数学代考|代数几何代写algebraic geometry代考| Computing with Polynomials

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• (Generalized) Linear Models 广义线性模型
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代考|代数几何代写algebraic geometry代考|Divisibility of Polynomials

We begin with some very basic facts about divisibility.
4.1.1 Proposition (Chinese Remainder Theorem) Let $I_{1}, \ldots, I_{n}$ be ideals in a commutative ring $R$ such that $1 \in I_{i}+I_{j}$ for all $i \neq j$. Then $I_{1} \cap \cdots \cap I_{n}=I_{1} \cdots \cdots I_{n}$ (the product of ideals is the ideal generated by $x_{1} \cdots x_{n}$ with $\left.x_{i} \in I_{i}\right)$ and the product of

projections
$$R /\left(I_{1} \cap \cdots \cap I_{n}\right) \rightarrow \prod_{i=1}^{n} R / I_{i}$$
is an isomorphism.
Proof It suffices to consider the case $n=2$ (then we can use induction). For $n=2$, we always have $I_{1} I_{2} \subseteq I_{1} \cap I_{2}$. To show the opposite inequality, let $1=a_{1}+a_{2}$ where $a_{i} \in I_{i}$. Then for $x \in I_{1} \cap I_{2}, x=x a_{1}+x a_{2} \in I_{1} I_{2}$. Now (4.1.1) is always injective since an element goes to 0 on the right hand side if and only if it is in every $I_{i}$. To show surjectivity for $n=2$, choosing $x_{1}, x_{2} \in R$, the element $x_{1} a_{2}+x_{2} a_{1}$ is congruent to $x_{i}$ modulo $I_{i}$ for $i=1,2$, which proves surjectivity.

An element $u \in R$ of a commutative ring is called a unit if there exists another element $u^{-1} \in R$ such that $u u^{-1}=1$. Let $R$ be an integral domain. An irreducible element is an element $x \in R$ which is not zero or a unit such that $y z=x$ implies that one of the elements $y, z$ is a unit. An integral domain $R$ is called a unique factorization domain (or UFD) if every element $x \in R$ which is not 0 or a unit factors uniquely into irreducible elements up to order and multiplication by units, i.e.
$$x=x_{1} \ldots x_{n}$$
where $x_{i}$ are irreducible, and whenever
$$x=y_{1} \ldots y_{m}$$
where $y_{i}$ are irreducible, we have $m=n$ and there exists a permutation $\sigma$ and units $u_{i}$ such that
$$x_{i}=u_{i} y_{\sigma(i)}$$
In a UFD, any set of elements $S$ has a greatest common divisor (GCD) which is an element $x$ dividing all elements of $S$ such that every other elements dividing all elements of $S$ divides $x$. The GCD is, of course, uniquely determined up to multiplication by a unit.
A particular type of example of a UFD is a principal ideal domain (or PID) which means an integral domain whose every ideal is principal (i.e. generated by a single element). In particular, then $R$ is Noetherian, which guarantees that a decomposition into irreducible elements exists. Then the principal ideal property guarantees that an irreducible element $a$ generates a prime ideal: if $x y \in(a)$ and $x \notin(a)$, then $(x, a)=(b)$ for some element $b$, but $b$ has to be a unit by irreducibility. Thus, $y \in(y x, y a) \subseteq(a)$.

数学代考|代数几何代写algebraic geometry代考|Gröbner Basis

We shall now prove that rings of polynomials over a Noetherian ring are Noetherian. In the special case of multivariable polynomials over a field, we can be a lot more explicit, with computational applications.
4.2.1 Theorem (Hilbert Basis Theorem) If a ring $R$ is a Noetherian, then so is the ring of polynomials $R[x]$.

COMMENT In this context, the term “basis” refers to a set of generators of an ideal, no linear independence is implied.

Proof Assume $R$ is Noetherian. Let $I \subseteq R[x]$ be an ideal. Then the top coefficients (i.e. coefficients of the highest power of $x$ ) of all the polynomials $f \in I$ form an ideal $J \subseteq R$ (since two nonzero polynomials of unequal degrees can be brought to the same degree by multiplying the polynomial of lesser degree by a power of $x$ ). By assumption, then, the ideal $J$ is generated by the top coefficients of some polynomials $f_{1}, \ldots, f_{n} \in I$.

Let $d$ be the maximum of the degrees of the polynomials $f_{1}, \ldots, f_{n}$. Then by construction, for any polynomial $g \in I$ of degree $\geq d$, there exist $a_{1}, \ldots a_{n} \in R$, $m_{1} \ldots, m_{n} \in \mathbb{N}{0}$ such the top coefficients of $g(x)$ and $a{1} f_{1}(x) x^{m_{1}}+\ldots a_{n} f_{n}(x) x^{m_{n}}$ coincide. By induction, then, there exists an $R[x]$-linear combination $g_{0}$ of the polynomials $f_{1}, \ldots, f_{n}$ such that $g(x)-g_{0}(x)$ is either 0 or is of degree $<d$.

Now consider for each fixed $i \in \mathbb{N}{0}$ the ideal $J{i} \subseteq R$ of all the top coefficients of all polynomials in $I$ of degree $i$. Then each of these ideals $J_{i}$ is finitely generated, so by taking finitely many polynomials $h_{1}, \ldots, h_{\ell}$ in $I$ of degrees $i=0, \ldots, d-1$ whose top coefficients are the generators of all the $J_{i}$ ‘s, $0 \leq i<d$, we see that every polynomial in $I$ of degree $<d$ is an $R$-linear combination of $h_{1}, \ldots, h_{\ell}$. Thus, we are done.

Next, we shall discuss the ring $k\left[x_{1}, \ldots, x_{n}\right]$ of polynomials in $n$ variables over a field $k$. Even though this ring is not a Euclidean domain for $n>1$ (because it is not a PIDthink, for example, of the ideal $\left(x_{1}, \ldots, x_{n}\right)$ ) there is a certain analog of the long division algorithm which allows us decide, for example, whether a polynomial is an element of a given ideal, or whether two ideals are the same.

By a monomial, we shall mean an expression of the form $x_{1}^{m_{1}} \ldots x_{n}^{m_{n}}$, i.e. equivalently, the $n$-tuple $a=\left(m_{1}, \ldots, m_{n}\right) \in \mathbb{N}{0}^{n}$, which are sometimes referred to as multidegrees. For what follows, we need to fix a monomial order. This means a total ordering $\geq$ on $n$-tuples of non-negative integers (i.e. for any two $n$-tuples $a, b$ we have $a \geq b$ or $b \geq a$ ) which satisfies the descending chain condition (or DCC), i.e. any sequence $a{1} \geq a_{2} \geq \ldots$ is eventually constant. A totally ordered set satisfying the DCC is also sometimes called well ordered. In addition, we require that for multidegrees $a, b$, $c$, if $a \geq b$, then $a+c \geq b+c$.
Note that this implies that the multidegree $(0, \ldots, 0)$ is the smallest (since otherwise, the DCC would be violated). This implies that when $m_{i} \leq p_{i}$ for all $i=1, \ldots n$, then $\left(m_{1}, \ldots, m_{n}\right) \leq\left(p_{1}, \ldots, p_{n}\right)$

数学代考|代数几何代写algebraic geometry代考|Nullstellensatz

4.3.2 Proposition Suppose $k$ is an algebraically closed field. Then every maximal ideal $I \subset k\left[x_{1}, \ldots, x_{n}\right]$ is of the form
$$\left(x_{1}-a_{1}, \ldots, x_{n}-a_{n}\right)$$
for some $a_{1}, \ldots, a_{n} \in k$.
Proof We will show that any ideal $I \subseteq k\left[x_{1}, \ldots, x_{n}\right]$ is contained in an ideal of the form (4.3.1). This is proved by induction on $n$. Suppose the statement is true with $n \geq 1$ replaced by any lower number. (For $n=1$, the assumption is vacuous.) Then there are two possibilities:

Case 1: The ideal $J=I \cap k\left[x_{n}\right]$ in $k\left[x_{n}\right]$ is non-zero. Then, since $k\left[x_{n}\right]$ is a PID, $J=$ $(f)$ is a principal ideal, and since $k$ is algebraically closed, $f$ factors into powers of linear factors $\left(x_{n}-b_{i}\right)^{\ell_{i}}, i=1, \ldots, m$. By the Chinese Remainder Theorem, $k\left[x_{1}, \ldots, x_{n}\right] / I$ is isomorphic to the product of the rings $k\left[x_{1}, \ldots, x_{n}\right] /\left(I+\left(x_{n}-b_{i}\right)^{\ell_{i}}\right)$. Then, for some $i$, $I+\left(x_{n}-b_{i}\right)^{\ell_{i}} \neq(1)$, but this implies $I+\left(x_{n}-b_{n}\right) \neq(1)$ (since an ideal whose radical is (1) is itself (1)). Therefore, we can pass to the ring $k\left[x_{1}, \ldots, x_{n}\right] /\left(x_{n}-b_{i}\right) \cong k\left[x_{1}, \ldots, x_{n-1}\right]$ and use the induction hypothesis.

Case $2: I \cap k\left[x_{n}\right]=(0)$. Therefore, if we set $R=k\left(x_{n}\right)\left[x_{1}, \ldots, x_{n-1}\right]$ (recall that $k(x)$ denotes the field of rational functions in $k$ in one variable), then $I, R \neq R$. Now we can apply the induction hypothesis to the ring of polynomials $\bar{R}=\overline{k\left(x_{n}\right)}\left[x_{1}, \ldots, x_{n-1}\right]$ where $\overline{k\left(x_{n}\right)}$ denotes the algebraic closure of $k\left(x_{n}\right)$. Thus, the ideal $I \cdot \bar{R}$ is contained in an ideal of the form $\left(x_{1}-b_{1}, \ldots, x_{n-1}-b_{n-1}\right)$ for $b_{i} \in \overline{k\left(x_{n}\right)}$. Thus, each $b_{i}$ is the root of a polynomial with coefficients in $k\left(x_{n}\right)$. Now since $k$ is algebraically closed, all of the coefficient polynomials factor into linear factors, and there are only finitely many values of $x_{n} \in k$ for which either the denominator or numerator of any of the coefficient polynomials is 0 . Since $k$ is algebraically closed, it is infinite, and we can choose an element $a_{n} \in k$ which is different from any of those values. Plugging in $x_{n}=a_{n}$, all the expressions for $b_{i}$ give meaningful formulas for elements $a_{i} \in k$. Then, the ideal $I$ is contained in (4.3.1).

数学代考|代数几何代写algebraic geometry代考|Divisibility of Polynomials

4.1.1 命题（中国剩余定理）让一世1,…,一世n是交换环中的理想R这样1∈一世一世+一世j对全部一世≠j. 然后一世1∩⋯∩一世n=一世1⋯⋯一世n（理想的乘积是由X1⋯Xn和X一世∈一世一世)和产品

R/(一世1∩⋯∩一世n)→∏一世=1nR/一世一世

X=X1…Xn

X=是1…是米

X一世=在一世是σ(一世)

UFD 的一个特定类型的示例是主理想域（或 PID），这意味着一个积分域，其每个理想都是主域（即由单个元素生成）。特别是，那么R是 Noetherian，它保证存在分解成不可约元素。那么主理想性质保证一个不可约元素一种产生一个素理想：如果X是∈(一种)和X∉(一种)， 然后(X,一种)=(b)对于某些元素b， 但b必须是不可约的单位。因此，是∈(是X,是一种)⊆(一种).

数学代考|代数几何代写algebraic geometry代考|Gröbner Basis

4.2.1 定理（希尔伯特基定理）如果环R是 Noetherian，那么多项式环也是R[X].

数学代考|代数几何代写algebraic geometry代考|Nullstellensatz

4.3.2 命题假设ķ是代数闭域。那么每个极大理想一世⊂ķ[X1,…,Xn]是形式
(X1−一种1,…,Xn−一种n)

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MATLAB代写

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