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# 数学代写|交换代数代写Commutative Algebra代考|Multivariate Factorization

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## 数学代写|交换代数代写Commutative Algebra代考|Multivariate Factorization

In this chapter we show how to reduce the factorization of multivariate polynomials in $K\left[x_1, \ldots, x_n\right]$ to the case of one variable. The idea is similar to the reduction of the factorization in $\mathbb{Z}[x]$ to the factorization in $\mathbb{Z} / p[x]$. We choose a so-called main variable, say $x_n$ and a suitable point $a=\left(a_1, \ldots, a_{n-1}\right) \in K^{n-1}$. Let $\mathfrak{m}a \in K\left[x_1, \ldots, x{n-1}\right]$ be the maximal ideal corresponding to $a$. We factorize $f\left(a, x_n\right)$ in $K\left[x_n\right]=\left(K\left[x_1, \ldots, x_{n-1}\right] / \mathfrak{m}a\right)\left[x_n\right]$ and use Hensel lifting to lift the factors to $\left(K\left[x_1, \ldots, x{n-1}\right] / \mathfrak{m}_a^N\right)\left[x_n\right]$ for sufficiently large $N$. We choose (unique) representatives of these liftings and combine them to obtain the true factors. Let us start with an example.
Example B.6.1.
\begin{aligned} f= & x^4+(-z+3) x^3+\left(z^3+(y-3) z-y^2-13\right) x^2 \ & +\left(-z^4+\left(y^2+3 y+15\right) z+6\right) x \ & +y z^4+2 z^3+z\left(-y^3-15 y\right)-2 y^2-30 . \end{aligned}
We chose $x$ as main variable, $a=(0,0), \mathfrak{m}_a=\langle y, z\rangle$ and factorize $f(x, 0,0)$. We obtain
$$f(x, 0,0)=x^4+3 x^3-13 x^2+6 x-30=g_1 \cdot h_1$$
with $g_1=x^2+2$ and $h_1=x^2+3 x-15$.
We want to lift the factorization $f=g_1 h_1\left(\bmod \mathfrak{m}_a\right)$ to $f=g_i h_i\left(\bmod \mathfrak{m}_a^i\right)$ for increasing $i$ (Hensel lifting).
We have
\begin{aligned} f-g_1 h_1\left(\bmod \mathfrak{m}_a^2\right) & =-z x^3-3 z x^2+15 z x \ & =-z x \cdot h_1 . \end{aligned}
For the Hensel lifting we obtain
$$h_2=h_1 \text { and } g_2=g_1-z x=x^2-z x+2 .$$
In the next step we have
\begin{aligned} f-g_2 h_2 \bmod \mathfrak{m}_a^3 & =\left(y z-y^2\right) x^2+3 y z x-15 y z-2 y^2 \ & =y z g_2-y^2 \cdot h_2 \end{aligned}

## 数学代写|交换代数代写Commutative Algebra代考|Absolute Factorization

Let $K$ be a field of characteristic $0, \bar{K}$ its algebraic closure and assume we are able to compute the multivariate factorization over algebraic extensions of $K$ (our main example is $K=\mathbb{Q}$ ). In this chapter we explain how to compute the absolute factorization of a polynomial $f \in K\left[x_1, \ldots, x_n\right]$, that is, to compute the irreducible factors (and their multiplicities) of $f$ in $\bar{K}\left[x_1, \ldots, x_n\right]$. To solve this problem we may assume that $f$ is irreducible in $K\left[x_1, \ldots, x_n\right]$.

There exist several approaches to solve this problem (cf. [59], the part written by Chèze and Galligo, or [42]). We concentrate on the algorithm implemented by G. Lecerf in SingULAR.

The idea of this algorithm is to find an algebraic field extension $K(\alpha)$ of $K$ and a smooth point of the affine variety $V(f)$ in $K(\alpha)^n$. Then an (absolutely) irreducible factor of $f$ will be defined over $K(\alpha)$ which can be computed by using the factorization over $K(\alpha)$ described in Section B.5. The idea is based on the following theorem:

Theorem B.7.1. Let $f \in K\left[x_1, \ldots, x_n\right]$ be irreducible and $a \in K^n$ a smooth point of $V(f) \subseteq \bar{K}^n$. Then $f$ is absolutely irreducible, i.e. irreducible in $\bar{K}\left[x_1, \ldots, x_n\right]$

Proof. Let $f=f_1 \cdot \ldots \cdot f_t$ be the factorization of $f$ in $\bar{K}\left[x_1, \ldots, x_n\right]$. We may assume that $f_1(a)=0$. Assume that $t>1$. This implies that $f_i \notin$ $K\left[x_1, \ldots, x_n\right]$ for all $i$. Now $a$ being a smooth point of $V(f)$ implies that $f_i(a) \neq 0$ for $i>1$. We may choose $\alpha \in \bar{K}$ such that $f_i \in K(\alpha)\left[x_1, \ldots, x_n\right]$ for all $i$. Since $f_1 \notin K\left[x_1, \ldots, x_n\right]$ there exist $\sigma \in \operatorname{Gal}_K(K(\alpha))$ such that $\sigma\left(f_1\right) \neq f_1$.

But $\sigma(f)=f=\sigma\left(f_1\right) \cdot \ldots \cdot \sigma\left(f_t\right)$ implies that there is $i \neq 1$ such that $\sigma\left(f_1\right)=c \cdot f_i$ for some non-zero constant $c \in K(\alpha)$.
This implies $0=\sigma\left(f_1(a)\right)=c \cdot f_i(a)$ which is a contradiction. We obtain $t=1$ and $f$ is absolutely irreducible.
If we apply the theorem to $K(\alpha)$ we obtain:
Corollary B.7.2. Let $f \in K\left[x_1, \ldots, x_n\right]$ be irreducible, $\alpha \in \bar{K}$ and $a \in$ $K(\alpha)^n$ a smooth point of $V(f) \subseteq \bar{K}^n$, then at least one absolutely irreducible factor of $f$ is defined over $K(\alpha)$.

For $f$ irreducible it is not difficult to find $\alpha$ and a smooth point of $V(f)$ in $K(\alpha)^n$ (use Lemma B.6.8). We deduce that for irreducible $f \in K\left[x_1, \ldots, x_n\right]$, $\operatorname{deg}_{x_n}(f)>0, f\left(a, x_n\right)$ is squarefree for almost all $a \in K^{n-1}$.

## 数学代写|交换代数代写Commutative Algebra代考|Multivariate Factorization

\begin{aligned} f= & x^4+(-z+3) x^3+\left(z^3+(y-3) z-y^2-13\right) x^2 \ & +\left(-z^4+\left(y^2+3 y+15\right) z+6\right) x \ & +y z^4+2 z^3+z\left(-y^3-15 y\right)-2 y^2-30 . \end{aligned}

$$f(x, 0,0)=x^4+3 x^3-13 x^2+6 x-30=g_1 \cdot h_1$$

\begin{aligned} f-g_1 h_1\left(\bmod \mathfrak{m}_a^2\right) & =-z x^3-3 z x^2+15 z x \ & =-z x \cdot h_1 . \end{aligned}

$$h_2=h_1 \text { and } g_2=g_1-z x=x^2-z x+2 .$$

\begin{aligned} f-g_2 h_2 \bmod \mathfrak{m}_a^3 & =\left(y z-y^2\right) x^2+3 y z x-15 y z-2 y^2 \ & =y z g_2-y^2 \cdot h_2 \end{aligned}

## MATLAB代写

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