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.
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.(edit) As pointed out in the comments, the last sentence of this answer is incorrect. Sadly, it has been too long for me to recall the details here and I cannot un-accept this answer =(