TITLE: Ideal defining the nilpotent cone of $\mathfrak{gl}_n(k)$
QUESTION [9 upvotes]: Let $k$ be an algebraically closed field, and let $\mathfrak{g}=\mathfrak{gl}_n(k)$.  Let $\mathcal{N}\subset\mathfrak{g}$ be the nilpotent cone, that is:
$$\mathcal{N}=\{A\in\mathfrak{g}\mid A^n=0\}=\{A\in\mathfrak{g}\mid \text{ch}(A)=x^n\}$$
where ch$(A)$ is the characteristic polynomial of the matrix $A$.  Let $X=(x_{ij})$ be a matrix of indeterminates, and notice that as an affine variety $\mathcal{N}\subset\mathbb{A}_k^{n^2}$ can be defined in two ways.  If $I$ is the ideal generated by the $n^2$ homogeneous polynomial equations of degree $n$ given by $X^n=0$, then $\mathcal{N}=Z(I)$.  Also, if $J$ is the ideal generated by the $n$ homogeneous equations of degrees $1,2,\ldots,n$ which define the non-leading coefficients of ch$(X)$, then $\mathcal{N}=Z(J)$.  Now it follows that $\sqrt{I}=\sqrt{J}$.
It is clear to me that $I$ is not radical because $x_{11}+x_{22}+\ldots+x_{nn}\in \sqrt I\setminus I$.  Is $J$ a radical ideal?  I believe it is, based on the comments in this question, but I can't seem to prove it.

REPLY [2 votes]: This is not a complete answer, but it might still be of use. Either I find the time to finish it later or someone else picks up where I leave off.
The $r$-th coefficient of the characteristic polynomial of matrix $A\in\mathbb A^{n\times n}\cong k^n \otimes k^n$ is the trace of $\wedge^r A$. That is, it is the sum of certain $r$-th minors of the matrix $A$. To be very precise,
$$\chi_r = \sum_{\substack{I\subseteq\{1,\ldots,n\}\\\\\# I=r}} \underset{\chi_{rI}}{\underbrace{\sum_{\pi\in\mathfrak{S}_I} \mathrm{sgn}(\pi) \cdot \prod_{i\in I} x_{i,\pi(i)}}}$$
is the polynomial computing the $r$-th coefficient of the characteristic polynomial. Observe that $\deg(\chi_r)=r$. Furthermore, the $\chi_{rI}$ are irreducible, multilinear and share no monomial. Hence, $\chi_r$ is irreducile. An ideal generated by irreducible, homogeneous, multilinear polynomials of pairwise different degrees should be radical.