= Solution
The <coordinate ring> is $\mathbb C[W]=\operatorname{Sym}(W^*)$. The induced action is contragredient on functions:
$$
(g\cdot p)(w)=p(\rho(g^{-1})w).
$$
It is a <group action> by degree-preserving algebra automorphisms, extending the dual action on linear <polynomial functions>. Its <polynomial invariant ring> is
$$
\mathbb C[W]^G=\{p\in\mathbb C[W]:g\cdot p=p\text{ for every }g\in G\}.
$$
For a monic <polynomial> $f(t)=\prod_{i=1}^d(t-r_i)$, its <polynomial discriminant> is $\operatorname{disc}(f)=\prod_{i<j}(r_i-r_j)^2$. This is symmetric in the roots, hence <polynomial> in the coefficients, and is zero exactly when a root is repeated.
For the <alternating group> action, let $e_1,\ldots,e_n$ be the <elementary symmetric polynomials> and put $\Delta=\prod_{i<j}(X_i-X_j)$. If $p$ is $A_n$-invariant and $\tau$ is any <transposition>, decompose
$$
p=p_++p_-,\qquad p_+=\frac{p+\tau p}{2},\quad p_-=\frac{p-\tau p}{2}.
$$
Normality and index two of $A_n$ show that $p_+$ is symmetric and $p_-$ transforms by the sign <character> of $S_n$. Any <alternating polynomial> vanishes when $X_i=X_j$, so every $X_i-X_j$ divides it. These pairwise nonassociate prime factors have product $\Delta$, so $p_-=\Delta q$ with $q$ symmetric. The <Fundamental theorem of symmetric polynomials> yields
$$
\mathbb C[X_1,\ldots,X_n]^{A_n}
=\mathbb C[e_1,\ldots,e_n]\oplus\Delta\mathbb C[e_1,\ldots,e_n].
$$
The sum is direct, since a <polynomial> that is both symmetric and alternating is zero in <characteristic zero>. Also $\Delta^2=D(e_1,\ldots,e_n)$, where $D$ is the discriminant <polynomial> of
$$
t^n-e_1t^{n-1}+e_2t^{n-2}-\cdots+(-1)^ne_n.
$$
Thus, more precisely than the requested quotient assertion,
$$
\boxed{\mathbb C[X_1,\ldots,X_n]^{A_n}
\cong\mathbb C[T_1,\ldots,T_n,Z]/(Z^2-D(T_1,\ldots,T_n)).}
$$
Surjectivity follows from the direct-sum expression. Divide any putative kernel element by the monic quadratic in $Z$; its remainder is $a(T)+Zb(T)$. The <direct sum> forces $a(e)=b(e)=0$, and <algebraic independence> of the $e_i$ gives $a=b=0$. Hence the displayed relation is the entire kernel. The argument also covers $n=2$, where $A_2$ is trivial.
Now let $G$ be finite with no nontrivial <linear characters>, and write $R=\mathbb C[W]$, $S=R^G$. The <polynomial ring> $R$ is a <unique factorization domain>. Factor a nonzero invariant $f$ in $R$; invariance permutes the associate classes of its irreducible factors and makes their exponents constant on each orbit. For an orbit $\mathcal O$, choose representatives and form its <orbit product of polynomial factors>
$$
q_{\mathcal O}=\prod_{p\in\mathcal O}p.
$$
For every $g$, $g\cdot q_{\mathcal O}=\theta(g)q_{\mathcal O}$ for a nonzero scalar $\theta(g)$. Applying two group elements proves that $\theta:G\to\mathbb C^\times$ is a homomorphism. The hypothesis forces $\theta=1$, so $q_{\mathcal O}\in S$.
This orbit product is prime in $S$. If it divides $ab$ with $a,b\in S$, one factor $p\in\mathcal O$ divides $a$ or $b$ in $R$. Invariance of that chosen <polynomial> makes every factor in the orbit divide it. Their product therefore divides it in $R$, and the quotient is invariant because numerator and denominator are invariant and cancellation is valid in the <integral domain> $R$. Thus $q_{\mathcal O}$ divides $a$ or $b$ in $S$. Every nonzero $f\in S$ is a scalar times a product of these prime orbit products, and the only units of $S$ are nonzero constants, as they are units in $R$. Therefore \b[$\mathbb C[W]^G$ is a <unique factorization domain>]. This gives a direct proof, without assuming a divisor-class-group theorem.
For a failure in <characteristic zero>, let the <cyclic group> of order two act on $\mathbb C^2$ by $-I$. Its coordinate invariants are
$$
S=\mathbb C[X^2,XY,Y^2]
\cong\mathbb C[a,b,c]/(ac-b^2).
$$
Every invariant <monomial> has even total degree: its exponents are either both even, or both odd, giving the indicated generators. Reducing powers of $b$ to at most one shows that $ac-b^2$ is the only relation, since $a^ic^j$ and $ba^ic^j$ map to distinct <monomials>. The three quadratic invariants are irreducible in $S$: each nonconstant invariant has degree at least two, so a product of two nonunits has degree at least four. They are pairwise nonassociate, but
$$
\boxed{X^2Y^2=(X^2)(Y^2)=(XY)^2}
$$
gives two different irreducible factorizations. Hence \b[this invariant ring is not a <unique factorization domain>]. The nontrivial sign <character> is precisely the kind of <character> excluded in the preceding theorem.
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