Symmetric polynomials in the Jucys-Murphy elements give exactly the centre

Part (i). Let $f$ be symmetric and let $T,T^{\mathrm{\prime }}$ be standard tableaux of the same shape $\lambda$. By [F3] the vectors $\mathrm{Cont}(T)$ and $\mathrm{Cont}(T^{\mathrm{\prime }})$ are permutations of the same multiset, so symmetry of $f$ gives $f(\mathrm{Cont}(T))=f(\mathrm{Cont}(T^{\mathrm{\prime }}))$; by [F1] the element $f(X)$ acts on every Young line of shape $\lambda$ by the same scalar, hence on the whole irreducible $V^{\lambda }$ by that scalar. By [F2] an element acting by scalars on every irreducible is central, so $f(X_1,\dots ,X_n)\in Z(\mathbb{C}[S_n])$.

Part (ii), coordinates and their distinctness. For a partition $\lambda ⊢n$ put $q_{\lambda }:=(p_1(\lambda ),\dots ,p_n(\lambda ))\in {\mathbb{C}}^n$, where $p_k(\lambda ):={\sum }_{x\in [\lambda ]}c(x{)}^k$ is the $k$-th power sum of the multiset of node contents. If $q_{\lambda }=q_{\mu }$, then $p_k(\lambda )=p_k(\mu )$ for $k\le n$, and Newton's identities of [F4] recursively express $e_k$ in terms of $p_1,\dots ,p_k$ over $\mathbb{C}$, so $e_k(\lambda )=e_k(\mu )$ for $k\le n$; the monic polynomial ${\prod }_{x\in [\lambda ]}(t-c(x))=t^n-e_1(\lambda )t^{n-1}+\cdots +(-1{)}^n{e}_n(\lambda )$ then equals ${\prod }_{x\in [\mu ]}(t-c(x))$, so the two content multisets coincide and $\lambda =\mu$ by [F3]. Hence the $p(n)$ points $q_{\lambda }$ are pairwise distinct.

Lagrange interpolation. Let $z\in Z(\mathbb{C}[S_n])$ act on $V^{\lambda }$ by ${\zeta }_{\lambda }$, as in [F2]. For each ordered pair $\lambda \mathrm{\ne }\mu$, let $j(\lambda ,\mu )$ be the least index with $q_{\lambda ,j}\mathrm{\ne }{q}_{\mu ,j}$; it exists by step 1.2. Define $F(y_1,\dots ,y_n):={\sum }_{\lambda ⊢n}{\zeta }_{\lambda }{\prod }_{\mu \mathrm{\ne }\lambda }\frac{y_{j(\lambda ,\mu )}-q_{\mu ,j(\lambda ,\mu )}}{q_{\lambda ,j(\lambda ,\mu )}-q_{\mu ,j(\lambda ,\mu )}}\mathrm{.}$ Every denominator is nonzero by its selection. The product indexed by $\lambda$ is $1$ at $q_{\lambda }$ and $0$ at every $q_{\nu }$ with $\nu \mathrm{\ne }\lambda$, because its factor indexed by $\mu =\nu$ vanishes there. Thus $F(q_{\lambda })={\zeta }_{\lambda }$ for every partition, including $n=1$, when the product is empty.

Substitution. By [F4] the power sums $p_1,\dots ,p_n$ in the variables $X_1,\dots ,X_n$ generate the symmetric polynomials; define $f$ to be the symmetric polynomial $F(p_1,\dots ,p_n)$ obtained by substituting $P_k\mapsto p_k$ in the polynomial $F(P_1,\dots ,P_n)$. Then $f$ is symmetric and $f(X_1,\dots ,X_n)$ acts on $\mathbb{C}{v}_T$ by $F(p_1(\lambda ),\dots ,p_n(\lambda ))=F(q_{\lambda })={\zeta }_{\lambda }$, where $\lambda =\mathrm{shape}(T)$ and $p_k(\lambda )={\sum }_j{c}_T(j{)}^k$ by [F3].

The difference $f(X_1,\dots ,X_n)-z$ acts by ${\zeta }_{\lambda }-{\zeta }_{\lambda }=0$ on every $V^{\lambda }$, hence is zero by [F2]; therefore $z=f(X_1,\dots ,X_n)$ with $f$ symmetric. With step 1.1 this proves both directions.