Write $A=(x_{ij})$ for the generic matrix (comprised of indeterminates) defined over $\mathbb Z[x_{11},\dots,x_{nn}]$. In their constructive commutative algebra book, Lombardi and Quitte write that the determinant of the family $(e_1,Ae_1,\dots,A^{n-1}e_1)$ is nonzero. Hence this is a cyclic basis whence the generic matrix is similar to the companion matrix of its characteristic polynomial over the fraction field of $\mathbb Z[x_{ij}]$. If I understand correctly, being non-zero in the polynomial domain ensures the localization at the determinant is injective and there we already have the similarity. On the other hand, the determinant is nonzero because it can be specialized to one which is nonzero.
Now, Chapter III proposition 5.3 "the generic matrix is diagonalizable" has me a little stuck. The proof is as follows.
The authors first claim the coefficients of the characteristic polynomial of $A$ are algebraically independent over $\mathbb Z$. They write "to realize this, it suffices to speicalize $A$ as the companion matrix of a generic monic polynomial". Does this mean passing to a basis in which the representation is the companion matrix of the characteristic polynomial? How does this prove the assertion?
They then claim the discriminant of the characteristic polynomial is nonzero. This makes sense because otherwise every split characteristic polynomial would have a repeated root, but it seems the remainder of the proof makes no mention of the algebraic independence of the roots. What am I missing?
Best Answer
Let me answer both of your explicitly asked questions in detail; if you have any further questions on the proof (which is indeed fast-going and slightly handwavy), please add them to your post.
Answer: Let $\mathbf{R}$ be the polynomial ring $\mathbb{Z}\left[c_0, c_1, \ldots, c_{n-1}\right]$ in $n$ new indeterminates $c_0, c_1, \ldots, c_{n-1}$ over $\mathbb{Z}$. Let $C$ be the $n\times n$-matrix \begin{align} \begin{pmatrix} 0 & 0 & \dots & 0 & -c_0 \\ 1 & 0 & \dots & 0 & -c_1 \\ 0 & 1 & \dots & 0 & -c_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & 1 & -c_{n-1} \end{pmatrix} \end{align} over $\mathbf{R}$. This matrix $C$ is the companion matrix of the polynomial $c_0 + c_1 T + c_2 T^2 + \cdots + c_{n-1} T^{n-1} + T^n \in \mathbf{R}\left[T\right]$ (in fact, I even stole the LaTeX code from the Wikipedia article). Now, let $\phi$ be the $\mathbb{Z}$-algebra homomorphism from $\mathbf{A}$ to $\mathbf{R}$ that sends each entry of the matrix $\left(a_{i,j}\right)$ to the corresponding entry of $C$ (that is, $a_{i,j}$ goes to $-c_{i-1}$ if $j = n$, goes to $1$ if $i = j+1$, and goes to $0$ otherwise). This $\phi$ is uniquely determined, due to the universal property of the polynomial ring $\mathbf{A}$. This homomorphism $\phi$ canonically induces a $\mathbb{Z}\left[T\right]$-algebra homomorphism $\phi\left[T\right] : \mathbf{A}\left[T\right] \to \mathbf{R}\left[T\right]$ (which simply applies $\phi$ to each coefficient independently). This latter homomorphism $\phi\left[T\right]$ sends the characteristic polynomial of the matrix $\left(a_{i,j}\right)$ to the characteristic polynomial of the matrix $C$ (since $\phi$ sends the matrix $\left(a_{i,j}\right)$ to the matrix $C$, and since each coefficient of the characteristic polynomial of a matrix is a universal polynomial in the entries of the matrix). In other words, $\phi\left[T\right]$ sends the polynomial $f\left(T\right)$ to the polynomial $c_0 + c_1 T + c_2 T^2 + \cdots + c_{n-1} T^{n-1} + T^n$. Thus, $\phi$ sends each coefficient $\left(-1\right)^{n-i}s_{n-i}$ of the former polynomial to the corresponding coefficient $c_i$ of the latter. Since the $c_0, c_1, \ldots, c_{n-1}$ are algebraically independent over $\mathbb{Z}$ (by their definition as distinct indeterminates!), we thus conclude that the $\left(-1\right)^n s_n, \left(-1\right)^{n-1} s_{n-1}, \ldots, -s_1$ are algebraically independent over $\mathbb{Z}$ as well (because any polynomial relation between them would be mapped by $\phi$ to a polynomial relation between $c_0, c_1, \ldots, c_{n-1}$, which would contradict the algebraic independence of the latter). Hence, the $s_1, s_2, \ldots, s_n$ are algebraically independent over $\mathbb{Z}$ (because changing the order of a bunch of elements and flipping some of their signs clearly cannot damage their algebraic independence).
Answer: We know that the coefficients $\pm s_1, \pm s_2, \ldots, \pm s_n$ of $f$ are algebraically independent. Thus, $f$ is "as good as" a generic monic polynomial of degree $n$ (in the sense that there is a ring isomorphism from the above-mentioned polynomial ring $\mathbf{R} = \mathbb{Z}\left[c_0, c_1, \ldots, c_{n-1}\right]$ to a subring of $\mathbf{A}$ that sends the generic monic polynomial $c_0 + c_1 T + c_2 T^2 + \cdots + c_{n-1} T^{n-1} + T^n$ to $f$). Thus, proving that the discriminant of $f$ is nonzero is equivalent to proving that the discriminant of the generic monic polynomial $c_0 + c_1 T + c_2 T^2 + \cdots + c_{n-1} T^{n-1} + T^n$ is nonzero. But the latter is easy: If the discriminant of the generic monic polynomial $c_0 + c_1 T + c_2 T^2 + \cdots + c_{n-1} T^{n-1} + T^n$ was zero, then the discriminant of every monic polynomial of degree $n$ would be $0$ (since it could be obtained by specializing the $c_0, c_1, \ldots, c_{n-1}$ in the discriminant of the generic monic polynomial); but this would contradict the fact that (for example) the monic polynomial $T^n - 1$ has nonzero discriminant.