= Solution
A <representation> over $F$ is a homomorphism $G\to\operatorname{GL}(V)$ 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 $\lambda=(\lambda_1,\lambda_2,\ldots)$ 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> $\lambda'$ has $\lambda'_j$ 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 $1,\ldots,n$. 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 put
$$
\kappa_t=\sum_{g\in C_t}\operatorname{sgn}(g)g,\qquad e_t=\kappa_t\{t\},\qquad S_F^\lambda=\operatorname{span}_F\{e_t:t\text{ has shape }\lambda\}.
$$
Thus $e_t$ 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 $v\in M_F^\lambda$,
$$
\kappa_tv=\langle v,e_t\rangle e_t.
$$
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 \b[$S_\mathbb Q^\lambda$ is absolutely <irreducible>], not merely <irreducible> over $\mathbb Q$.
To distinguish shapes, the cancellation argument above works for a $\mu$-tabloid too. If $\kappa_t\{u\}\ne0$, its first $r$ rows contain at most $\min(r,\lambda'_j)$ entries from column $j$, so
$$
\sum_{i\le r}\mu_i\le\sum_j\min(r,\lambda'_j)=\sum_{i\le r}\lambda_i.
$$
Thus $\lambda$ 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 $\lambda\unrhd\mu$, 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. Define
$$
R_F^\lambda=S_F^\lambda\cap(S_F^\lambda)^\perp,\qquad D_F^\lambda=S_F^\lambda/R_F^\lambda.
$$
The <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 $\langle e_s,e_t\rangle$, common <tabloids> are acted on freely by <permutations> of equal-length rows. Such a row <permutation> has sign $\operatorname{sgn}(\pi)^j$ in both <polytabloids>, so its contribution to the product of coefficients is unchanged. Each orbit has size $\prod_jm_j!$. Thus that integer divides every pairing. On the other hand, reverse each row of $t$ to obtain $t^*$. A <tabloid> common to $e_t$ and $e_{t^*}$ 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. Hence
$$
\langle e_t,e_{t^*}\rangle=\prod_j(m_j!)^j.
$$
These two products have exactly the same prime divisors. All pairings vanish modulo $p$ precisely when some $m_j\ge p$. Therefore $D_F^\lambda\ne0$ exactly for $p$-regular $\lambda$.
For such a shape, the displayed pairing gives $\kappa_te_{t^*}=h e_t$ with $h\ne0$, 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 $\lambda\unrhd\mu$ 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 identity
$$
\prod_{j\ge1}(1+t^j+\cdots+t^{(p-1)j})
=\prod_{j\ge1}\frac{1-t^{pj}}{1-t^j}
=\prod_{p\nmid j}(1-t^j)^{-1}
$$
shows 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,
$$
\boxed{\operatorname{Irr}(FS_n)=\{D_F^\lambda:\lambda\vdash n\text{ is }p\text{-regular}\}.}
$$
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.
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