{\lambda }\times D_F^{\lambda }\to F$ by $\bar{\beta }(x+R_F^{\lambda },y+R_F^{\lambda })={\beta }_F(x,y)$. This is well defined: replacing $x$ by $x+r$ and $y$ by $y+s$ with $r,s\in R_F^{\lambda }\subseteq (S_F^{\lambda }{)}^{\perp }$ changes the value by ${\beta }_F(r,y)+{\beta }_F(x,s)+{\beta }_F(r,s)=0$. The form $\bar{\beta }$ is symmetric and $S_n$-invariant, inherited from ${\beta }_F$ by [F2], and it is nondegenerate: if $\bar{\beta }(x+R_F^{\lambda },y+R_F^{\lambda })=0$ for all $y\in S_F^{\lambda }$, then $x\in (S_F^{\lambda }{)}^{\perp }\cap S_F^{\lambda }=R_F^{\lambda }$. Hence the $F$-linear map $\bar{\phi }:D_F^{\lambda }\to (D_F^{\lambda }{)}^{\ast }$, $\bar{\phi }(\xi )=\bar{\beta }(\xi ,\cdot )$, is injective; since ${\dim }_F{D}_F^{\lambda }={\dim }_F(D_F^{\lambda }{)}^{\ast }$ by step 1.2, it is an isomorphism of vector spaces, and for $\sigma \in S_n$ and $\eta \in D_F^{\lambda }$, $\bar{\phi }(\sigma \xi )(\eta )=\bar{\beta }(\sigma \xi ,\eta )=\bar{\beta }(\xi ,{\sigma }^{-1}\eta )=(\sigma \cdot \bar{\phi }(\xi ))(\eta ),$ so $\bar{\phi }$ is $F[S_n]$-linear and $D_F^{\lambda }\cong (D_F^{\lambda }{)}^{\ast }$; this includes the case $D_F^{\lambda }=0$, where both sides are zero.

Suppose $D_F^{\lambda }\mathrm{\ne }0$, so $R_F^{\lambda }\mathrm{\ne }{S}_F^{\lambda }$. By step 2.1 every proper submodule of $S_F^{\lambda }$ lies in $R_F^{\lambda }$, while $R_F^{\lambda }$ is itself a proper submodule by assumption; hence $R_F^{\lambda }$ contains every proper submodule and is therefore the unique maximal submodule of $S_F^{\lambda }$. By [F6] the module radical equals the intersection of the maximal submodules, so $\mathrm{rad}(S_F^{\lambda })=R_F^{\lambda }$; in particular $D_F^{\lambda }=S_F^{\lambda }\mathrm{/}\mathrm{rad}(S_F^{\lambda })$ is the head of $S_F^{\lambda }$. If $W\le D_F^{\lambda }$ is a submodule and $N\le S_F^{\lambda }$ is its preimage, then $R_F^{\lambda }\le N$ and either $N=S_F^{\lambda }$, giving $W=D_F^{\lambda }$, or $N$ is proper, in which case $N\le R_F^{\lambda }$ by step 2.1 and hence $N=R_F^{\lambda }$, giving $W=0$; thus $D_F^{\lambda }$ is simple.

Let $E\mathrm{/}F$ be a field extension and suppose $D_F^{\lambda }\mathrm{\ne }0$. By step 2.2, $D_E^{\lambda }\cong E{\otimes }_F{D}_F^{\lambda }$ is nonzero. The arguments of steps 1.1, 2.1 and 3.1 apply verbatim with $F$ replaced by the field $E$: [F3] holds over every field, [F2] holds over every commutative ring, and the alternative of step 1.1 and its consequence step 2.1 use only those facts, so $R_E^{\lambda }$ is the unique maximal submodule of $S_E^{\lambda }$ and $D_E^{\lambda }$ is simple. As $E\mathrm{/}F$ was arbitrary, $D_F^{\lambda }$ is absolutely irreducible.

Step 1.1 is the James submodule theorem. If $D_F^{\lambda }=0$ then the consequences are vacuous, and $D_F^{\lambda }=0\cong (D_F^{\lambda }{)}^{\ast }$ is self-dual; if $D_F^{\lambda }\mathrm{\ne }0$, then step 3.1 gives the unique maximal submodule, the identification $R_F^{\lambda }=\mathrm{rad}(S_F^{\lambda })$ and the simple head, step 2.3 gives self-duality, and step 4.1 gives absolute irreducibility. For $n=0$ and $\lambda =\mathrm{\varnothing }$ there is one tabloid and one polytabloid with ${\beta }_F(e,e)=1$ and $C_t=\{1\}$, so $R_F^{\lambda }=0$, $D_F^{\lambda }\cong F$ is the trivial module, simple and absolutely irreducible, and all assertions hold; the argument above never divides by a group order or uses positivity.