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For a Young tableau $t$, let $R_t$ permute the entries within its rows and $C_t$ within its columns. We use the Young symmetrizer conventionReversing the order gives another usual realization of the same irreducible polynomial module. On the tensor power $T=V^{\otimes n}$, the actions areThey commute because applying $g$ to every factor commutes with permuting the factors.

We state the permitted combinatorial input explicitly. A Young symmetrizer satisfies $h_t^2=H_{\lambda }{h}_t$, where $H_{\lambda }$ is the nonzero hook product of a partition. Thus $e_t=H_{\lambda }^{-1}{h}_t$ is a primitive idempotent. The standard-tableau decomposition of the right regular module is $\mathbb{C}{S}_n={⨁}_{\lambda ,t\mathrm{standard}}{e}_t\mathbb{C}{S}_n$. Tensor this right-module direct sum with the left module $T$. The map $e_t\mathbb{C}{S}_n{\otimes }_{\mathbb{C}{S}_n}T\to e_{t}T$ sending $e_{t}a\otimes v$ to $e_{t}av$ is an isomorphism, with inverse $w\mapsto e_t\otimes w$. Therefore the Young-symmetrizer tensor decomposition isThis is a direct sum of $GL(V)$-modules; individual summands need not be $S_n$-invariant.

We also state the allowed Schur algebra result, namely Schur–Weyl duality: the two actions are mutual commutants, andwhere the $D_{\lambda }(V)$ are pairwise nonisomorphic irreducible homogeneous polynomial representations of degree $n$. We also use the standard Schur algebra equivalence between its modules and homogeneous degree-$n$ polynomial representations, so these exhaust the irreducibles in that category. A primitive idempotent $e_t$ has one-dimensional image on $S^{\lambda }$ and zero image on the other simple factors, so $h_{t}T\cong D_{\lambda }(V)$. This identifies the requested irreducibles. They classify the polynomial degree-$n$ representations in this tensor power, not all rational representations of every degree.

The length bound for a Schur module is $D_{\lambda }(V)\ne 0$ if and only if $\ell (\lambda )\le m$. A column of length greater than $m$ antisymmetrizes more than $m$ vectors and gives zero. Conversely, for $\ell (\lambda )\le m$, fill every tensor position in row $i$ with the $i$th basis vector of $V$. Row symmetrization multiplies it by ${\prod }_i{\lambda }_i!$, and column antisymmetrization is nonzero because the vectors within each column are distinct basis vectors.

A rational representation of an algebraic group is a regular morphism into the general linear group of its representation space. For $G{L}_m$, its matrix entries belong to $\mathbb{C}[g_{ij},\det (g{)}^{-1}]$; rational here permits determinant denominators but not arbitrary poles on $G{L}_m$. A one-dimensional rational character of ${\mathbb{C}}^{\times }$ is a Laurent polynomial $p(z)$ with $p(zw)=p(z)p(w)$ and $p(1)=1$. Comparing Laurent coefficients shows that just one monomial occurs and its coefficient is $1$, hence $p(z)=z^r$ for $r\in \mathbb{Z}$.

Restrict a one-dimensional rational character of $G{L}_m$ to its diagonal torus. The same argument in several variables gives ${\prod }_i{x}_i^{r_i}$. Conjugation by permutation matrices makes all $r_i$ equal. On every diagonalizable invertible matrix it consequently agrees with $(\det g{)}^r$. The allowed Zariski-density statement, and equality of regular functions on a dense subset, giveThese are the one-dimensional rational characters of the general linear group.

For complete reducibility of rational GL and SL representations, use the compact-group averaging argument. Average any positive definite Hermitian inner product over $SU(m)$ using normalized Haar measure. The orthogonal complement of an invariant subspace is then $SU(m)$-invariant. Differentiating makes it invariant under $\mathfrak{su}(m)$ and therefore under its complex span ${\mathfrak{sl}}_m(\mathbb{C})$. The elementary unipotent matrices $\exp (t{E}_{ij})$ generate $S{L}_m(\mathbb{C})$, so the complement is $S{L}_m$-invariant. This proves complete reducibility. Averaging over $U(m)$ similarly gives complete reducibility for rational $G{L}_m$ representations.

Here is an explicit rational extension from SL to GL. Decompose the representation space by the finite scalar center of $S{L}_m$ into subspaces $W_k$ on which $\zeta I$ acts as ${\zeta }^k$, with $0\le k<m$ and ${\zeta }^m=1$. These subspaces are invariant. For $g\in G{L}_m$, choose $a$ with $a^m=\det g$, set $s=a^{-1}g\in S{L}_m$, and defineChanging $a$ to $\zeta a$ changes $s$ to ${\zeta }^{-1}s$, so the two factors cancel. It is a homomorphism, since scalar roots multiply up to the same harmless factor, and it restricts to $\rho$ on $S{L}_m$.

It is rational as well. Every matrix coefficient of $\rho {|}_{W_k}$ has a polynomial representative on $S{L}_m$. Averaging that representative over the finite scalar center selects its homogeneous parts $P_d$ with $d\equiv k(\mathrm{mod}m)$. Substitution in the extension gives ${\sum }_d(\det g{)}^{(k-d)/m}{P}_d(g)$, a regular function on $G{L}_m$. Each $S{L}_m$-invariant subspace decomposes into its $W_k$ parts, and the extension acts on each part by a scalar times an $S{L}_m$ action. Thus these subspaces are also invariant under the chosen extension. Consequently $\rho$ is irreducible if and only if this ${\rho }^{\prime }$ is irreducible. For $m=1$, $S{L}_1$ is trivial and the trivial extension supplies the same conclusion.

For a box $(i,j)$ of a Young diagram, its hook of a Young diagram contains that box, all boxes to its right in its row, and all boxes below it in its column. Its hook length is $h_{ij}={\lambda }_i-j+{\lambda }_j^{\prime }-i+1$. The hook graph of a partition is the diagram with each box labeled by its hook length. Write $H_{\lambda }={\prod }_{(i,j)\in \lambda }{h}_{ij}$ for the hook product of a partition. The hook-length formula is

Pad $\lambda$ to $m$ rows and use the beta set of a partition ${\ell }_i={\lambda }_i+m-i$, so ${\ell }_1>\cdots >{\ell }_m\ge 0$ and ${\sum }_i{\ell }_i=n+d$, where $d=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. We prove the beta-set hook-product identity row by row. For $1\le j\le {\lambda }_i$, put $q_j=m+j-{\lambda }_j^{\prime }-1$; then $h_{ij}={\ell }_i-q_j$. The $q_j$ are distinct integers in $[0,{\ell }_i-1]$, and none equals a beta number. Indeed, if $k\le {\lambda }_j^{\prime }$ then ${\ell }_k\ge q_j+1$, whereas if $k>{\lambda }_j^{\prime }$ then ${\ell }_k\le q_j-1$. Exactly $m-i$ beta numbers lie below ${\ell }_i$, so the remaining ${\lambda }_i$ integers in that interval are precisely these $q_j$. ThereforeThis gives the equivalent Specht module dimension expression $f^{\lambda }=n!\Delta (\ell )/{\prod }_i{\ell }_i!$.

The standard Young tableaux of shape $\lambda$ form the dimension count for the Specht module. The largest entry $n$ must occupy a removable corner. Deleting it bijects tableaux with the disjoint union of standard tableaux of the shapes obtained by deleting a corner, provingSet a term to zero whenever $\lambda -e_i$ is not a partition: this includes equal adjacent rows and an attempted deletion from a zero row. The empty diagram has dimension $1$.

For an algebraic proof that the proposed formula has the same recurrence, establish the Vandermonde shift identityIts left side is an alternating polynomial in the $x_i$, because permuting the variables permutes the summands and changes the sign of every Vandermonde factor. It is therefore divisible by $\Delta (x)$. The quotient is symmetric in $x$ and homogeneous of total degree one in $(x,t)$, so has the form $a{\sum }_i{x}_i+bt$. At $t=0$, $a=1$. Differentiating in $t$ at zero and using the Euler theorem for homogeneous functions gives ${\sum }_i{x}_i{\partial }_i\Delta =d\Delta$, so $b=d$. This proves the identity as a polynomial identity, including repeated coordinates.

Take $x=\ell$ and $t=-1$. Since ${\sum }_i{\ell }_i=n+d$, it gives ${\sum }_i{\ell }_i\Delta (\ell -e_i)=n\Delta (\ell )$. For $n\ge 1$, this is exactlyThe summand is the proposed dimension for $\lambda -e_i$; if two beta numbers collide its Vandermonde is zero, and if ${\ell }_i=0$ its coefficient is zero, so no negative factorial is needed. For the empty partition, $\ell =(m-1,\dots ,0)$ and $\Delta (\ell )={\prod }_{k=0}^{m-1}k!$, giving initial value $1$. Induction now proves the hook-length formula from the tableau recurrence.

For the final sum, tuples with repeated coordinates contribute zero. Sorting each distinct nonnegative tuple with sum $\left(\genfrac{}{}{0ex}{}{m+1}{2}\right)$ gives one beta set of a partition of size $m$, since subtracting the staircase removes $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ from the sum. Conversely every partition of $m$, padded to $m$ rows, supplies exactly $m!$ ordered tuples, with the same squared summand. Hence the square-sum identity for shifted partition coordinates isThe middle equality uses the Artin–Wedderburn theorem for the group algebra and the complete classification of its simple modules by Specht modules. It explains why the last identity is a representation-dimension count rather than an accidental cancellation.

For a Young tableau $t$, let $r_t$ sum its row permutations and $c_t$ be the signed sum of its column permutations in the group algebra of the symmetric group. The product $h_t=c_t{r}_t$ is a Young symmetrizer. The other order is another conventional realization. It satisfies $h_t^2=H_{\lambda }{h}_t$, where $H_{\lambda }$ is the hook product of a partition, so division by that scalar gives a primitive idempotent in characteristic zero. Acting on a tensor power constructs a Schur module.