Specht modules classify the irreducibles over every field of characteristic zero

Fix $\lambda ⊢n$. For every field $E$ the standard polytabloids are an $E$-basis of $S_E^{\lambda }$, so ${\dim }_E{S}_E^{\lambda }=f^{\lambda }\ge 1$ and $S_E^{\lambda }\mathrm{\ne }0$. If $K\mathrm{/}E$ is a field extension, the coefficient inclusion $S_E^{\lambda }\to S_K^{\lambda }$ is $E$-linear and sends $e_t$ to $e_t$, so it induces a $K$-linear map $K{\otimes }_E{S}_E^{\lambda }\to S_K^{\lambda }$ with $1\otimes e_t\mapsto e_t$; this map carries the $K$-basis $(1\otimes e_t{)}_{t\text{ standard}}$ of $K{\otimes }_E{S}_E^{\lambda }$ to the $K$-basis $(e_t{)}_{t\text{ standard}}$ of $S_K^{\lambda }$ and is compatible with the actions because both actions are given by the same formulas on tabloids. Hence $K{\otimes }_E{S}_E^{\lambda }\cong S_K^{\lambda }$; in particular $S_{\mathbb{C}}^{\lambda }\cong \mathbb{C}{\otimes }_{\mathbb{Q}}{S}_{\mathbb{Q}}^{\lambda }$ and $F{\otimes }_{\mathbb{Q}}{S}_{\mathbb{Q}}^{\lambda }\cong S_F^{\lambda }$.

Let $V,W$ be finite-dimensional $E{S}_n$-modules and $K\mathrm{/}E$ a field extension. Choose $E$-bases and identify ${\mathrm{Hom}}_E(V,W)$ with the space of $k\times m$ matrices, $m={\dim }_{E}V$, $k={\dim }_{E}W$; an element is $E{S}_n$-linear if and only if ${\rho }_W(\sigma )A=A{\rho }_V(\sigma )$ for all $\sigma \in S_n$, a finite system of linear equations with coefficients in $E$. Its solution space is the kernel of a matrix with entries in $E$, and Gaussian elimination over $E$ shows that the rank of such a matrix, hence the dimension of the solution space, is the same over $E$ and over $K$; the $K$-solutions are exactly the $K$-span of the $E$-solutions. In the chosen coordinates this says that the natural map $K{\otimes }_E{\mathrm{Hom}}_{E{S}_n}(V,W)\to {\mathrm{Hom}}_{K{S}_n}(K{\otimes }_{E}V,K{\otimes }_{E}W)$ is an isomorphism.

Taking $E=\mathbb{Q}$, $K=\mathbb{C}$ in $1.2$ with $V=W=S_{\mathbb{Q}}^{\lambda }$ and using $S_{\mathbb{C}}^{\lambda }\cong \mathbb{C}{\otimes }_{\mathbb{Q}}{S}_{\mathbb{Q}}^{\lambda }$ gives ${\dim }_{\mathbb{Q}}{\mathrm{End}}_{\mathbb{Q}{S}_n}(S_{\mathbb{Q}}^{\lambda })={\dim }_{\mathbb{C}}{\mathrm{End}}_{\mathbb{C}{S}_n}(S_{\mathbb{C}}^{\lambda })=1$ by [F3, F6]; as the scalars lie in ${\mathrm{End}}_{\mathbb{Q}{S}_n}(S_{\mathbb{Q}}^{\lambda })$, that endomorphism ring is $\mathbb{Q}$. For $\lambda \mathrm{\ne }\mu$, $1.2$ gives $\mathbb{C}{\otimes }_{\mathbb{Q}}{\mathrm{Hom}}_{\mathbb{Q}{S}_n}(S_{\mathbb{Q}}^{\lambda },S_{\mathbb{Q}}^{\mu })\cong {\mathrm{Hom}}_{\mathbb{C}{S}_n}(S_{\mathbb{C}}^{\lambda },S_{\mathbb{C}}^{\mu })=0$, so ${\mathrm{Hom}}_{\mathbb{Q}{S}_n}(S_{\mathbb{Q}}^{\lambda },S_{\mathbb{Q}}^{\mu })=0$.

Suppose $0\mathrm{\ne }U\subseteq S_{\mathbb{Q}}^{\lambda }$ is a proper $\mathbb{Q}{S}_n$-submodule. By [F4, F5] there is a $\mathbb{Q}{S}_n$-submodule $V\mathrm{\ne }0$ with $S_{\mathbb{Q}}^{\lambda }=U\oplus V$, and the projection $\pi$ onto $U$ along $V$ is a $\mathbb{Q}{S}_n$-endomorphism that is not a scalar: $\pi$ acts as $1$ on $U$ and as $0$ on $V$. This contradicts step 2.1, so $S_{\mathbb{Q}}^{\lambda }$ is irreducible.

Applying step 1.2 with $E=\mathbb{Q}$, $K=F$ and step 2.1 gives ${\mathrm{End}}_{F{S}_n}(S_F^{\lambda })\cong F{\otimes }_{\mathbb{Q}}{\mathrm{End}}_{\mathbb{Q}{S}_n}(S_{\mathbb{Q}}^{\lambda })=F$ and ${\mathrm{Hom}}_{F{S}_n}(S_F^{\lambda },S_F^{\mu })=0$ for $\lambda \mathrm{\ne }\mu$.

Let $W$ be a simple $\mathbb{Q}{S}_n$-module, so $W\mathrm{\ne }0$. For $0\mathrm{\ne }w\in W$ the map $\mathbb{Q}{S}_n\to W$, $x\mapsto xw$, is a nonzero $\mathbb{Q}{S}_n$-linear map onto the nonzero submodule $W$, so $W$ is a quotient of the finite-dimensional module $\mathbb{Q}{S}_n$ and ${\dim }_{\mathbb{Q}}W<\mathrm{\infty }$. Then $W_{\mathbb{C}}:=\mathbb{C}{\otimes }_{\mathbb{Q}}W$ is a nonzero finite-dimensional complex $S_n$-module, so by [F3] it has an irreducible submodule isomorphic to $S_{\mathbb{C}}^{\lambda }$ for some $\lambda ⊢n$; hence ${\mathrm{Hom}}_{\mathbb{C}{S}_n}(S_{\mathbb{C}}^{\lambda },W_{\mathbb{C}})\mathrm{\ne }0$. By step 1.2 and step 1.1 this is $\mathbb{C}{\otimes }_{\mathbb{Q}}{\mathrm{Hom}}_{\mathbb{Q}{S}_n}(S_{\mathbb{Q}}^{\lambda },W)$, so ${\mathrm{Hom}}_{\mathbb{Q}{S}_n}(S_{\mathbb{Q}}^{\lambda },W)\mathrm{\ne }0$. A nonzero map from the irreducible module $S_{\mathbb{Q}}^{\lambda }$ (step 3.1) to the simple module $W$ has zero kernel and image $W$, hence is an isomorphism [F6]; thus every simple $\mathbb{Q}{S}_n$-module is isomorphic to some $S_{\mathbb{Q}}^{\lambda }$.

Suppose $0\mathrm{\ne }U\subseteq S_F^{\lambda }$ is a proper $F{S}_n$-submodule. By [F4, F5] there is a complement $V\mathrm{\ne }0$, $S_F^{\lambda }=U\oplus V$, and the projection $\pi$ onto $U$ along $V$ lies in ${\mathrm{End}}_{F{S}_n}(S_F^{\lambda })$ and is not a scalar, since it acts as $1$ on $U\mathrm{\ne }0$ and as $0$ on $V\mathrm{\ne }0$. This contradicts step 3.2, so $S_F^{\lambda }$ is irreducible.

By [F4, F5] the finite-dimensional $\mathbb{Q}{S}_n$-module $\mathbb{Q}{S}_n$ is a finite direct sum of simple modules, and by step 4.1 each is isomorphic to some $S_{\mathbb{Q}}^{\lambda }$; collecting isomorphic summands and using the pairwise non-isomorphism from step 2.1 gives $\mathbb{Q}{S}_n\cong {⨁}_{\lambda ⊢n}(S_{\mathbb{Q}}^{\lambda }{)}^{m_{\lambda }}$ for multiplicities $m_{\lambda }\ge 0$. Each $m_{\lambda }\ge 1$: by step 3.1 and step 4.1 the module $S_{\mathbb{Q}}^{\lambda }$ is simple, and for $0\mathrm{\ne }v\in S_{\mathbb{Q}}^{\lambda }$ the surjection $\mathbb{Q}{S}_n\to S_{\mathbb{Q}}^{\lambda }$, $x\mapsto xv$, splits by [F4], exhibiting $S_{\mathbb{Q}}^{\lambda }$ as a direct summand of $\mathbb{Q}{S}_n$.

The canonical map $F{\otimes }_{\mathbb{Q}}\mathbb{Q}{S}_n\to F{S}_n$, $c\otimes [\sigma ]\mapsto c[\sigma ]$, sends the $F$-basis $(1\otimes [\sigma ]{)}_{\sigma \in S_n}$ of $F{\otimes }_{\mathbb{Q}}\mathbb{Q}{S}_n$ to the basis $([\sigma ]{)}_{\sigma \in S_n}$ of $F{S}_n$ and is compatible with multiplication by [F5]; hence it is an isomorphism of $F{S}_n$-modules. Applying $F{\otimes }_{\mathbb{Q}}-$ to step 5.1 and using [F7] together with step 1.1 gives $F{S}_n\cong F{\otimes }_{\mathbb{Q}}\mathbb{Q}{S}_n\cong {⨁}_{\lambda ⊢n}(F{\otimes }_{\mathbb{Q}}{S}_{\mathbb{Q}}^{\lambda }{)}^{m_{\lambda }}\cong {⨁}_{\lambda ⊢n}(S_F^{\lambda }{)}^{m_{\lambda }}$ with all $m_{\lambda }\ge 1$.

Let $W$ be a simple $F{S}_n$-module, so $W\mathrm{\ne }0$. For $0\mathrm{\ne }w\in W$ the map $F{S}_n\to W$, $x\mapsto xw$, is a nonzero $F{S}_n$-linear map onto $W$; hence ${\dim }_{F}W<\mathrm{\infty }$ and $W$ is a quotient of $F{S}_n$. By step 6.1, $F{S}_n\cong {⨁}_{\lambda ⊢n}(S_F^{\lambda }{)}^{m_{\lambda }}$ with $m_{\lambda }\ge 1$, so some component $(S_F^{\lambda }{)}^{m_{\lambda }}\to W$ of the quotient map is nonzero, and then some single copy $S_F^{\lambda }\to W$ is nonzero. The module $S_F^{\lambda }$ is irreducible (step 4.2), so this nonzero map is an isomorphism [F6]; hence $W\cong S_F^{\lambda }$.

Steps 1.1, 3.1 and 4.2 show that every $S_F^{\lambda }$ is nonzero and irreducible over $F$ (and over $\mathbb{Q}$, with the complex case being [F3]); steps 2.1 and 3.2 show that distinct partitions give modules with no nonzero homomorphism between them, and steps 4.1 and 7.1 show over $\mathbb{Q}$ and over $F$ that every irreducible module is isomorphic to one of them. Finally, a finite-dimensional $F{S}_n$-module decomposes into simple submodules by [F4, F5] applied repeatedly (each proper submodule splits off a nonzero complement, and the dimension strictly decreases), and each simple summand is isomorphic to some $S_F^{\lambda }$ by step 7.1. Thus the $S_F^{\lambda }$ are a complete irredundant list of irreducible $F{S}_n$-modules, and over $F=\mathbb{C}$ this is the published complex classification [F3].