# 数学代写|交换代数代写commutative algebra代考|Decomposition of Polynomials into Products of Irreducible Factors

#### Doug I. Jones

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## 数学代写|交换代数代写commutative algebra代考|Decomposition of Polynomials into Products of Irreducible Factors

8.14 Lemma Let $\mathbf{K}$ be a discrete field. The polynomials of $\mathbf{K}[X]$ can be decomposed into products of irreducible factors if and only if we have an algorithm to compute the zeros in $\mathbf{K}$ of an arbitrary polynomial of $\mathbf{K}[X]$.

D The second condition is a priori weaker since it amounts to determining the factors of degree 1 for some polynomial of $\mathbf{K}[X]$. Assume this condition is satisfied. To know whether there exists a decomposition $f=g h$ with $g$ and $h$ monic of fixed degrees $>0$, we apply Kronecker’s theorem. We obtain for each coefficient of $g$ and $h$ a finite number of possibilities (they are the zeros of monic polynomials that we can explicitly express according to the coefficients of $f$ ).
8.15 Proposition In $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$ the polynomials can be decomposed into products of irreducible factors. A nonconstant polynomial of $\mathbb{Z}[X]$ is irreducible in $\mathbb{Z}[X]$ if and only if it is primitive and irreducible in $\mathbb{Q}[X]$.

D For $\mathbb{Q}[X]$ we apply Lemma 8.14. We must therefore show that we know how to determine the rational zeros of a monic polynomial $f$ with rational coefficients. We can even assume that the coefficients of $f$ are integral. The elementary theory of divisibility in $\mathbb{Z}$ shows then that if $a / b$ is a zero of $f, a$ must divide the leading coefficient and $b$ the constant coefficient of $f$; there is therefore only a finite number of tests to execute.

For $\mathbb{Z}[X]$, a primitive polynomial $f$ being given, we want to know if there exists a decomposition $f=g h$ with $g$ and $h$ of fixed degrees $>0$. We can assume $f(0) \neq 0$. We apply Kronecker’s theorem. A product $g_0 h_j$ for instance must be a zero in $\mathbb{Z}$ of a monic polynomial $q_{0, j}$ of $\mathbb{Z}[T]$ that we can compute. In particular, $g_0 h_j$ must divide $q_{0, j}(0)$, which only leaves a finite number of possibilities for $h_j$.

## 数学代写|交换代数代写commutative algebra代考|Rings of Integers of a Number Field

If $\mathbf{K}$ is a number field its ring of integers is the integral closure of $\mathbb{Z}$ in $\mathbf{K}$.
8.17 Proposition and definition (Discriminant of a number field) Let $\mathbf{K}$ be a number field and $\mathbf{Z}$ its ring of integers.

An element $y$ of $\mathbf{K}$ is in $\mathbf{Z}$ if and only if $\operatorname{Min}_{\mathbb{Q}, y}(X) \in \mathbb{Z}[X]$.

We have $\mathbf{K}=\left(\mathbb{N}^*\right)^{-1} \mathbf{Z}$.

Assume that $\mathbf{K}=\mathbb{Q}[x]$ with $x \in \mathbf{Z}$. Let $f(X)=\operatorname{Min}_{\mathbb{Q}, x}(X)$ be in $\mathbb{Z}[X]$ and $\Delta^2$ be the greatest square factor of $\operatorname{disc}_X f$.
Then, $\mathbb{Z}[x] \subseteq \mathbf{Z} \subseteq \frac{1}{\Delta} \mathbb{Z}[x]$

The ring $\mathbf{Z}$ is a free $\mathbb{Z}$-module of rank $[\mathbf{K}: \mathbf{Q}]$.

The integer $\operatorname{Disc}_{\mathbf{Z} / \mathbb{Z}}$ is well-defined. We call it the discriminant of the number field $\mathbf{K}$.
D 1. Results from Lemma 8.10 (Kronecker’s theorem).

Let $y \in \mathbf{K}$ and $g(X) \in \mathbb{Z}[X]$ be a nonzero polynomial that annihilates $y$. If $a$ is the leading coefficient of $g$, ay is integral over $\mathbb{Z}$.

Let $\mathbf{A}=\mathbb{Z}[x]$ and $n=[\mathbf{K}: \mathbf{Q}]$. Let $z \in \mathbf{Z}$, which we as $h(x) / \delta$ with $\delta \in \mathbb{N}^*$, $\langle\delta\rangle+\mathrm{c}(h)=\langle 1\rangle$ and $\operatorname{deg} h<n$. We have $\mathbf{A}+\mathbb{Z} z \subseteq \frac{1}{\delta} \mathbf{A}$ and it thus suffices to prove that $\delta^2$ divides $\operatorname{disc}X(f)$. The ring $\mathbf{A}$ is a free $\mathbb{Z}$-module of rank $n$, with the basis $\mathcal{B}_0=\left(1, x, \ldots, x^{n-1}\right)$. Proposition 5.10 gives $$\operatorname{Disc}{\mathbf{A} / \mathbb{Z}}=\operatorname{disc}{\mathbf{A} / \mathbb{Z}}\left(\mathcal{B}_0\right)=\operatorname{disc}{\mathbf{K} / \mathbb{Q}}\left(\mathcal{B}_0\right)=\operatorname{disc}_X f$$

# 交换代数代考

## 数学代写|交换代数代写commutative algebra代考|Decomposition of Polynomials into Products of Irreducible Factors

8.14引理设$\mathbf{K}$为离散域。当且仅当我们有一种算法来计算$\mathbf{K}[X]$的任意多项式在$\mathbf{K}$中的零时，$\mathbf{K}[X]$的多项式可以分解为不可约因子的乘积。

8.15 In命题 $\mathbb{Z}[X]$ 和 $\mathbb{Q}[X]$ 多项式可以分解成不可约因子的乘积。的非常数多项式 $\mathbb{Z}[X]$ 是不可约的 $\mathbb{Z}[X]$ 当且仅当它是基本且不可约的 $\mathbb{Q}[X]$．

D对于$\mathbb{Q}[X]$，我们应用引理8.14。因此，我们必须证明我们知道如何确定具有有理数系数的单多项式$f$的有理数零。我们甚至可以假设$f$的系数是积分的。由$\mathbb{Z}$的可整除性的基本理论可知，如果$a / b$是$f, a$的零，则必须整除$f$的前导系数和$b$的常系数;因此，要执行的测试数量是有限的。

## 数学代写|交换代数代写commutative algebra代考|Rings of Integers of a Number Field

8.17命题与定义(数域的判别式)设$\mathbf{K}$为一个数域，$\mathbf{Z}$为它的整数环。

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