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Let $k$ be a field of characteristic $0$, let $\mathbf X=[X_{ij}]_{1\le i,j\le n} $ be a square matrix of indeterminates where $n\ge 2$. Consider the polynomial $f(\mathbf X)=\text{det}(\mathbf X)\in k[\mathbf X]$. Consider its Jacobian ideal $\text{Jac}(f)=\left(\dfrac{\partial f}{\partial X_{ij}}\right)_{1\le i, j\le n}\in k[\mathbf X]$.

My question is: Does $\text{Jac}(f)$ have height at least $3$ ?

Of course when $n=2$, this is seen by direct computation, but what about higher values of $n$ ?

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    $\begingroup$ The Jacobian of the determinant is generated by the $(n-1)$-minors, so geometrically corresponds to the set of matrices of rank $\le n-2$. This has codimension 4 in the space of all matrices, for any $n$ $\endgroup$
    – math54321
    Mar 30, 2023 at 19:36

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