= Solution
Let $\lambda=(\lambda_1\ge\lambda_2\ge\cdots)$ be a <partition of an integer> $n$. Its <Young diagram> consists of cells $(i,j)$ with $1\le j\le\lambda_i$, and its <conjugate partition> has parts $\lambda'_j=\#\{i:\lambda_i\ge j\}$. A numbered <Young tableau> fills these cells bijectively with $1,\ldots,n$. Its row and column stabilizers $R_t,C_t$ permute entries within each row and each column respectively. A <tabloid> $\{t\}$ remembers the sets in the ordered rows, forgetting their internal order. The <Young subgroup> $R_t\cong\prod_i S_{\lambda_i}$ is its stabilizer, so the <Young permutation module> is
$$
M^\lambda=F\{\text{tabloids of shape }\lambda\}
\cong\operatorname{Ind}_{R_t}^{S_n}F.
$$
Here the group acts by permuting numbered entries and extending linearly. A <submodule> is an invariant linear subspace. Give the <tabloid> basis its orthonormal symmetric <bilinear form>, denoted $\langle\ ,\ \rangle$. This form is nondegenerate and invariant under the group. Define the <column antisymmetrizer> and <polytabloid> by
$$
b_t=\sum_{g\in C_t}\operatorname{sgn}(g)g\in FS_n,
\qquad e_t=b_t\{t\},
\qquad S^\lambda=\operatorname{span}_F\{e_t\}.
$$
Thus $S^\lambda$ is the <Specht module>, generated by any one $e_t$, because the group is transitive on numbered <Young tableaux>. The <orthogonal complement> $(S^\lambda)^\perp$ consists of vectors pairing to zero with every <polytabloid>.
The key calculation is
$$
\boxed{b_t v=\langle v,e_t\rangle e_t\quad(v\in M^\lambda).}
$$
To prove it, first take $v=\{u\}$. If two entries of one column of $t$ occur in one row of $u$, their <transposition> $s$ fixes $\{u\}$; pairing the terms indexed by $g$ and $gs$ cancels $b_t\{u\}$. This also works in <characteristic two>, where the paired equal terms have sum zero. Otherwise each row of $u$ contains at most one entry from each column of $t$. The first row has $\lambda_1$ entries and there are $\lambda_1$ columns, so it uses every column once. Removing those entries, the second row must use exactly the columns of remaining positive height, and induction does the same for all rows. Consequently there is a unique $h\in C_t$ with $\{u\}=\{ht\}$. Uniqueness follows from $R_t\cap C_t=1$. In that case
$$
b_t\{u\}=\operatorname{sgn}(h)e_t,
\qquad \langle\{u\},e_t\rangle=\operatorname{sgn}(h).
$$
Linearity proves the identity.
Now let $U\le M^\lambda$ be a <submodule>. If $U$ is not contained in $(S^\lambda)^\perp$, choose $v\in U$ and $t$ with $\langle v,e_t\rangle\ne0$. The displayed identity puts $e_t$ in $U$, so its group translates put all of $S^\lambda$ in $U$. This proves the <James submodule theorem> without a semisimplicity assumption.
Put $R=S^\lambda\cap(S^\lambda)^\perp$. If $R\ne S^\lambda$, every proper <submodule> of $S^\lambda$ lies in $R$, by the same theorem. Hence $R$ is its unique maximal <submodule>, and $D^\lambda=S^\lambda/R$ is <irreducible>. If $R=S^\lambda$, the quotient is zero. This conclusion remains true after every <field extension> $E/F$: the integral <tabloid> construction becomes $M^\lambda_E=E\otimes_FM^\lambda$, and the radical of the restricted form is the kernel of its Gram matrix, so it commutes with extension of scalars. A nonzero Gram entry stays nonzero in $E$. Applying the theorem over $E$ makes $E\otimes_FD^\lambda$ <irreducible>. Thus a nonzero quotient is absolutely <irreducible>.
The precise nonvanishing criterion is
$$
\boxed{D^\lambda\ne0\iff p=0\text{ or }\lambda\text{ is a }p\text{-regular partition}.}
$$
A <regular partition> has fewer than $p$ rows of each positive length. Here is the integral pairing argument behind this condition. Write $m_j$ for the number of rows of length $j$. Expanding two <polytabloids> into <tabloids>, simultaneous <permutations> of equal-length rows <partition of an integer> all contributions to their pairing into free orbits of size $\prod_jm_j!$; the signs agree within each orbit because the same row <permutation> is applied in every relevant column on both sides. Therefore every pairing is divisible by $\prod_jm_j!$. If some $m_j\ge p$, the entire restricted form is zero modulo $p$. Conversely, reverse the order of the entries in each row of a tableau to obtain $t^*$. Counting the matching column <permutations> in the two expansions gives, up to the harmless overall sign,
$$
\langle e_t,e_{t^*}\rangle=\pm\prod_j(m_j!)^j.
$$
Indeed the $m_j$ equal rows of length $j$ can be matched independently in each of their $j$ positions; after these choices the decreasing row lengths force every other matching. For a <regular partition> this integer is nonzero modulo $p$, so the form and its radical quotient are nonzero. In <characteristic zero> it is nonzero for every shape. This supplies the requested justification of the criterion.
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