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Consider the ring $R=\mathbb C[a_{ij}]$ which is the free polynomial ring of $n^2$ variables $a_{ij}$ over the field of complex numbers $\mathbb C$. Set matrix $A=[a_{ij}]_{n\times n}$ and $f(\lambda):=\det(\lambda I-A)=\lambda^n+\sum_{k=0}^{n-1}f_k\lambda^k$ be its characteristic polynomial.

Let the ideal $J=(f_0,\cdots,f_{n-1})\subset R$. Is $J$ a prime ideal?

What I have done is the following:

Take a look at the algebraic set $Z=V_{\mathbb C^{n\times n}}\ (J)\subset M_n(\mathbb C)$, the set of all nilpotent matrices, one can show that $Z$ is irreducible. Consider the Jordan block of size $n$ and eigenvalue $0$, $B=J_n(0)$. Check the map $\mathrm{GL}_n(\mathbb C)\to M_n(\mathbb C)$ given by $P\mapsto PBP^{-1}$. The image of this map is nilpotent matrices with Jordan canonical form $B$, since the set of these matrices is Zariski dense in $Z$, so the closure of the image is $Z$. By the fact that $\mathrm{GL}_n(\mathbb C)$ is irreducible, we know the image is irreducible, so $Z$ is irreducible too.

It remains to show that $J$ is a radical ideal: by $I(Z)=\sqrt J$ if $J=\sqrt J$ then it is prime.

For $n=2$, this can be done by hand. When $n=3$, By a SageMath check:

sage: R.<a11, a12, a13, a21, a22, a23, a31, a32, a33> = PolynomialRing(QQ,9)
....: I = R * [a13*a22*a31 - a12*a23*a31- a13*a21*a32 + a11*a23*a32
      + a12*a21*a33 - a11*a22*a33, - a12*a21 + a11*a22 - a13*a31 - 
      a23*a32 + a11*a33 + a22*a33, -a11 - a22 - a33]
....: I.is_prime()
True
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  • 2
    $\begingroup$ Welcome to Mathematics SE. Take a tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$
    – cpiegore
    Apr 18, 2023 at 4:20
  • $\begingroup$ See here for an affirmative answer $\endgroup$
    – math54321
    Apr 18, 2023 at 18:11
  • $\begingroup$ @math54321 It seems to me that the answer you're referring to only shows irreducibility of the (topological) vanishing locus $Z=V_{\mathbb{C}^{n\times n}}(J)$, and doesn't address whether the $J$ in question is radical or not. $\endgroup$
    – imtrying46
    Apr 21, 2023 at 9:38
  • $\begingroup$ @imtrying46: This comment explicitly states that the ideal $J$ is radical $\endgroup$
    – math54321
    Apr 21, 2023 at 15:26

1 Answer 1

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(Since this question has been asked and highly upvoted multiple times over different posts, with no definitive answer as of yet, I thought it worthwhile to give a full, detailed answer.)

As shown in the question, the ideal $J$ generated by (non-leading) coefficients of the (generic) characteristic polynomial defines an irreducible variety, namely the set of nilpotent matrices. Since this variety has dimension $n^2 - n$ and $J$ has $n$ generators, $J$ is a complete intersection. I will address the question of radicality, for complete intersections in general. $\DeclareMathOperator{\n}{\mathfrak{n}} \DeclareMathOperator{\m}{\mathfrak{m}} \DeclareMathOperator{\codim}{codim}$

For a Noetherian ring, there are $2$ aspects to being reduced. The first is being generically reduced ($\iff$ Serre's condition $R_0 =$ regular in codimension zero), and the second is having no embedded primes ($\iff$ Serre's condition $S_1 =$ Cohen-Macaulay in codimension one).

For complete intersections, the Cohen-Macaulay condition comes for free. So one just needs to check regularity in codimension zero. Usually for finitely generated $k$-algebras this is done via the Jacobian criterion, but there is another approach, which I think deserves to be more well known.

What does it mean for a finitely generated $k$-algebra $R = k[X_i]/I$ to be generically reduced? In geometric terms (for $k$ algebraically closed): the Zariski tangent space at a general point $p \in V(I)$ should have dimension equal to that of $V(I)$. Restated algebraically: if $\m_p$ is the maximal ideal of $R$ corresponding to $p$, then $\dim_k \m_p/\m_p^2 \le \dim R$ (the other inequality always holds). Now $\m_p = \n_p/I$ for some maximal ideal $\n_p$ in $k[X_i]$, and

$$\m_p/\m_p^2 = (\n_p/I)/(\n_p/I)^2 = (\n_p/I)/((\n_p^2 + I)/I)\cong \n_p/(\n_p^2 + I) \cong (\n_p/\n_p^2)/((\n_p^2 + I)/\n_p^2)$$

as $k = R/\m_p = k[X_i]/\n_p$-vector spaces. This proves the following:


$\textbf{Theorem}$. Let $S = k[X_i]$ be a polynomial ring over an algebraically closed field $k$, $I = (f_1, \ldots, f_c) \subseteq S$ an ideal, and $p \in V(I)$ a general point, with maximal ideal $\n_p \subseteq S$. Then $I$ is generically reduced if and only if $\dim_k (\n_p^2+I)/\n_p^2 \ge \codim I$. In particular, if $c = \codim I$ (i.e. $I$ is a complete intersection), then $I$ is radical if and only if $f_1, \ldots, f_c$ are $k$-linearly independent mod $\n_p^2$, i.e.

\begin{equation} \sum_{i=1}^c a_i f_i \in \n_p^2, \; a_i \in k \implies a_1 = \ldots = a_c = 0. \end{equation}


Next, to apply this to the ideal $J$ in question: consider strictly upper-triangular matrices of the form

\begin{bmatrix} 0 & \ast & 0 & \ldots & 0 \\ 0 & 0 & \ast & \ldots & 0 \\ \vdots & & \ddots & & \vdots \\ 0 & 0 & \ldots & 0 & \ast \\ 0 & 0 & \ldots & 0 & 0 \\ \end{bmatrix}

with $\ast \in k^\times$ nonzero: these are regular nilpotent matrices, which - up to conjugation by $GL_n$ - are dense in the set of all nilpotent matrices. Thus it suffices to check the theorem at such points $p$. As it happens, this can be done in a very clean combinatorial way: for such a point $p$,

$$\n_p = (X_{i,i+1} - b_i \mid 1 \le i \le n-1) + (X_{i,j} \mid 1 \le i, j \le n, j \ne i+1) \subseteq k[X_{ij}]$$

for some $b_1, \ldots, b_{n-1} \in k^\times$. If $A = (X_{ij})$ is the generic $n \times n$ matrix, and $f_i \in J$ is the coefficient of $\lambda^i$ in the characteristic polynomial of $A$, then (up to sign) $f_i$ is the trace of $\wedge^{n-i} A$, which is the sum of all principal $(n-i)$-minors of $A$ (e.g. $f_0 = \det(A)$, $f_{n-1} = \operatorname{tr}(A)$). Now mod $\n_p^2$, $f_i$ is a linear combination of $X_{n-i,1}, \ldots, X_{n,i+1}$ (namely the variables on the subdiagonal of $A$ of length $i+1$), where each coefficient is an $(n-1-i)$-fold product of $b$'s, hence is nonzero. (To see this, consider the expansion of $f_0 = \det A$: the only term which survives mod $\n_p^2$ is $X_{n,1} \prod_{i=1}^{n-1} X_{i,i+1}$, which is congruent to $X_{n,1} \prod_{i=1}^{n-1} b_i$. Finally, a principal minor of $A$ is nonzero mod $\n_p^2$ if and only if the row/column indices form an interval, in which case the corresponding principal submatrix of $p$ is again regular nilpotent, and the same reasoning with $f_0$ applies to the submatrix.)

In particular, $f_i$ mod $\n_p^2$ are linear forms, with disjoint supports for distinct $i$. This implies that $f_0, \ldots, f_{n-1}$ are linearly independent mod $\n_p^2$, so by the theorem above, $J$ is radical.


Additional remarks:

(i) It's useful to view radicality/primality criteria in a Noetherian ring from the viewpoint of primary decomposition. If $I = Q_1 \cap \ldots \cap Q_s$ is a minimal primary decomposition, and $P_i := \sqrt{Q_i}$ are the associated primes of $I$, with $P_1, \ldots, P_r$ the minimal primes, then:

  1. $V(I)$ is irreducible iff $r = 1$, i.e. $I$ has a unique minimal prime

  2. $I$ is generically reduced iff $Q_i = P_i$ for all $1 \le i \le r$

  3. $I$ has no embedded primes iff $r = s$

It follows immediately that $I$ is radical if and only if (2) and (3) hold, and prime if and only if (1)-(3) hold.

(ii) Throughout the discussion on generic reducedness: if $V(I)$ is not irreducible, one should take multiple general points, one in each irreducible component of $V(I)$.

(iii) Although the stated theorem relies on $k$ being algebraically closed, the result about primeness of $J$ holds over any field (since base change to the algebraic closure is faithfully flat, and $J$ is defined over $\mathbb{Z}$).

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