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Let $U$ be a submodule of the Young permutation module $M^{\lambda }$ over any field. Then eitherfor the tabloid bilinear form. The key identity is $b_{t}u=⟨u,e(t)⟩e(t)$: if one pairing is nonzero, the cyclic generator $e(t)$ and hence the whole Specht module lies in $U$.

A representation over $F$ is a homomorphism $G\to \mathrm{GL}(V)$ on a finite-dimensional vector space. It is irreducible if it has no nonzero proper invariant subspace, and has absolute irreducibility of a group representation if it remains irreducible after every field extension. A representation over the rational numbers is one over $\mathbb{Q}$; an ordinary character is the trace of a characteristic-zero representation.

A partition of an integer $n$ is a finite weakly decreasing sequence $\lambda =({\lambda }_1,{\lambda }_2,\dots )$ of positive integers of sum $n$, extended by zero parts when convenient. Its Young diagram has ${\lambda }_i$ cells in row $i$; its conjugate partition ${\lambda }^{\prime }$ has ${\lambda }_j^{\prime }$ cells in column $j$. Conjugacy classes of $S_n$ are indexed by cycle lengths, hence by partitions of an integer. The number of ordinary irreducible characters equals the number of conjugacy classes; we now construct that many mutually distinct representations over the rational numbers.

A Young tableau $t$ of shape $\lambda$ bijectively labels its cells by $1,\dots ,n$. Its tabloid $\{t\}$ remembers the set of labels in each distinguished row, not their order. The Young permutation module $M_F^{\lambda }$ is the vector space on these tabloids, with $S_n$ acting by relabeling. Its tabloid bilinear form makes the tabloid basis orthonormal. Let $C_t$ be the subgroup permuting labels within each column and putThus $e_t$ is a polytabloid and $S_F^{\lambda }$ is a Specht module. Relabeling takes $e_t$ to $e_{gt}$, so any polytabloid generates the module. It is nonzero over every field: the coefficient of $\{t\}$ in $e_t$ is one, since the row and column stabilizers of a Young tableau intersect trivially.

Here is the elementary identity behind irreducibility. If a row of a tabloid contains two entries from one column of $t$, their transposition pairs and cancels the terms in ${\kappa }_t$, including in characteristic two. Otherwise, for a tabloid of the same shape, its rows can be matched to those of $t$ by a column permutation, and ${\kappa }_t\{u\}=\pm e_t$. Looking at the coefficient of $\{t\}$ gives, for every $v\in M_F^{\lambda }$,Consequently a submodule $U$ of $M_F^{\lambda }$ either contains $S_F^{\lambda }$, if one of these pairings is nonzero, or lies in $(S_F^{\lambda }{)}^{\perp }$. This proves the James submodule theorem used below.

Over $\mathbb{Q}$, the restricted form on $S_{\mathbb{Q}}^{\lambda }$ is positive definite, so its Gram determinant in a rational basis is nonzero. Extending scalars to any characteristic-zero field preserves that nonzero determinant and hence keeps the restricted form a nondegenerate bilinear form. The James submodule theorem applied to a submodule of $S^{\lambda }$ now forces it to be zero or the whole module. Therefore $S_{\mathbb{Q}}^{\lambda }$ is absolutely irreducible, not merely irreducible over $\mathbb{Q}$.

To distinguish shapes, the cancellation argument above works for a $\mu$-tabloid too. If ${\kappa }_t\{u\}\ne 0$, its first $r$ rows contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$, soThus $\lambda$ dominates $\mu$. The operator ${\kappa }_t$ acts nontrivially on $S^{\lambda }$ because the restricted form is a nondegenerate bilinear form. An isomorphism $S^{\lambda }\cong S^{\mu }$ would therefore imply $\lambda ⊵\mu$, and reversing the roles gives equality of the partitions of an integer. Counting conjugacy classes proves exhaustion. All matrices in a rational polytabloid basis have rational entries, giving the requested representations over the rational numbers.

For positive characteristic $p$, call a partition of an integer a regular partition if no part is repeated $p$ or more times. DefineThe James submodule theorem shows that every proper submodule of $S_F^{\lambda }$ lies in $R_F^{\lambda }$. Whenever the restricted form is nonzero, $R_F^{\lambda }$ is proper and its quotient is simple.

We verify exactly when this happens. Let $m_j$ be the number of rows of length $j$. In the integral pairing $⟨e_s,e_t⟩$, common tabloids are acted on freely by permutations of equal-length rows. Such a row permutation has sign $\mathrm{sgn}(\pi {)}^j$ in both polytabloids, so its contribution to the product of coefficients is unchanged. Each orbit has size ${\prod }_j{m}_j!$. Thus that integer divides every pairing. On the other hand, reverse each row of $t$ to obtain $t^{\ast }$. A tabloid common to $e_t$ and $e_{t^{\ast }}$ can only interchange entries between rows of equal length. There are $m_j!$ independent choices in each of their $j$ columns, and the two signs agree. HenceThese two products have exactly the same prime divisors. All pairings vanish modulo $p$ precisely when some $m_j\ge p$. Therefore $D_F^{\lambda }\ne 0$ exactly for $p$-regular $\lambda$.

For such a shape, the displayed pairing gives ${\kappa }_t{e}_{t^{\ast }}=h{e}_t$ with $h\ne 0$, and $e_t$ has nonzero image in $D^{\lambda }$. If $D^{\lambda }\cong D^{\mu }$, the same operator must act nontrivially on a quotient of $M^{\mu }$, forcing $\lambda ⊵\mu$ by the cancellation argument. Interchanging the shapes proves $\lambda =\mu$.

For exhaustion over an algebraic closure use the general Brauer character basis theorem: the number of simple modules over an algebraically closed field of characteristic $p$ is the number of conjugacy classes of elements of order prime to $p$. Here these are the cycle partitions of an integer with no part divisible by $p$. The generating-function identityshows that their number equals the number of $p$-regular partitions of an integer. We have already constructed that many distinct simple modules, so they exhaust all simples. The form, its radical and its quotient commute with field extension, and the same simplicity proof holds after every extension. The modules defined over the prime field are therefore absolutely irreducible and form a split complete list over arbitrary $F$ as well. In summary,For characteristic zero the list is instead all $S_F^{\lambda }$. In positive characteristic a Specht module itself need not be irreducible; the radical quotient is essential.

For integer partitions $\lambda ,\mu$ of $n$, padded by zero parts, the dominance order on partitions isA $\lambda$-tableau fills the Young diagram of $\lambda$ with $1,\dots ,n$, each once. A tabloid $[t]$ remembers the sets of entries in each row, ignoring their order within that row. The Young permutation module $M^{\lambda }=\mathbb{C}{X}_{\lambda }$ has these tabloids as basis, with $S_n$ acting by permuting the entries.

The row and column stabilizers $R_t,C_t$ independently permute entries within the rows and columns of $t$. The column antisymmetrizer is ${\kappa }_t={\sum }_{c\in C_t}\mathrm{sgn}(c)c$, and $e_t={\kappa }_t[t]$ is the associated polytabloid. Since $C_{gt}=g{C}_t{g}^{-1}$ and conjugation preserves permutation sign,Thus the Specht module is the submoduleEvery tableau is $gt$ for a fixed $t$, so any single $e_t$ generates it under $S_n$. Also $C_t\cap R_t=\{1\}$, since a permutation preserving both rows and columns must fix every cell. The tabloids $[ct]$ for $c\in C_t$ are consequently distinct, and $e_t$ has coefficient $1$ at $[t]$; in particular $e_t\ne 0$.

We use the following precise column antisymmetrizer facts. If some row of a $\mu$-tableau $s$ contains two entries from one column of $t$, the transposition interchanging them pairs equal tabloids with opposite signs in ${\kappa }_t[s]$, so ${\kappa }_t[s]=0$. Otherwise each row of $s$ meets each column of $t$ at most once. The first $r$ rows of $s$ therefore contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$ of $t$, whenceThis states and explains dominance from a nonzero column antisymmetrizer: ${\kappa }_t[s]\ne 0$ implies $\lambda ⊵\mu$.

When $\mu =\lambda$, the nonzero column antisymmetrizer criterion says more: either ${\kappa }_t[s]=0$, or a column permutation takes $[s]$ to $[t]$, and ${\kappa }_t[s]=\pm e_t$. Thus ${\kappa }_t{M}^{\lambda }=\mathbb{C}{e}_t$. This follows by matching the row positions within each column when all intersections have size at most one; the equal row and column sizes force the Ferrers incidence pattern.

For the inner product, use the positive Hermitian extension of the orthonormal tabloid basis, linear in its first argument. It has the supplied self-adjointness of ${\kappa }_t$. Its orthogonal complement of $S^{\lambda }$ agrees with that obtained from the complex tabloid bilinear form, since the spanning polytabloids have real coefficients. Because the coefficient of $[t]$ in $e_t$ is $1$, self-adjointness and the one-dimensional image give the useful identityIndeed, the scalar multiplying $e_t$ is the $[t]$-coefficient of ${\kappa }_{t}u$, which is $⟨{\kappa }_{t}u,[t]⟩=⟨u,{\kappa }_t[t]⟩=⟨u,e_t⟩$.

The James submodule theorem says that for every $F{S}_n$-submodule $U\le M^{\lambda }$, eitherwhere orthogonality is taken with respect to the tabloid bilinear form.

Fix a $\lambda$-tableau $t$. Part a shows that for every tabloid $\{s\}$, the vector $b_t\{s\}$ is either zero or a signed copy of the polytabloid $e(t)$. Comparing the coefficient of $\{t\}$ gives the precise identityIf $U⊈(S^{\lambda }{)}^{\perp }$, choose $u\in U$ and a tableau $t$ with $⟨u,e(t)⟩\ne 0$. Since $U$ is a submodule, the identity puts $e(t)$ in $U$. Every polytabloid of shape $\lambda$ is an $S_n$-translate of $e(t)$, so their span $S^{\lambda }$ lies in $U$. If no such $u,t$ exist, then by definition $U\subseteq (S^{\lambda }{)}^{\perp }$. This proves the theorem over the arbitrary field $F$.

Because $C(t^{\prime })=R(t)$,Applying $\theta$ and using $re(u)=\mathrm{sgn}(r)e(u)$ givesThus $m={\sum }_{r\in R(t)}re(t)$. Every $r$ fixes the tabloid $\{t\}$, while $e(t)$ has coefficient one at $\{t\}$. Invariance of the tabloid bilinear form now yields

The assumed one-dimensional-image property givesfor some $c\in \mathbb{F}$. The coefficient of $\{t\}$ in $e(t)$ is one, soThe tabloid bilinear form is invariant, and the involution on the group algebra fixes the Column antisymmetrizer of a Young tableau because inversion preserves sign. Thereforewhich proves the formula.

If some $a_j\ge p$, then $a_j!=0$ in $\mathbb{F}$. Part b(ii) makes every pairing $⟨e(t),e(u)⟩$ zero, so the Tabloid bilinear form vanishes identically on $S^{\lambda }$ and $S^{\lambda }\subseteq (S^{\lambda }{)}^{\perp }$.

Conversely, if $\lambda$ is $p$-regular, every $a_j!$ is nonzero. Fact 1 supplies tableaux $t,u$ withThus the restriction of the form is not identically zero and $S^{\lambda }⊈(S^{\lambda }{)}^{\perp }$.