= Solution
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> convention
$$
r_t=\sum_{\sigma\in R_t}\sigma,\qquad c_t=\sum_{\tau\in C_t}\operatorname{sgn}(\tau)\tau,\qquad h_t=c_tr_t.
$$
Reversing the order gives another usual realization of the same irreducible <polynomial> <module>. On the <tensor power> $T=V^{\otimes n}$, the actions are
$$
\sigma(v_1\otimes\cdots\otimes v_n)=v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)},\qquad
g(v_1\otimes\cdots\otimes v_n)=gv_1\otimes\cdots\otimes gv_n.
$$
They 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 $\mathbb CS_n=\bigoplus_{\lambda,t\ \mathrm{standard}}e_t\mathbb CS_n$. Tensor this right-module direct sum with the left <module> $T$. The map $e_t\mathbb CS_n\otimes_{\mathbb CS_n}T\to e_tT$ sending $e_ta\otimes v$ to $e_tav$ is an isomorphism, with inverse $w\mapsto e_t\otimes w$. Therefore the <Young-symmetrizer tensor decomposition> is
$$
T=\bigoplus_{\lambda,t\ \mathrm{standard}}e_tT=\bigoplus_{\lambda,t\ \mathrm{standard}}h_tT.
$$
This 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>, and
$$
T\cong\bigoplus_{\lambda\vdash n,\ \ell(\lambda)\le m}S^\lambda\otimes D_\lambda(V),
$$
where 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 $h_tT\cong D_\lambda(V)$. 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 \b[$D_\lambda(V)\ne0$ 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 $GL_m$, its matrix entries belong to $\mathbb C[g_{ij},\det(g)^{-1}]$; rational here permits <determinant> denominators but not arbitrary poles on $GL_m$. 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 $GL_m$ 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, give
$$
\boxed{\chi(g)=(\det g)^r,\qquad r\in\mathbb Z.}
$$
These 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 $\exp(tE_{ij})$ generate $SL_m(\mathbb C)$, so the complement is $SL_m$-invariant. This proves complete reducibility. Averaging over $U(m)$ similarly gives complete reducibility for rational $GL_m$ representations.
Here is an explicit <rational extension from SL to GL>. Decompose the representation space by the finite scalar center of $SL_m$ 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 $g\in GL_m$, choose $a$ with $a^m=\det g$, set $s=a^{-1}g\in SL_m$, and define
$$
\rho'(g)|_{W_k}=a^k\rho(s)|_{W_k}.
$$
Changing $a$ 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 $SL_m$.
It is rational as well. Every matrix coefficient of $\rho|_{W_k}$ has a <polynomial> representative on $SL_m$. Averaging that representative over the finite scalar center selects its homogeneous parts $P_d$ with $d\equiv k\pmod m$. Substitution in the extension gives $\sum_d(\det g)^{(k-d)/m}P_d(g)$, a <regular function> on $GL_m$. Each $SL_m$-invariant subspace decomposes into its $W_k$ parts, and the extension acts on each part by a scalar times an $SL_m$ action. Thus these subspaces are also invariant under the chosen extension. Consequently \b[$\rho$ is irreducible if and only if this $\rho'$ is irreducible]. For $m=1$, $SL_1$ is trivial and the trivial extension supplies the same conclusion.
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