# 数学代写|交换代数代写commutative algebra代考|Definition of the Discriminant of a Monic Polynomial

#### Doug I. Jones

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## 数学代写|交换代数代写commutative algebra代考|Definition of the Discriminant of a Monic Polynomial

We define the discriminant of a univariate monic polynomial $f$ over a commutative ring A starting with the case where $f$ is the generic monic polynomial of degree $n$ :
$$f(T)=T^n-S_1 T^{n-1}+S_2 T^{n-2}+\cdots+(-1)^n S_n \in \mathbb{Z}\left[S_1, \ldots, S_n\right][T]=\mathbb{Z}[S][T] .$$
We can write $f(T)=\prod_i\left(T-X_i\right)$ in $\mathbb{Z}\left[X_1, \ldots, X_n\right]$ (Corollary 1.6), and we set
$$\operatorname{disc}T(f)=(-1)^{n(n-1) / 2} \prod{i=1}^n f^{\prime}\left(X_i\right)=\prod_{1 \leqslant i<j \leqslant n}\left(X_i-X_j\right)^2 .$$
As this polynomial in the $X_i$ ‘s is clearly variable permutation invariant, there exists a unique polynomial in the $S_i$ ‘s, $D_n\left(S_1, \ldots, S_n\right) \in \mathbb{Z}[S]$, which is equal to $\operatorname{disc}T(f)$. In short, the auxiliary variables $X_i$ can indeed vanish. Then, for a “concrete” polynomial $$g(T)=T^n-s_1 T^{n-1}+s_2 T^{n-2}+\cdots+(-1)^n s_n \in \mathbf{A}[T],$$ we define $\operatorname{disc}_T(g)=D_n\left(s_1, \ldots, s_n\right)$. Naturally, if it happens that $g(T)=\prod{i=1}^n\left(T-b_i\right)$ in a ring $\mathbf{B} \supseteq \mathbf{A}$, we would then $\operatorname{obtain~}^{\operatorname{disc}T}(g)=\prod{1 \leqslant i<j \leqslant n}\left(b_i-b_j\right)^2$ by evaluating the formula (1). In particular, by using the universal splitting algebra we could directly define the discriminant by this formula.

A monic polynomial is said to be separable when its discriminant is invertible.

Diagonalization of the Matrices on a Ring
Let us first recall that if $f \in \mathbf{A}[T]$, a zero of $f$ in an $\mathbf{A}$-algebra $\mathbf{B}$ (given by a homomorphism $\varphi: \mathbf{A} \rightarrow \mathbf{B})$ is a $y \in \mathbf{B}$ which annihilates the polynomial $f^{\varphi}$, the image of $f$ in $\mathbf{B}[T]$.
In addition, the zero $y$ is said to be simple if $f^{\prime}(y) \in \mathbf{B}^{\times}$(we also say that it is a simple root of $f$ ).
Here, we are interested in the diagonalizations of matrices on an arbitrary commutative ring, when the characteristic polynomial is separable.
First of all, we have the classical “Kernels’ Lemma” II-4.8.
Next is a generalization of the theorem which states (in the discrete field case) that a simple zero of the characteristic polynomial defines a proper subspace of dimension 1.

## 数学代写|交换代数代写commutative algebra代考|The Generic Matrix is Diagonalizable

Consider $n^2$ indeterminates $\left(a_{i, j}\right){i, j \in \llbracket 1 . . n \rrbracket}$ and let $A$ be the corresponding matrix (it has coefficients in $\left.\mathbf{A}=\mathbb{Z}\left[\left(a{i, j}\right)\right]\right)$.
5.3 Proposition The generic matrix $A$ is diagonalizable over a ring $\mathbf{B}$ containing $\mathbb{Z}\left[\left(a_{i, j}\right)\right]=\mathbf{A}$.

Det $f(T)=T^n-s_1 T^{n-1}+\cdots+(-1)^n s_n$ be the characteristic polynomial of $A$. Then the coefficients $s_i$ are algebraically independent over $\mathbb{Z}$. To realize this, it suffices to specialize $A$ as the companion matrix of a generic monic polynomial.
In particular, the discriminant $\Delta=\operatorname{disc}(f)$ is nonzero in the integral ring $\mathbf{A}$. Then consider the ring $\mathbf{A}1=\mathbf{A}[1 / \Delta] \supseteq \mathbf{A}$ and the universal splitting algebra $\mathbf{C}=\operatorname{Adu}{\mathbf{A}1, f}$. Let the $x_i$ be the elements of $\mathbf{C}$ such that $f(T)=\prod_i\left(T-x_i\right)$. Finally, apply Proposition 5.2. If we want to obtain a diagonalizable matrix, we invert for instance $a=\prod_i \operatorname{det}\left(\left(A-x_i \mathbf{I}_n\right){1 . . n-1,1 . . n-1}\right)$. This is an element of $\mathbf{A}$ and it suffices to convince ourselves that it is nonzero by exhibiting a particular matrix, for example the companion matrix of the polynomial $X^n-1$.
Ultimately, consider $\mathbf{A}2=\mathbf{A}[1 /(a \Delta)] \supseteq \mathbf{A}$ and take $\mathbf{B}=\mathrm{Adu}{\mathbf{A}_2, f} \supseteq \mathbf{A}_2$.
The strength of the previous result, “which makes life considerably easier” is illustrated in the following two subsections.

# 交换代数代考

## 数学代写|交换代数代写commutative algebra代考|Definition of the Discriminant of a Monic Polynomial

$$f(T)=T^n-S_1 T^{n-1}+S_2 T^{n-2}+\cdots+(-1)^n S_n \in \mathbb{Z}\left[S_1, \ldots, S_n\right][T]=\mathbb{Z}[S][T] .$$

$$\operatorname{disc}T(f)=(-1)^{n(n-1) / 2} \prod{i=1}^n f^{\prime}\left(X_i\right)=\prod_{1 \leqslant i<j \leqslant n}\left(X_i-X_j\right)^2 .$$

## 数学代写|交换代数代写commutative algebra代考|The Generic Matrix is Diagonalizable

5.3命题一般矩阵$A$在含有$\mathbb{Z}\left[\left(a_{i, j}\right)\right]=\mathbf{A}$的环$\mathbf{B}$上是可对角的。

Det $f(T)=T^n-s_1 T^{n-1}+\cdots+(-1)^n s_n$为$A$的特征多项式。那么系数$s_i$在$\mathbb{Z}$上是代数无关的。为了实现这一点，将$A$专门化为一般单多项式的伴随矩阵就足够了。

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