Show that the coinvariant ring is self-injective. Let $I$ be the ideal of $\mathbb{C}[x_1,\ldots,x_n]$ generated by all symmetric functions of positive degree.

I want to show that the coinvariant ring $C:=\mathbb{C}[x_1,\ldots,x_n]/I$ is self-injective.

It is well-known that $C$ has the Artin basis $$\{x_1^{i_1}\cdots x_n^{i_n}\mid 1\leq\forall k\leq n, 0\leq i_k\leq n-k\}.$$
We define a $\mathbb{C}$-linear map $\varphi:C\rightarrow\mathbb{C}$ by sending $f\in C$ to the coordinate of $f$ corresponding to $x_1^{n-1}x_2^{n-2}\cdots x_{n-1}$ with respect to the Artin basis. And then we define a $\mathbb{C}$-bilinear form $\Phi:C\times C\rightarrow\mathbb{C}$ by sending $(f,g)\in C\times C$ to $\varphi(fg)\in \mathbb{C}$.

To show that $C$ is self-injective, it suffices to prove that $\Phi$ is nondegenerate. To prove this, it is sufficient that for every homogeneous polynomial $f\in\mathbb{C}[x_1,\ldots,x_n]\setminus I$, there exists a homogeneous polynomial $g\in\mathbb{C}[x_1,\ldots,x_n]$ such that $\mathrm{deg}(f)+\mathrm{deg}(g)=n(n-1)/2$ and $fg\notin I$.

How to show the last statement?
 A: Given any non-zero $f$ in $C$, take some basis vector $v$ appearing with non-zero coefficient in $f$. Then there is a unique basis vector $w$ such that $vw=x_1^{n-1}x_2^{n-2}\cdots x_{n-1}$. Thus $\varphi(fw)\neq0$, and hence $f$ is not in the kernel of the bilinear form $\Phi$. Thus the kernel is zero, and $\Phi$ is non-degenerate.
EDIT: sorry, as in the comment below, it’s not quite that simple. We first order all monomials using the lexicographic ordering. Given $f$ non-zero, we take the smallest basis element $v$ occurring in $f$, and the element $w$ as above.
To see that this really works, we note that every monomial larger than those in the Artin basis lies in the ideal $I$.
For example, for $n=4$ we have $x_1^4,x_1^3x_2^3\in I$. The first follows from considering the four elements of $I$
$$ x_1^3(x_1+x_2+x_3+x_4) $$
$$ x_1^2(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4) $$
$$ x_1(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4) $$
$$ x_1x_2x_3x_4. $$
For the second we use
$$ x_1^3x_2^2(x_1+x_2+x_3+x_4) $$
$$ x_1^2x_2(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4) $$
together with $x_1^4$ and $x_1x_2x_3x_4$.
From this we observe that the socle of $C$ is generated by the (image of) $\alpha=x_1^{n-1}x_2^{n-2}\cdots x_{n-1}$, which is why we take this basis element in the definition of $\varphi$.
Now if $v’>v$ are basis elements and $vw=\alpha$, then $v’w>vw$, so lies in $I$. So the coefficient of $\alpha$ in $fw$ is precisely the coefficient of $v$, so non-zero.
