# Is the ideal generated by coefficients of characteristic polynomial of a matrix prime?

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

• 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. Apr 18, 2023 at 4:20
• See here for an affirmative answer Apr 18, 2023 at 18:11
• @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. Apr 21, 2023 at 9:38
• @imtrying46: This comment explicitly states that the ideal $J$ is radical Apr 21, 2023 at 15:26

(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.

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

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.

(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 $$\mathbb{Z}$$).