多元多项式在域上的运算和GRÖBNER基

9.6 POLYNOMIALS IN SEVERAL VARIABLES OVER A FIELD AND GRÖBNER BASES

9.6 多元多项式在域上的运算和格罗布纳基

In this section we consider polynomials in many variables, present some basic computational tools, and indicate some applications. The results of this section are not required in Chapters 10 through 14. Additional applications will be given in Chapter 15.

在这一节中，我们考虑多个变量的多项式，介绍一些基本的计算工具，并指出一些应用。本节的结果在第10章到第14章中不是必需的。在第15章中将给出额外的应用。

We proved in Section 2 that a polynomial ring $F\left\lbrack x\right\rbrack$ in a variable $x$ over a field $F$ is a Euclidean Domain,and Corollary 8 showed that the polynomial ring $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ is a U.F.D. However it follows from Corollary 8 in Section 8.2 that the latter ring is not a P.I.D. unless $n = 1$ . Our first result below shows that ideals in such polynomial rings, although not necessarily principal, are always finitely generated. General rings with this property are given a special name:

我们在第二章证明了变量 $F\left\lbrack x\right\rbrack$ 上关于域 $x$ 的多项式环 $F$ 是一个欧几里得域，并且推论8表明该多项式环 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 是一个唯一分解整环（U.F.D.）。然而，从8.2节中的推论8可以得出，除非 $n = 1$ ，否则后者不是一个主理想整环（P.I.D.）。下面我们的第一个结果表明，此类多项式环中的理想，尽管不一定为主理想，但总是有限生成的。具有这种性质的一般环被赋予了一个特殊名称：

Definition. A commutative ring $R$ with 1 is called Noetherian if every ideal of $R$ is finitely generated.

定义。如果一个带有单位元 $R$ 的交换环是每个理想都是有限生成的，则该环被称为诺特环。

Noetherian rings will be studied in greater detail in Chapters 15 and 16. In this section we develop some of the basic theory and resulting algorithms for working with (finitely generated) ideals in $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ .

诺特环将在第15章和第16章中详细研究。在本节中，我们开发了一些基本理论以及用于处理 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 中的（有限生成的）理想的相关算法。

As we saw in Section 1,a polynomial ring in $n$ variables can be considered as a polynomial ring in one variable with coefficients in a polynomial ring in $n - 1$ variables. By following this inductive approach—as we did in Theorem 7 and Corollary 8—we can deduce that $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ is Noetherian from the following more general result.

正如我们在第一章中看到的，$n$ 个变量的多项式环可以被视为一个变量的多项式环，其系数在 $n - 1$ 个变量的多项式环中。通过遵循这种归纳方法——正如我们在定理7和推论8中所做的那样——我们可以从以下更一般的结果推断出 $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ 是诺特环。

Theorem 21. (Hilbert’s Basis Theorem) If $R$ is a Noetherian ring then so is the polynomial ring $R\left\lbrack x\right\rbrack$ .

定理21。（希尔伯特基定理）如果 $R$ 是一个诺特环，那么多项式环 $R\left\lbrack x\right\rbrack$ 也是诺特环。

Proof: Let $I$ be an ideal in $R\left\lbrack x\right\rbrack$ and let $L$ be the set of all leading coefficients of the elements in $I$ . We first show that $L$ is an ideal of $R$ ,as follows. Since $I$ contains the zero polynomial, $0 \in L$ . Let $f = a{x}^{d} + \cdots$ and $g = b{x}^{e} + \cdots$ be polynomials in $I$ of degrees $d,e$ and leading coefficients $a,b \in R$ . Then for any $r \in R$ either ${ra} - b$ is zero or it is the leading coefficient of the polynomial $r{x}^{e}f - {x}^{d}g$ . Since the latter polynomial is in $I$ we have ${ra} - b \in L$ ,which shows $L$ is an ideal of $R$ . Since $R$ is assumed Noetherian,the ideal $L$ in $R$ is finitely generated,say by ${a}_{1},{a}_{2},\ldots ,{a}_{n} \in R$ . For each $i = 1,\ldots ,n$ let ${f}_{i}$ be an element of $I$ whose leading coefficient is ${a}_{i}$ . Let ${e}_{i}$ denote the degree of ${f}_{i}$ ,and let $N$ be the maximum of ${e}_{1},{e}_{2},\ldots ,{e}_{n}$ .

证明：设 $I$ 是 $R\left\lbrack x\right\rbrack$ 中的一个理想，设 $L$ 为 $I$ 中所有元素的首项系数的集合。我们首先证明 $L$ 是 $R$ 的一个理想，如下所示。由于 $I$ 包含零多项式，$0 \in L$。设 $f = a{x}^{d} + \cdots$ 和 $g = b{x}^{e} + \cdots$ 是 $I$ 中次数分别为 $d,e$ 且首项系数分别为 $a,b \in R$ 的多项式。那么对于任意的 $r \in R$，要么 ${ra} - b$ 是零，要么它是多项式 $r{x}^{e}f - {x}^{d}g$ 的首项系数。由于后者多项式在 $I$ 中，因此我们有 ${ra} - b \in L$，这表明 $L$ 是 $R$ 的一个理想。由于假设 $R$ 是诺特环，$R$ 中的理想 $L$ 是有限生成的，设生成元为 ${a}_{1},{a}_{2},\ldots ,{a}_{n} \in R$。对于每个 $i = 1,\ldots ,n$，设 ${f}_{i}$ 是 $I$ 中首项系数为 ${a}_{i}$ 的一个元素。设 ${e}_{i}$ 表示 ${f}_{i}$ 的次数，$N$ 是 ${e}_{1},{e}_{2},\ldots ,{e}_{n}$ 中的最大值。

For each $d \in \{ 0,1,\ldots ,N - 1\}$ ,let ${L}_{d}$ be the set of all leading coefficients of polynomials in $I$ of degree $d$ together with 0 . A similar argument as that for $L$ shows each ${L}_{d}$ is also an ideal of $R$ ,again finitely generated since $R$ is Noetherian. For each nonzero ideal ${L}_{d}$ let ${b}_{d,1},{b}_{d,2},\ldots ,{b}_{d,{n}_{d}} \in R$ be a set of generators for ${L}_{d}$ ,and let ${f}_{d,i}$ be a polynomial in $I$ of degree $d$ with leading coefficient ${b}_{d,i}$ .

对于每个 $d \in \{ 0,1,\ldots ,N - 1\}$，设 ${L}_{d}$ 为所有次数为 $d$ 的多项式的首项系数的集合，包括 0。类似于 $L$ 的论证表明每个 ${L}_{d}$ 也是 $R$ 的一个理想，由于 $R$ 是诺特环，因此也是有限生成的。对于每个非零理想 ${L}_{d}$，设 ${b}_{d,1},{b}_{d,2},\ldots ,{b}_{d,{n}_{d}} \in R$ 为 ${L}_{d}$ 的一组生成元，设 ${f}_{d,i}$ 为 $I$ 中次数为 $d$ 且首项系数为 ${b}_{d,i}$ 的一个多项式。

We show that the polynomials ${f}_{1},\ldots ,{f}_{n}$ together with all the polynomials ${f}_{d,i}$ for all the nonzero ideals ${L}_{d}$ are a set of generators for $I$ ,i.e.,that

我们证明，多项式 ${f}_{1},\ldots ,{f}_{n}$ 以及所有非零理想 ${L}_{d}$ 的所有多项式 ${f}_{d,i}$ 构成了 $I$ 的一组生成元，即

By construction,the ideal ${I}^{\prime }$ on the right above is contained in $I$ since all the generators were chosen in $I$ . If ${I}^{\prime } \neq I$ ,there exists a nonzero polynomial $f \in I$ of minimum degree with $f \notin {I}^{\prime }$ . Let $d = \deg f$ and let $a$ be the leading coefficient of $f$ .

根据构造，上述右侧的理想 ${I}^{\prime }$ 包含在 $I$ 中，因为所有生成元都是从 $I$ 中选择的。如果 ${I}^{\prime } \neq I$ ，则存在一个最小次数的非零多项式 $f \in I$ ，使得 $f \notin {I}^{\prime }$ 。设 $d = \deg f$ ，并设 $a$ 为 $f$ 的首项系数。

Suppose first that $d \geq N$ . Since $a \in L$ we may write $a$ as an $R$ -linear combination of the generators of $L : a = {r}_{1}{a}_{1} + \cdots + {r}_{n}{a}_{n}$ . Then $g = {r}_{1}{x}^{d - {e}_{1}}{f}_{1} + \cdots + {r}_{n}{x}^{d - {e}_{n}}{f}_{n}$ is an element of ${I}^{\prime }$ with the same degree $d$ and the same leading coefficient $a$ as $f$ . Then $f - g \in I$ is a polynomial in $I$ of smaller degree than $f$ . By the minimality of $f$ ,we must have $f - g = 0$ ,so $f = g \in {I}^{\prime }$ ,a contradiction.

首先假设 $d \geq N$ 。由于 $a \in L$ ，我们可以将 $a$ 写成 $R$ -线性组合的生成元。那么 $g = {r}_{1}{x}^{d - {e}_{1}}{f}_{1} + \cdots + {r}_{n}{x}^{d - {e}_{n}}{f}_{n}$ 是 ${I}^{\prime }$ 的一个元素，具有与 $f$ 相同的次数和相同的首项系数 $a$ 。那么 $f - g \in I$ 是 $I$ 中次数小于 $f$ 的多项式。由于 $f$ 的极小性，我们必须有 $f - g = 0$ ，因此 $f = g \in {I}^{\prime }$ ，产生矛盾。

Suppose next that $d < N$ . In this case $a \in {L}_{d}$ for some $d < N$ ,and so we may write $a = {r}_{1}{b}_{d,1} + \cdots + {r}_{{n}_{d}}{b}_{{n}_{d}}$ for some ${r}_{i} \in R$ . Then $g = {r}_{1}{f}_{d,1} + \cdots + {r}_{{n}_{d}}{f}_{{n}_{d}}$ is a polynomial in ${I}^{\prime }$ with the same degree $d$ and the same leading coefficient $a$ as $f$ ,and we have a contradiction as before.

接下来假设 $d < N$ 。在这种情况下 $a \in {L}_{d}$ 对于某个 $d < N$ ，因此我们可以写出 $a = {r}_{1}{b}_{d,1} + \cdots + {r}_{{n}_{d}}{b}_{{n}_{d}}$ 对于某个 ${r}_{i} \in R$ 。那么 $g = {r}_{1}{f}_{d,1} + \cdots + {r}_{{n}_{d}}{f}_{{n}_{d}}$ 是 ${I}^{\prime }$ 中的一个多项式，具有与 $f$ 相同的次数和相同的首项系数 $a$ ，我们像之前一样得到矛盾。

It follows that $I = {I}^{\prime }$ is finitely generated,and since $I$ was arbitrary,this completes the proof that $R\left\lbrack x\right\rbrack$ is Noetherian.

这意味着 $I = {I}^{\prime }$ 是有限生成的，由于 $I$ 是任意的，这完成了 $R\left\lbrack x\right\rbrack$ 是诺特环的证明。

Since a field is clearly Noetherian, Hilbert's Basis Theorem and induction immediately give:

由于一个域显然是诺特的，希尔伯特基定理和归纳法立即给出：

Corollary 22. Every ideal in the polynomial ring $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ with coefficients from a field $F$ is finitely generated.

推论 22. 多项式环 $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ 中带有来自域 $F$ 的系数的每个理想都是有限生成的。

If $I$ is an ideal in $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ generated by a (possibly infinite) set $\mathcal{S}$ of polynomials,Corollary 22 shows that $I$ is finitely generated,and in fact $I$ is generated by a finite number of the polynomials from the set $\mathcal{S}$ (cf. Exercise 1).

如果 $I$ 是由 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 中的一组（可能是无限的）多项式 $\mathcal{S}$ 生成的理想，推论 22 显示 $I$ 是有限生成的，实际上 $I$ 是由集合 $\mathcal{S}$ 中的有限个多项式生成的（参见练习 1）。

As the proof of Hilbert’s Basis Theorem shows, the collection of leading coefficients of the polynomials in an ideal $I$ in $R\left\lbrack x\right\rbrack$ forms an extremely useful ideal in $R$ that can be used to understand $I$ . This suggests studying "leading terms" in $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ more generally (and somewhat more intrinsically). To do this we need to specify a total ordering on the monomials, since without some sort of ordering we cannot in general tell which is the “leading” term of a polynomial. We implicitly chose such an ordering in the inductive proof of Corollary 22- we first viewed a polynomial $f$ as a polynomial in ${x}_{1}$ with coefficients in $R = F\left\lbrack {{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ ,say,then viewed its "leading coefficient" in $F\left\lbrack {{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ as a polynomial in ${x}_{2}$ with coefficients in $F\left\lbrack {{x}_{3},\ldots ,{x}_{n}}\right\rbrack$ , etc. This is an example of a lexicographic monomial ordering on the polynomial ring $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ which is defined by first declaring an ordering of the variables,for example ${x}_{1} > {x}_{2} > \cdots > {x}_{n}$ and then declaring that the monomial term $A{x}_{1}^{{a}_{1}}{x}_{2}^{{a}_{2}}\cdots {x}_{n}^{{a}_{n}}$ with exponents $\left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right)$ has higher order than the monomial term $B{x}_{1}^{{b}_{1}}{x}_{2}^{{b}_{2}}\cdots {x}_{n}^{{b}_{n}}$ with exponents $\left( {{b}_{1},{b}_{2},\ldots ,{b}_{n}}\right)$ if the first component where the $n$ -tuples differ has ${a}_{i} > {b}_{i}$ . This is analogous to the ordering used in a dictionary (hence the name), where the letter “a” comes before “b” which in turn comes before “c”, etc., and then “aardvark” comes before “abacus” (although the ‘word’ ${a}^{2} = {aa}$ comes before $a$ in the lexicographical order). Note that the ordering is only defined up to multiplication by units (elements of ${F}^{ \times }$ ) and that multiplying two monomials by the same nonzero monomial does not change their ordering. This can be formalized in general.

正如希尔伯特基定理的证明所显示，理想 $I$ 在 $R\left\lbrack x\right\rbrack$ 中的多项式首项系数的集合在 $R$ 中构成了一个非常有用的理想，可以用来理解 $I$ 。这提示我们更一般地（并且 somewhat 更本质地）研究 $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ 中的“首项”。为此，我们需要在单项式上指定一个全序，因为没有某种排序，我们通常无法判断多项式的“首项”是什么。我们在推论 22 的归纳证明中隐式地选择了这样的排序 - 我们首先将多项式 $f$ 视为在 ${x}_{1}$ 中的多项式，其系数在 $R = F\left\lbrack {{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ 中，比如说，然后将其在 $F\left\lbrack {{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ 中的“首项系数”视为在 ${x}_{2}$ 中的多项式，其系数在 $F\left\lbrack {{x}_{3},\ldots ,{x}_{n}}\right\rbrack$ 中，等等。这是多项式环 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 上字典序单项式排序的一个例子，它通过首先声明变量的排序，例如 ${x}_{1} > {x}_{2} > \cdots > {x}_{n}$ ，然后声明如果第一个分量中 $n$ -元组不同的单项式项 $A{x}_{1}^{{a}_{1}}{x}_{2}^{{a}_{2}}\cdots {x}_{n}^{{a}_{n}}$ 的指数 $\left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right)$ 大于单项式项 $B{x}_{1}^{{b}_{1}}{x}_{2}^{{b}_{2}}\cdots {x}_{n}^{{b}_{n}}$ 的指数 $\left( {{b}_{1},{b}_{2},\ldots ,{b}_{n}}\right)$ ，则该单项式项具有更高阶。这类似于字典中使用的排序（因此得名），其中字母“a”排在“b”之前，而“b”又排在“c”之前，等等，然后“aardvark”排在“abacus”之前（尽管在字典序中，'词' ${a}^{2} = {aa}$ 排在 $a$ 之前）。请注意，该排序仅在乘以单位（${F}^{ \times }$ 的元素）时定义，并且将两个单项式乘以相同的非零单项式不会改变它们的排序。这可以在一般情况下形式化。

Definition. A monomial ordering is a well ordering " $\geq$ " on the set of monomials that satisfies $m{m}_{1} \geq m{m}_{2}$ whenever ${m}_{1} \geq {m}_{2}$ for monomials $m,{m}_{1},{m}_{2}$ . Equivalently, a monomial ordering may be specified by defining a well ordering on the $n$ -tuples $\alpha = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \in {\mathbb{Z}}^{n}$ of multidegrees of monomials $A{x}_{1}^{{a}_{1}}\cdots {x}_{n}^{{a}_{n}}$ that satisfies $\alpha + \gamma \geq \beta + \gamma$ if $\alpha \geq \beta .$

定义。单项式排序是一个良序 " $\geq$ " 定义在单项式集合上，它满足 $m{m}_{1} \geq m{m}_{2}$ 当 ${m}_{1} \geq {m}_{2}$ 对于单项式 $m,{m}_{1},{m}_{2}$ 。等价地，单项式排序可以通过在 $n$ -元组 $\alpha = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \in {\mathbb{Z}}^{n}$ 的单项式的多重度上定义一个良序来实现，该良序满足 $A{x}_{1}^{{a}_{1}}\cdots {x}_{n}^{{a}_{n}}$ 如果 $\alpha + \gamma \geq \beta + \gamma$

It is easy to show for any monomial ordering that $m \geq 1$ for every monomial $m$ (cf. Exercise 2). It is not difficult to show, using Hilbert’s Basis Theorem, that any total ordering on monomials which for every monomial $m$ satisfies $m \geq 1$ and $m{m}_{1} \geq m{m}_{2}$ whenever ${m}_{1} \geq {m}_{2}$ ,is necessarily a well ordering (hence a monomial ordering)-this equivalent set of axioms for a monomial ordering may be easier to verify. For simplicity we shall limit the examples to the particularly easy and intuitive lexicographic ordering, but it is important to note that there are useful computational advantages to using other monomial orderings in practice. Some additional commonly used monomial orderings are introduced in the exercises.

对于任何单项式排序，很容易证明 $m \geq 1$ 对于每个单项式 $m$ （参见练习2）。使用希尔伯特基定理，不难证明，对于每个单项式 $m$ ，任何满足 $m \geq 1$ 和 $m{m}_{1} \geq m{m}_{2}$ 当 ${m}_{1} \geq {m}_{2}$ 的单项式的全序，必然是一个良序（因此是一个单项式排序）- 这组单项式排序的等价公理可能更容易验证。为了简单起见，我们将例子限制在特别简单且直观的字典序上，但重要的是要注意，在实践中使用其他单项式排序具有有用的计算优势。一些其他常用的单项式排序在练习中介绍。

As mentioned, once we have a monomial ordering we can define the leading term of a polynomial:

如前所述，一旦我们有了单项式排序，我们就可以定义多项式的前项：

Definition. Fix a monomial ordering on the polynomial ring $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ .

定义。在多项式环 $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ 上固定一个单项式排序。

(1) The leading term of a nonzero polynomial $f$ in $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ ,denoted ${LT}\left( f\right)$ ,is the monomial term of maximal order in $f$ and the leading term of $f = 0$ is 0. Define the multidegree of $f$ ,denoted $\partial \left( f\right)$ ,to be the multidegree of the leading term of $f$ .

（1）非零多项式 $f$ 在 $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ 中的首项，记作 ${LT}\left( f\right)$ ，是 $f$ 中最高阶的单项式项，且 $f = 0$ 的首项为0。定义 $f$ 的多重度，记作 $\partial \left( f\right)$ ，为 $f$ 的首项的多重度。

(2) If $I$ is an ideal in $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ ,the ideal of leading terms,denoted ${LT}\left( I\right)$ , is the ideal generated by the leading terms of all the elements in the ideal, i.e., ${LT}\left( I\right) = \left( {{LT}\left( f\right) \mid f \in I}\right) .$

(2) 如果 $I$ 是 $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ 中的一个理想，那么首项的理想，记作 ${LT}\left( I\right)$ ，是由该理想中所有元素的首项生成的理想，即 ${LT}\left( I\right) = \left( {{LT}\left( f\right) \mid f \in I}\right) .$。

The leading term and the multidegree of a polynomial clearly depend on the choice of the ordering. For example ${LT}\left( {{2xy} + {y}^{3}}\right) = {2xy}$ with multidegree(1,1)if $x > y$ , but ${LT}\left( {{2xy} + {y}^{3}}\right) = {y}^{3}$ with multidegree(0,3)if $y > x$ . In particular,the leading term of a polynomial need not be the term of largest total degree. Similarly, the ideal of leading terms ${LT}\left( I\right)$ of an ideal $I$ in general depends on the ordering used. Note also that the multidegree of a polynomial satisfies $\partial \left( {fg}\right) = \partial f + \partial g$ when $f$ and $g$ are nonzero,and that in this case ${LT}\left( {fg}\right) = {LT}\left( f\right) + {LT}\left( g\right)$ (cf. Exercise 2).

多项式的首项和多重度显然依赖于排序的选择。例如 ${LT}\left( {{2xy} + {y}^{3}}\right) = {2xy}$ 在 $x > y$ 时具有多重度(1,1)，但在 ${LT}\left( {{2xy} + {y}^{3}}\right) = {y}^{3}$ 时具有多重度(0,3)。特别是，多项式的首项不必是总次数最大的项。类似地，一个理想 $I$ 的首项理想 ${LT}\left( I\right)$ 通常也依赖于所用的排序。还应注意，当 $f$ 和 $g$ 不为零时，多项式的多重度满足 $\partial \left( {fg}\right) = \partial f + \partial g$，并且在这种情况下 ${LT}\left( {fg}\right) = {LT}\left( f\right) + {LT}\left( g\right)$（参见图练习2）。

The ideal ${LT}\left( I\right)$ is by definition generated by monomials. Such ideals are called monomial ideals and are typically much easier to work with than generic ideals. For example, a polynomial is contained in a monomial ideal if and only if each of its monomial terms is a multiple of one of the generators for the ideal (cf. Exercise 10).

理想 ${LT}\left( I\right)$ 由定义上是单项式生成的。这样的理想被称为单项式理想，通常比一般理想更容易处理。例如，一个多项式包含在单项式理想中当且仅当它的每一个单项式项是理想生成元之一的倍数（参见图练习10）。

It was important in the proof of Hilbert's Basis Theorem to have all of the leading terms of the ideal $I$ . If $I = \left( {{f}_{1},\ldots ,{f}_{m}}\right)$ ,then ${LT}\left( I\right)$ contains the leading terms ${LT}\left( {f}_{1}\right) ,\ldots ,{LT}\left( {f}_{m}\right)$ of the generators for $I$ by definition. Since ${LT}\left( I\right)$ is an ideal,it contains the ideal generated by these leading terms:

在证明希尔伯特基定理时，拥有理想 $I$ 的所有首项是很重要的。如果 $I = \left( {{f}_{1},\ldots ,{f}_{m}}\right)$ ，那么 ${LT}\left( I\right)$ 根据定义包含 $I$ 的生成元的首项 ${LT}\left( {f}_{1}\right) ,\ldots ,{LT}\left( {f}_{m}\right)$。由于 ${LT}\left( I\right)$ 是一个理想，它包含由这些首项生成的理想：

The first of the following examples shows that the ideal ${LT}\left( I\right)$ of leading terms can in general be strictly larger than the ideal generated just by the leading terms of some generators for $I$ .

下面示例中的第一个表明，首项理想 ${LT}\left( I\right)$ 通常可以严格大于仅由 $I$ 的某些生成元的首项生成的理想。

Examples

示例

(1) Choose the lexicographic ordering $x > y$ on $F\left\lbrack {x,y}\right\rbrack$ . The leading terms of the polynomials ${f}_{1} = {x}^{3}y - x{y}^{2} + 1$ and ${f}_{2} = {x}^{2}{y}^{2} - {y}^{3} - 1$ are ${LT}\left( {f}_{1}\right) = {x}^{3}y$ (so the multidegree of $\left. {{f}_{1}\text{is}\partial \left( {f}_{1}\right) = \left( {3,1}\right) }\right)$ and ${LT}\left( {f}_{2}\right) = {x}^{2}{y}^{2}$ (so $\partial \left( {f}_{2}\right) = \left( {2,2}\right)$ ). If $I = \left( {{f}_{1},{f}_{2}}\right)$ is the ideal generated by ${f}_{1}$ and ${f}_{2}$ then the leading term ideal ${LT}\left( I\right)$ contains ${LT}\left( {f}_{1}\right) = {x}^{3}y$ and ${LT}\left( {f}_{2}\right) = {x}^{2}{y}^{2}$ ,so $\left( {{x}^{3}y,{x}^{2}{y}^{2}}\right) \subseteq {LT}\left( I\right)$ . Since

(1) 在 $x > y$ 上选择字典序 $F\left\lbrack {x,y}\right\rbrack$ 。多项式 ${f}_{1} = {x}^{3}y - x{y}^{2} + 1$ 和 ${f}_{2} = {x}^{2}{y}^{2} - {y}^{3} - 1$ 的首项是 ${LT}\left( {f}_{1}\right) = {x}^{3}y$（因此 $\left. {{f}_{1}\text{is}\partial \left( {f}_{1}\right) = \left( {3,1}\right) }\right)$ 的多重度以及 ${LT}\left( {f}_{2}\right) = {x}^{2}{y}^{2}$（所以 $\partial \left( {f}_{2}\right) = \left( {2,2}\right)$ ）。如果 $I = \left( {{f}_{1},{f}_{2}}\right)$ 是由 ${f}_{1}$ 和 ${f}_{2}$ 生成的理想，那么首项理想 ${LT}\left( I\right)$ 包含 ${LT}\left( {f}_{1}\right) = {x}^{3}y$ 和 ${LT}\left( {f}_{2}\right) = {x}^{2}{y}^{2}$，所以 $\left( {{x}^{3}y,{x}^{2}{y}^{2}}\right) \subseteq {LT}\left( I\right)$ 。由于

we see that $g = x + y$ is an element of $I$ and so the ideal ${LT}\left( I\right)$ also contains the leading term ${LT}\left( g\right) = x$ . This shows that ${LT}\left( I\right)$ is strictly larger than $\left( {{LT}\left( {f}_{1}\right) ,{LT}\left( {f}_{2}\right) }\right)$ , since every element in $\left( {{LT}\left( {f}_{1}\right) ,{LT}\left( {f}_{2}\right) }\right) = \left( {{x}^{3}y,{x}^{2}{y}^{2}}\right)$ has total degree at least 4 . We shall see later that in this case ${LT}\left( I\right) = \left( {x,{y}^{4}}\right)$ .

我们看到 $g = x + y$ 是 $I$ 的一个元素，因此理想 ${LT}\left( I\right)$ 也包含首项 ${LT}\left( g\right) = x$ 。这表明 ${LT}\left( I\right)$ 严格大于 $\left( {{LT}\left( {f}_{1}\right) ,{LT}\left( {f}_{2}\right) }\right)$ ，因为 $\left( {{LT}\left( {f}_{1}\right) ,{LT}\left( {f}_{2}\right) }\right) = \left( {{x}^{3}y,{x}^{2}{y}^{2}}\right)$ 中的每个元素的总次数至少为4。稍后我们将看到，在这种情况下 ${LT}\left( I\right) = \left( {x,{y}^{4}}\right)$ 。

(2) With respect to the lexicographic ordering $y > x$ ,the leading terms of ${f}_{1}$ and ${f}_{2}$ in the previous example are ${LT}\left( {f}_{1}\right) = - x{y}^{2}$ (which one could write as $- {y}^{2}x$ to emphasize the chosen ordering) and ${LT}\left( {f}_{2}\right) = - {y}^{3}$ . We shall see later that in this ordering ${LT}\left( I\right) = \left( {{x}^{4},y}\right)$ ,which is a different ideal than the ideal ${LT}\left( I\right)$ obtained in the previous example using the ordering $x > y$ ,and is again strictly larger than $\left( {{LT}\left( {f}_{1}\right) ,{LT}\left( {f}_{2}\right) }\right)$ .

(2) 关于字典序 $y > x$ ，前一个例子中 ${f}_{1}$ 和 ${f}_{2}$ 的首项是 ${LT}\left( {f}_{1}\right) = - x{y}^{2}$（可以写成 $- {y}^{2}x$ 来强调所选的排序）和 ${LT}\left( {f}_{2}\right) = - {y}^{3}$ 。稍后我们将看到，在这种排序中 ${LT}\left( I\right) = \left( {{x}^{4},y}\right)$ ，与前一个例子中使用排序 $x > y$ 得到的理想 ${LT}\left( I\right)$ 不同，且再次严格大于 $\left( {{LT}\left( {f}_{1}\right) ,{LT}\left( {f}_{2}\right) }\right)$ 。

(3) Choose any ordering on $F\left\lbrack {x,y}\right\rbrack$ and let $f = f\left( {x,y}\right)$ be any nonzero polynomial. The leading term of every element of the principal ideal $I = \left( f\right)$ is then a multiple of the leading term of $f$ ,so in this case ${LT}\left( I\right) = \left( {{LT}\left( f\right) }\right)$ .

(3) 在 $F\left\lbrack {x,y}\right\rbrack$ 上选择任意排序，并让 $f = f\left( {x,y}\right)$ 是任意非零多项式。主理想 $I = \left( f\right)$ 的每个元素的首项都是 $f$ 的首项的倍数，所以在这种情况下 ${LT}\left( I\right) = \left( {{LT}\left( f\right) }\right)$ 。

In the case of one variable, leading terms are used in the Division Algorithm to reduce one polynomial $g$ modulo another polynomial $f$ to get a unique remainder $r$ ,and this remainder is 0 if and only if $g$ is contained in the ideal(f). Since $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ is not a Euclidean Domain if $n \geq 2$ (since it is not a P.I.D.),the situation is more complicated for polynomials in more than one variable. In the first example above, neither ${f}_{1}$ nor ${f}_{2}$ divides $g$ in $F\left\lbrack {x,y}\right\rbrack$ (by degree considerations,for example),so attempting to first divide $g$ by one of ${f}_{1}$ or ${f}_{2}$ and then by the other to try to reduce $g$ modulo the ideal $I$ would produce a (nonzero) “remainder” of $g$ itself. In particular, this would suggest that $g = y{f}_{1} - x{f}_{2}$ is not an element of the ideal $I$ even though it is. The reason the polynomial $g$ of degree 1 can be a linear combination of the two polynomials ${f}_{1}$ and ${f}_{2}$ of degree 4 is that the leading terms in $y{f}_{1}$ and $x{f}_{2}$ cancel in the difference,and this is reflected in the fact that ${LT}\left( {f}_{1}\right)$ and ${LT}\left( {f}_{2}\right)$ are not sufficient to generate ${LT}\left( I\right)$ . A set of generators for an ideal $I$ in $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ whose leading terms generate the leading terms of all the elements in $I$ is given a special name.

在一个变量的情况下，首项在除法算法中被用来将一个多项式 $g$ 模另一个多项式 $f$ 化简得到一个唯一的余数 $r$ ，当且仅当 $g$ 包含在理想(f)中时，这个余数为0。由于 $F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack$ 不是一个欧几里得整环（因为它不是一个主理想整环），所以在多个变量的多项式中情况更为复杂。在上述第一个例子中，${f}_{1}$ 和 ${f}_{2}$ 都不能整除 $g$ 在 $F\left\lbrack {x,y}\right\rbrack$ 中（例如通过次数考虑），所以尝试先除以 ${f}_{1}$ 或 ${f}_{2}$ 中的一个，然后再除以另一个来尝试将 $g$ 化简模理想 $I$ 将会产生一个（非零的）“余数”，即 $g$ 本身。特别是，这会暗示 $g = y{f}_{1} - x{f}_{2}$ 不是理想 $I$ 的元素，即使它是。多项式 $g$ 的次数为1可以是两个次数为4的多项式 ${f}_{1}$ 和 ${f}_{2}$ 的线性组合，是因为 $y{f}_{1}$ 和 $x{f}_{2}$ 的首项在相减时抵消了，这反映在 ${LT}\left( {f}_{1}\right)$ 和 ${LT}\left( {f}_{2}\right)$ 不足以生成 ${LT}\left( I\right)$ 的事实中。对于理想 $I$ 在 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 中，其首项生成所有元素首项的生成集有一个特殊的名称。

Definition. A Gröbner basis for an ideal $I$ in the polynomial ring $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ is a finite set of generators $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ for $I$ whose leading terms generate the ideal of all leading terms in $I$ ,i.e.,

定义。多项式环 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 中理想 $I$ 的一个Gröbner基是 $I$ 的有限生成集 $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ ，其首项生成所有 $I$ 中首项的理想，即，

Remark: Note that a Gröbner “basis” is in fact a set of generators for $I$ (that depends on the choice of ordering),i.e.,every element in $I$ is a linear combination of the generators, and not a basis in the sense of vector spaces (where the linear combination would be unique, cf. Sections 10.3 and 11.1). Although potentially misleading, the terminology “Gröbner basis” has been so widely adopted that it would be hazardous to introduce a different nomenclature.

注意：Gröbner“基”实际上是一组生成元 $I$（这取决于排序的选择），即 $I$ 中的每个元素都是生成元的线性组合，而不是在向量空间的意义上（在那里线性组合将是唯一的，参见第10.3节和11.1节）的基。尽管这种术语可能具有误导性，但由于“Gröbner基”被广泛采用，引入不同的命名将是危险的。

One of the most important properties of a Gröbner basis (proved in Theorem 23 following) is that every polynomial $g$ can be written uniquely as the sum of an element in $I$ and a remainder $r$ obtained by a general polynomial division. In particular,we shall see that $g$ is an element of $I$ if and only if this remainder $r$ is 0 . While there is a similar decomposition in general, we shall see that if we do not use a Gröbner basis the uniqueness is lost (and we cannot detect membership in $I$ by checking whether the remainder is 0 ) because there are leading terms not accounted for by the leading terms of the generators.

Gröbner基最重要的性质之一（在接下来的定理23中证明）是每个多项式 $g$ 都可以唯一地写成一个在 $I$ 中的元素和一个由一般多项式除法得到的余数 $r$ 的和。特别地，我们将看到，当且仅当这个余数 $r$ 为0时，$g$ 是 $I$ 的一个元素。尽管在一般情况下也存在类似的分解，但我们将看到，如果我们不使用Gröbner基，这种唯一性就会丢失（我们无法通过检查余数是否为0来检测是否属于 $I$），因为存在一些首项不由生成元的首项解释。

We first use the leading terms of polynomials defined by a monomial ordering on $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ to extend the one variable Division Algorithm to a noncanonical polynomial division in several variables. Recall that for polynomials in one variable, the usual Division Algorithm determines the quotient $q\left( x\right)$ and remainder $r\left( x\right)$ in the equation $f\left( x\right) = q\left( x\right) g\left( x\right) + r\left( x\right)$ by successively testing whether the leading term of the dividend $f\left( x\right)$ is divisible by the leading term of $g\left( x\right)$ : if ${LT}\left( f\right) = a\left( x\right) {LT}\left( g\right)$ , the monomial term $a\left( x\right)$ is added to the quotient and the process is iterated with $f\left( x\right)$ replaced by the dividend $f\left( x\right) - a\left( x\right) g\left( x\right)$ ,which is of smaller degree since the leading terms cancel (by the choice of $a\left( x\right)$ ). The process terminates when the leading term of the divisor $g\left( x\right)$ no longer divides the leading term of the dividend,leaving the remainder $r\left( x\right)$ . We can extend this to division by a finite number of polynomials in several variables simply by allowing successive divisions, resulting in a remainder and several quotients, as follows.

我们首先使用由单项式排序定义的多项式的前导项 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 来将单变量除法算法扩展到多个变量的非典型多项式除法。回顾一下，对于单变量多项式，通常的除法算法通过逐次测试被除数的前导项 $f\left( x\right)$ 是否可以被除数的前导项 $g\left( x\right)$ 整除，来确定商 $q\left( x\right)$ 和余数 $r\left( x\right)$ 在方程 $f\left( x\right) = q\left( x\right) g\left( x\right) + r\left( x\right)$ 中：如果 ${LT}\left( f\right) = a\left( x\right) {LT}\left( g\right)$ ，则将单项式项 $a\left( x\right)$ 加到商上，并用被除数 $f\left( x\right) - a\left( x\right) g\left( x\right)$ （由于前导项相消）替换 $f\left( x\right)$ ，这个过程迭代进行，直到除数的前导项 $g\left( x\right)$ 不再整除被除数的前导项，留下余数 $r\left( x\right)$。我们可以通过允许连续除法，将此扩展到由几个变量组成的有限多项式的除法，从而得到一个余数和几个商，如下所示。

General Polynomial Division

通用多项式除法

Fix a monomial ordering on $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ ,and suppose ${g}_{1},\ldots ,{g}_{m}$ is a set of nonzero polynomials in $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ . If $f$ is any polynomial in $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ ,start with a set of quotients ${q}_{1},\ldots ,{q}_{m}$ and a remainder $r$ initially all equal to 0 and successively test whether the leading term of the dividend $f$ is divisible by the leading terms of the divisors ${g}_{1},\ldots ,{g}_{m}$ ,in that order. Then

在 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 上固定一个单项式排序，并假设 ${g}_{1},\ldots ,{g}_{m}$ 是 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 中的一组非零多项式。如果 $f$ 是 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 中的任何多项式，从一组初始值都等于0的商 ${q}_{1},\ldots ,{q}_{m}$ 和余数 $r$ 开始，逐次测试被除数 $f$ 的前导项是否可以依次被除数 ${g}_{1},\ldots ,{g}_{m}$ 的前导项整除。

i. If ${LT}\left( f\right)$ is divisible by ${LT}\left( {g}_{i}\right)$ ,say, ${LT}\left( f\right) = {a}_{i}{LT}\left( {g}_{i}\right)$ ,add ${a}_{i}$ to the quotient ${q}_{i}$ , replace $f$ by the dividend $f - {a}_{i}{g}_{i}$ (a polynomial with lower order leading term), and reiterate the entire process.

i. 如果 ${LT}\left( f\right)$ 可以被 ${LT}\left( {g}_{i}\right)$ 整除，比如说 ${LT}\left( f\right) = {a}_{i}{LT}\left( {g}_{i}\right)$，将 ${a}_{i}$ 加到商 ${q}_{i}$ 上，用被除数 $f - {a}_{i}{g}_{i}$ 替换 $f$（一个前导项阶数较低的多项式），然后重新进行整个过程。

ii. If the leading term of the dividend $f$ is not divisible by any of the leading terms ${LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right)$ ,add the leading term of $f$ to the remainder $r$ ,replace $f$ by the dividend $f - {LT}\left( f\right)$ (i.e.,remove the leading term of $f$ ),and reiterate the entire process.

如果被除数 $f$ 的首项不能被任何除数的首项 ${LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right)$ 整除，则将 $f$ 的首项加到余数 $r$ 上，用被除数 $f - {LT}\left( f\right)$ （即移除 $f$ 的首项）替换 $f$，并重复整个过程。

The process terminates (cf. Exercise 3) when the dividend is 0 and results in a set of quotients ${q}_{1},\ldots ,{q}_{m}$ and a remainder $r$ with

当被除数为 0 时，该过程终止（参见练习 3），并得到一组商 ${q}_{1},\ldots ,{q}_{m}$ 和余数 $r$ ，其满足

Each ${q}_{i}{g}_{i}$ has multidegree less than or equal to the multidegree of $f$ and the remainder $r$ has the property that no nonzero term in $r$ is divisible by any of the leading terms ${LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right)$ (since only terms with this property are added to $r$ in (ii)).

每个 ${q}_{i}{g}_{i}$ 的多重度小于或等于 $f$ 的多重度，且余数 $r$ 具有如下性质：$r$ 中没有任何非零项可以被任何首项 ${LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right)$ 整除（因为在 (ii) 中只有具有这种性质的项被加到 $r$ 中）。

Examples

示例

Fix the lexicographic ordering $x > y$ on $F\left\lbrack {x,y}\right\rbrack$ .

在 $F\left\lbrack {x,y}\right\rbrack$ 上固定字典序 $x > y$ 。

(1) Suppose $f = {x}^{3}{y}^{3} + 3{x}^{2}{y}^{4}$ and $g = x{y}^{4}$ . The leading term of $f$ is ${x}^{3}{y}^{3}$ ,which is not divisible by (the leading term of) $g$ ,so ${x}^{3}{y}^{3}$ is added to the remainder $r$ (so now $\left. {r = {x}^{3}{y}^{3}}\right)$ and $f$ is replaced by $f - {LT}\left( f\right) = 3{x}^{2}{y}^{4}$ and we start over. Since $3{x}^{2}{y}^{4}$ is divisible by ${LT}\left( g\right) = x{y}^{4}$ ,with quotient $a = {3x}$ ,we add ${3x}$ to the quotient $q$ (so $q = {3x}$ ),and replace $3{x}^{2}{y}^{4}$ by $3{x}^{2}{y}^{4} - {aLT}\left( g\right) = 0$ ,at which point the process terminates. The result is the quotient $q = {3x}$ and remainder $r = {x}^{3}{y}^{3}$ and

(1) 假设 $f = {x}^{3}{y}^{3} + 3{x}^{2}{y}^{4}$ 和 $g = x{y}^{4}$ 。$f$ 的首项是 ${x}^{3}{y}^{3}$，它不能被（$g$ 的首项）整除，因此 ${x}^{3}{y}^{3}$ 被加到余数 $r$ 上（现在 $\left. {r = {x}^{3}{y}^{3}}\right)$ ，$f$ 被替换为 $f - {LT}\left( f\right) = 3{x}^{2}{y}^{4}$ 并重新开始）。由于 $3{x}^{2}{y}^{4}$ 可以被 ${LT}\left( g\right) = x{y}^{4}$ 整除，商为 $a = {3x}$，我们将 ${3x}$ 加到商 $q$ 上（因此 $q = {3x}$），并将 $3{x}^{2}{y}^{4}$ 替换为 $3{x}^{2}{y}^{4} - {aLT}\left( g\right) = 0$，此时过程终止。结果是商 $q = {3x}$ 和余数 $r = {x}^{3}{y}^{3}$。

Note that if we had terminated at the first step because the leading term of $f$ is not divisible by the leading term of $g$ (which terminates the Division Algorithm for polynomials in one variable),then we would have been left with a ‘remainder’ of $f$ itself,even though ’more’ of $f$ is divisible by $g$ . This is the reason for step 2 in the division process (which is not necessary for polynomials in one variable).

请注意，如果我们因为在 $f$ 的首项不能被 $g$ 的首项整除（这会终止单变量多项式的除法算法）而在第一步就终止了，那么我们将得到的‘余数’是 $f$ 本身，尽管 $f$ 的‘更多部分’是可以被 $g$ 整除的。这就是除法过程中第二步的原因（这对于单变量多项式来说并不是必要的）。

(2) Let $f = {x}^{2} + x - {y}^{2} + y$ ,and suppose ${g}_{1} = {xy} + 1$ and ${g}_{2} = x + y$ . In the first iteration, the leading term ${x}^{2}$ of $f$ is not divisible by the leading term of ${g}_{1}$ ,but is divisible by the leading term of ${g}_{2}$ ,so the quotient ${q}_{2}$ is $x$ and the dividend $f$ is replaced by the dividend $f - x{g}_{2} = - {xy} + x - {y}^{2} + y$ . In the second iteration,the leading term of $- {xy} + x - {y}^{2} + y$ is divisible by ${LT}\left( {g}_{1}\right)$ ,with quotient -1,so ${q}_{1} = - 1$ and the dividend is replaced by $\left( {-{xy} + x - {y}^{2} + y}\right) - \left( {-1}\right) {g}_{1} = x - {y}^{2} + y + 1$ . In the third iteration,the leading term of $x - {y}^{2} + y + 1$ is not divisible by the leading term of ${g}_{1}$ , but is divisible by the leading term of ${g}_{2}$ ,with quotient 1,so 1 is added to ${q}_{2}$ (which is now $\left. {{q}_{2} = x + 1}\right)$ and the dividend becomes $\left( {x - {y}^{2} + y + 1}\right) - \left( 1\right) \left( {g}_{2}\right) = - {y}^{2} + 1$ . The leading term is now $- {y}^{2}$ ,which is not divisible by either ${LT}\left( {g}_{1}\right) = {xy}$ or ${LT}\left( {g}_{2}\right) = x$ , so $- {y}^{2}$ is added to the remainder $r$ (which is now $- {y}^{2}$ ) and the dividend becomes simply 1. Finally,1 is not divisible by either ${LT}\left( {g}_{1}\right)$ or ${LT}\left( {g}_{2}\right)$ ,so is added to the remainder (so $r$ is now $- {y}^{2} + 1$ ),and the process terminates. The result is

(2) 设 $f = {x}^{2} + x - {y}^{2} + y$，并且假设 ${g}_{1} = {xy} + 1$ 和 ${g}_{2} = x + y$。在第一次迭代中，首项 ${x}^{2}$ 的 $f$ 不能被 ${g}_{1}$ 的首项整除，但可以被 ${g}_{2}$ 的首项整除，因此商 ${q}_{2}$ 是 $x$，而被除数 $f$ 被替换为被除数 $f - x{g}_{2} = - {xy} + x - {y}^{2} + y$。在第二次迭代中，$- {xy} + x - {y}^{2} + y$ 的首项可以被 ${LT}\left( {g}_{1}\right)$ 整除，商为 -1，所以 ${q}_{1} = - 1$，被除数被替换为 $\left( {-{xy} + x - {y}^{2} + y}\right) - \left( {-1}\right) {g}_{1} = x - {y}^{2} + y + 1$。在第三次迭代中，$x - {y}^{2} + y + 1$ 的首项不能被 ${g}_{1}$ 的首项整除，但可以被 ${g}_{2}$ 的首项整除，商为 1，因此 1 被加到 ${q}_{2}$ 上（现在为 $\left. {{q}_{2} = x + 1}\right)$），被除数变为 $\left( {x - {y}^{2} + y + 1}\right) - \left( 1\right) \left( {g}_{2}\right) = - {y}^{2} + 1$。此时的首项是 $- {y}^{2}$，它既不能被 ${LT}\left( {g}_{1}\right) = {xy}$ 整除，也不能被 ${LT}\left( {g}_{2}\right) = x$ 整除，所以 $- {y}^{2}$ 被加到余数 $r$ 上（现在为 $- {y}^{2}$），被除数变为简单的 1。最后，1 既不能被 ${LT}\left( {g}_{1}\right)$ 整除，也不能被 ${LT}\left( {g}_{2}\right)$ 整除，所以被加到余数上（因此 $r$ 现在是 $- {y}^{2} + 1$），过程终止。结果是

(3) Let $f = {x}^{2} + x - {y}^{2} + y$ as in the previous example and interchange the divisors ${g}_{1}$ and ${g}_{2} : {g}_{1} = x + y$ and ${g}_{2} = {xy} + 1$ . In this case an easy computation gives

(3) 设 $f = {x}^{2} + x - {y}^{2} + y$ 如前例所示，并且交换除数 ${g}_{1}$ 和 ${g}_{2} : {g}_{1} = x + y$ 以及 ${g}_{2} = {xy} + 1$。在这种情况下，一个简单的计算给出

showing that the quotients ${q}_{i}$ and the remainder $r$ are in general not unique and depend on the order of the divisors ${g}_{1},\ldots ,{g}_{m}$ .

这表明商 ${q}_{i}$ 和余数 $r$ 通常不是唯一的，并且依赖于除数 ${g}_{1},\ldots ,{g}_{m}$ 的顺序。

The computation in Example 3 shows that the polynomial $f = {x}^{2} + x - {y}^{2} + y$ is an element of the ideal $I = \left( {x + y,{xy} + 1}\right)$ since the remainder obtained in this case was 0 (in fact $f$ is just a multiple of the first generator). In Example 2,however,the same polynomial resulted in a nonzero remainder $- {y}^{2} + 1$ when divided by ${xy} + 1$ and $x + y$ ,and it was not at all clear from that computation that $f$ was an element of $I$ .

示例3中的计算表明，多项式 $f = {x}^{2} + x - {y}^{2} + y$ 是理想 $I = \left( {x + y,{xy} + 1}\right)$ 的一个元素，因为在此情况下得到的余数为0（实际上 $f$ 只不过是第一个生成子的倍数）。然而，在示例2中，同一个多项式在被 ${xy} + 1$ 和 $x + y$ 除时得到了非零余数 $- {y}^{2} + 1$，并且从那次计算中根本无法清楚地看出 $f$ 是 $I$ 的一个元素。

The next theorem shows that if we use a Gröbner basis for the ideal $I$ then these difficulties do not arise: we obtain a unique remainder, which in turn can be used to determine whether a polynomial $f$ is an element of the ideal $I$ .

下一个定理表明，如果我们使用理想 $I$ 的Gröbner基，那么这些困难就不会出现：我们得到一个唯一的余数，这个余数又可以用来确定多项式 $f$ 是否是理想 $I$ 的一个元素。

Theorem 23. Fix a monomial ordering on $R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ and suppose $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ is a Gröbner basis for the nonzero ideal $I$ in $R$ . Then

定理23。在 $R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 上固定一个单项式排序，并假设 $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ 是非零理想 $I$ 在 $R$ 上的一个Gröbner基。那么

(1) Every polynomial $f \in R$ can be written uniquely in the form

（1）每个多项式 $f \in R$ 都可以唯一地写成如下形式

where ${f}_{I} \in I$ and no nonzero monomial term of the ’remainder’ $r$ is divisible by any of the leading terms ${LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right)$ .

其中 ${f}_{I} \in I$ 并且 '余数' $r$ 的任何非零单项式项都不能被 ${LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right)$ 的任何首项整除。

(2) Both ${f}_{I}$ and $r$ can be computed by general polynomial division by ${g}_{1},\ldots ,{g}_{m}$ and are independent of the order in which these polynomials are used in the division.

（2）${f}_{I}$ 和 $r$ 都可以通过对 ${g}_{1},\ldots ,{g}_{m}$ 进行一般多项式除法来计算，并且与在除法中使用这些多项式的顺序无关。

(3) The remainder $r$ provides a unique representative for the coset of $f$ in the quotient ring $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /I$ . In particular, $f \in I$ if and only if $r = 0$ .

（3）余数 $r$ 为 $f$ 在商环 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /I$ 中的陪集提供了一个唯一的代表。特别是，当且仅当 $f \in I$ 时，$r = 0$。

Proof: Letting ${f}_{I} = \mathop{\sum }\limits_{{i = 1}}^{m}{q}_{i}{g}_{i} \in I$ in the general polynomial division of $f$ by ${g}_{1},\ldots ,{g}_{m}$ immediately gives a decomposition $f = {f}_{l} + r$ for any generators ${g}_{1},\ldots ,{g}_{m}$ . Suppose now that $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ is a Gröbner basis,and $f = {f}_{I} + r = {f}_{I}^{\prime } + {r}^{\prime }$ . Then $r - {r}^{\prime } = {f}_{I}^{\prime } - {f}_{I} \in I$ ,so its leading term ${LT}\left( {r - {r}^{\prime }}\right)$ is an element of ${LT}\left( I\right)$ ,which is the ideal $\left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right)$ since $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ is a Gröbner basis for $I$ . Every element in this ideal is a sum of multiples of the monomial terms ${LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right)$ , so is a sum of terms each of which is divisible by one of the ${LT}\left( {g}_{i}\right)$ . But both $r$ and ${r}^{\prime }$ ,hence also $r - {r}^{\prime }$ ,are sums of monomial terms none of which is divisible by ${LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right)$ ,which is a contradiction unless $r - {r}^{\prime } = 0$ . It follows that $r = {r}^{\prime }$ is unique,hence so is ${f}_{I} = f - r$ ,which proves (1).

证明：在 ${f}_{I} = \mathop{\sum }\limits_{{i = 1}}^{m}{q}_{i}{g}_{i} \in I$ 对 $f$ 除以 ${g}_{1},\ldots ,{g}_{m}$ 的一般多项式除法中，立即得到一个分解 $f = {f}_{l} + r$ 对于任何生成元 ${g}_{1},\ldots ,{g}_{m}$ 。现在假设 $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ 是一个Gröbner基，且 $f = {f}_{I} + r = {f}_{I}^{\prime } + {r}^{\prime }$ 。那么 $r - {r}^{\prime } = {f}_{I}^{\prime } - {f}_{I} \in I$ ，因此它的首项 ${LT}\left( {r - {r}^{\prime }}\right)$ 是 ${LT}\left( I\right)$ 的一个元素，这是理想 $\left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right)$ ，因为 $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ 是 $I$ 的Gröbner基。这个理想中的每个元素都是多项式项 ${LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right)$ 的倍数的和，所以也是项的和，每个项都可以被 ${LT}\left( {g}_{i}\right)$ 之一整除。但是 $r$ 和 ${r}^{\prime }$ ，因此还有 $r - {r}^{\prime }$ ，都是多项式项的和，没有任何项可以被 ${LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right)$ 整除，除非 $r - {r}^{\prime } = 0$ 。因此，$r = {r}^{\prime }$ 是唯一的，因此 ${f}_{I} = f - r$ 也是唯一的，这证明了（1）。

We have already seen that ${f}_{I}$ and $r$ can be computed algorithmically by polynomial division,and the uniqueness in (1) implies that $r$ is independent of the order in which the polynomials ${g}_{1},\ldots ,{g}_{m}$ are used in the division. Similarly ${f}_{I} = \mathop{\sum }\limits_{{i = 1}}^{m}{q}_{i}{g}_{i}$ is uniquely determined (even though the individual quotients ${q}_{i}$ are not in general unique),which gives (2).

我们已经看到 ${f}_{I}$ 和 $r$ 可以通过多项式除法算法计算，并且（1）中的唯一性意味着 $r$ 与多项式 ${g}_{1},\ldots ,{g}_{m}$ 在除法中使用的顺序无关。同样，${f}_{I} = \mathop{\sum }\limits_{{i = 1}}^{m}{q}_{i}{g}_{i}$ 也是唯一确定的（尽管单个商 ${q}_{i}$ 通常不是唯一的），这给出了（2）。

The first statement in (3) is immediate from the uniqueness in (1). If $r = 0$ ,then $f = {f}_{I} \in I$ . Conversely,if $f \in I$ ,then $f = f + 0$ together with the uniqueness of $r$ implies that $r = 0$ ,and the final statement of the theorem follows.

（3）中的第一个陈述直接来自（1）中的唯一性。如果 $r = 0$ ，那么 $f = {f}_{I} \in I$ 。反过来说，如果 $f \in I$ ，那么 $f = f + 0$ 以及 $r$ 的唯一性意味着 $r = 0$ ，定理的最后一个陈述随之得出。

As previously mentioned, the importance of Theorem 23, and one of the principal uses of Gröbner bases,is the uniqueness of the representative $r$ ,which allows effective computation in the quotient ring $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /I$ .

如前所述，定理23的重要性以及Groebner基的主要用途之一是代表元素的唯一性 $r$ ，这使得在商环 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /I$ 中进行有效计算成为可能。

We next prove that a set of polynomials in an ideal whose leading terms generate all the leading terms of an ideal is in fact a set of generators for the ideal itself (and so is a Gröbner basis-in some works this is tal-:n as the definition of a Gröbner basis), and this shows in particular that a Gröbner basis always exists.

我们接下来证明，一个多项式集合在某个理想中，如果其首项生成该理想的所有首项，那么这个多项式集合实际上是该理想的生成集（因此它是一个Groebner基——在某些作品中有时将此作为Groebner基的定义），特别是这表明Groebner基总是存在的。

Proposition 24. Fix a monomial ordering on $R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ and let $I$ be a nonzero ideal in $R$ .

命题24. 在 $R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 上固定一个单项式排序，并令 $I$ 为 $R$ 中的非零理想。

(1) If ${g}_{1},\ldots ,{g}_{m}$ are any elements of $I$ such that ${LT}\left( I\right) = \left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right)$ , then $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ is a Gröbner basis for $I$ .

(1) 如果 ${g}_{1},\ldots ,{g}_{m}$ 是 $I$ 中的任意元素，且满足 ${LT}\left( I\right) = \left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right)$ ，那么 $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ 是 $I$ 的Groebner基。

(2) The ideal $I$ has a Gröbner basis.

(2) 理想 $I$ 有一个Groebner基。

Proof: Suppose ${g}_{1},\ldots ,{g}_{m} \in I$ with ${LT}\left( I\right) = \left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right)$ . We need to see that ${g}_{1},\ldots ,{g}_{m}$ generate the ideal $I$ . If $f \in I$ ,use general polynomial division to write $f = \mathop{\sum }\limits_{{i = 1}}^{m}{q}_{i}{g}_{i} + r$ where no nonzero term in the remainder $r$ is divisible by any ${LT}\left( {g}_{i}\right)$ . Since $f \in I$ ,also $r \in I$ ,which means ${LT}\left( r\right)$ is in ${LT}\left( I\right)$ . But then ${LT}\left( r\right)$ would be divisible by one of ${LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right)$ ,which is a contradiction unless $r = 0$ . Hence $f = \mathop{\sum }\limits_{{i = 1}}^{m}{q}_{i}{g}_{i}$ and ${g}_{1},\ldots ,{g}_{m}$ generate $I$ ,so are a Gröbner basis for $I$ ,which proves (1).

证明：假设 ${g}_{1},\ldots ,{g}_{m} \in I$ ，其中 ${LT}\left( I\right) = \left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right)$ 。我们需要看到 ${g}_{1},\ldots ,{g}_{m}$ 生成理想 $I$ 。如果 $f \in I$ ，使用多项式除法的一般方法来写 $f = \mathop{\sum }\limits_{{i = 1}}^{m}{q}_{i}{g}_{i} + r$ ，其中余数 $r$ 中的非零项不能被任何 ${LT}\left( {g}_{i}\right)$ 整除。由于 $f \in I$ ，同样 $r \in I$ ，这意味着 ${LT}\left( r\right)$ 在 ${LT}\left( I\right)$ 中。但是，如果 ${LT}\left( r\right)$ 能被 ${LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right)$ 之一整除，就会产生矛盾，除非 $r = 0$ 。因此 $f = \mathop{\sum }\limits_{{i = 1}}^{m}{q}_{i}{g}_{i}$ ，${g}_{1},\ldots ,{g}_{m}$ 生成 $I$ ，所以它们是 $I$ 的Groebner基，这证明了(1)。

For (2),note that the ideal ${LT}\left( I\right)$ of leading terms of any ideal $I$ is a monomial ideal generated by all the leading terms of the polynomials in $I$ . By Exercise 1 a finite number of those leading terms suffice to generate ${LT}\left( I\right)$ ,say ${LT}\left( I\right) = \left( {{LT}\left( {h}_{1}\right) ,\ldots ,{LT}\left( {h}_{k}\right) }\right)$ for some ${h}_{1},\ldots ,{h}_{k} \in I$ . By (1),the polynomials ${h}_{1},\ldots ,{h}_{k}$ are a Gröbner basis of $I$ , completing the proof.

对于（2），请注意任何理想 ${LT}\left( I\right)$ 的首项理想是一个由 $I$ 中多项式的所有首项生成的单项式理想。由练习1可知，这些首项中的有限个数足以生成 ${LT}\left( I\right)$，比如说 ${LT}\left( I\right) = \left( {{LT}\left( {h}_{1}\right) ,\ldots ,{LT}\left( {h}_{k}\right) }\right)$ 对于某些 ${h}_{1},\ldots ,{h}_{k} \in I$。由（1）可知，多项式 ${h}_{1},\ldots ,{h}_{k}$ 是 $I$ 的一个Gröbner基，从而完成了证明。

Proposition 24 proves that Gröbner bases always exist. We next prove a criterion that determines whether a given set of generators of an ideal $I$ is a Gröbner basis, which we then use to provide an algorithm to find a Gröbner basis. The basic idea is very simple: additional elements in ${LT}\left( I\right)$ can arise by taking linear combinations of generators that cancel leading terms,as we saw in taking $y{f}_{1} - x{f}_{2}$ in the first example in this section. We shall see that obtaining new leading terms from generators in this simple manner is the only obstruction to a set of generators being a Gröbner basis.

命题24证明了Gröbner基总是存在的。接下来，我们证明一个准则，用于确定一个给定理想的生成集 $I$ 是否为Gröbner基，然后我们使用这个准则提供一个找到Gröbner基的算法。基本思想非常简单：${LT}\left( I\right)$ 中的额外元素可以通过取生成元的线性组合来消除首项，正如我们在本节第一个例子中看到取 $y{f}_{1} - x{f}_{2}$ 一样。我们将看到，以这种方式从生成元获得新的首项是生成元集合成为Gröbner基的唯一障碍。

In general,if ${f}_{1},{f}_{2}$ are two polynomials in $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ and $M$ is the monic least common multiple of the monomial terms ${LT}\left( {f}_{1}\right)$ and ${LT}\left( {f}_{2}\right)$ then we can cancel the leading terms by taking the difference

一般地，如果 ${f}_{1},{f}_{2}$ 是 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 中的两个多项式，$M$ 是单项式 ${LT}\left( {f}_{1}\right)$ 和 ${LT}\left( {f}_{2}\right)$ 的最小公倍数，那么我们可以通过取差来消去首项。

The next lemma shows that these elementary linear combinations account for all cancellation in leading terms of polynomials of the same multidegree.

下一个引理表明，这些基本的线性组合解释了相同多重度多项式首项的所有消去。

Lemma 25. Suppose ${f}_{1},\ldots ,{f}_{m} \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ are polynomials with the same multidegree $\alpha$ and that the linear combination $h = {a}_{1}{f}_{1} + \cdots + {a}_{m}{f}_{m}$ with constants ${a}_{i} \in F$ has strictly smaller multidegree. Then

引理25。假设 ${f}_{1},\ldots ,{f}_{m} \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 是具有相同多重度 $\alpha$ 的多项式，并且线性组合 $h = {a}_{1}{f}_{1} + \cdots + {a}_{m}{f}_{m}$ （系数为 ${a}_{i} \in F$）具有严格较小的多重度。那么

Proof: Write ${f}_{i} = {c}_{i}{f}_{i}^{\prime }$ where ${c}_{i} \in F$ and ${f}_{i}^{\prime }$ is a monic polynomial of multidegree $\alpha$ . We have

证明：写出 ${f}_{i} = {c}_{i}{f}_{i}^{\prime }$，其中 ${c}_{i} \in F$ 并且 ${f}_{i}^{\prime }$ 是一个具有多重度 $\alpha$ 的单项式多项式。我们有

Note that ${f}_{i - 1}^{\prime } - {f}_{i}^{\prime } = S\left( {{f}_{i - 1},{f}_{i}}\right)$ . Then since $h$ and each ${f}_{i - 1}^{\prime } - {f}_{i}^{\prime }$ has multidegree strictly smaller than $\alpha$ ,we have ${a}_{1}{c}_{1} + \cdots + {a}_{m}{c}_{m} = 0$ ,so the last term on the right hand side is 0 and the lemma follows.

注意 ${f}_{i - 1}^{\prime } - {f}_{i}^{\prime } = S\left( {{f}_{i - 1},{f}_{i}}\right)$ 。然后由于 $h$ 且每个 ${f}_{i - 1}^{\prime } - {f}_{i}^{\prime }$ 的重数都严格小于 $\alpha$ ，因此我们有 ${a}_{1}{c}_{1} + \cdots + {a}_{m}{c}_{m} = 0$ ，所以右手边的最后一项是 0 ，从而得出该引理。

The next proposition shows that a set of generators ${g}_{1},\ldots ,{g}_{m}$ is a Gröbner basis if there are no new leading terms among the differences $S\left( {{g}_{i},{g}_{j}}\right)$ not already accounted for by the ${g}_{i}$ . This result provides the principal ingredient in an algorithm to construct a Gröbner basis.

下一个命题表明，如果生成集合 ${g}_{1},\ldots ,{g}_{m}$ 在差值 $S\left( {{g}_{i},{g}_{j}}\right)$ 之间没有新的首项，而这些首项尚未由 ${g}_{i}$ 解释，那么这个生成集合是一个 Gröbner 基。这个结果为构造 Gröbner 基的算法提供了主要成分。

For a fixed monomial ordering on $R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ and ordered set of polynomials $G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ in $R$ ,write $f \equiv r{\;\operatorname{mod}\;G}$ if $r$ is the remainder obtained by general polynomial division of $f \in R$ by ${g}_{1},\ldots ,{g}_{m}$ (in that order).

对于在 $R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 上固定的单项式排序和 $R$ 中有序的多项式集合 $G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ ，如果 $r$ 是通过对 $f \in R$ 进行一般多项式除法得到的余数（按此顺序），则写作 $f \equiv r{\;\operatorname{mod}\;G}$ 。

Proposition 26. (Buchberger’s Criterion) Let $R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ and fix a monomial ordering on $R$ . If $I = \left( {{g}_{1},\ldots ,{g}_{m}}\right)$ is a nonzero ideal in $R$ ,then $G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ is a Gröbner basis for $I$ if and only if $S\left( {{g}_{i},{g}_{j}}\right) \equiv 0{\;\operatorname{mod}\;G}$ for $1 \leq i < j \leq m$ .

命题 26（Buchberger 准则）设 $R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 并在 $R$ 上固定一个单项式排序。如果 $I = \left( {{g}_{1},\ldots ,{g}_{m}}\right)$ 是 $R$ 中的非零理想，那么 $G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ 是 $I$ 的 Gröbner 基当且仅当 $S\left( {{g}_{i},{g}_{j}}\right) \equiv 0{\;\operatorname{mod}\;G}$ 对于所有 $1 \leq i < j \leq m$ 。

Proof: If $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ is a Gröbner basis for $I$ ,then $S\left( {{g}_{i},{g}_{j}}\right) \equiv 0{\;\operatorname{mod}\;G}$ by Theorem 23 since each $S\left( {{g}_{i},{g}_{j}}\right)$ is an element of $I$ .

证明：如果 $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ 是 $I$ 的 Gröbner 基，那么根据定理 23，$S\left( {{g}_{i},{g}_{j}}\right) \equiv 0{\;\operatorname{mod}\;G}$ ，因为每个 $S\left( {{g}_{i},{g}_{j}}\right)$ 都是 $I$ 的元素。

Suppose now that $S\left( {{g}_{i},{g}_{j}}\right) \equiv 0{\;\operatorname{mod}\;G}$ for $1 \leq i < j \leq m$ and take any element $f \in I$ . To see that $G$ is a Gröbner basis we need to see that $\left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right)$ contains ${LT}\left( f\right)$ . Since $f \in I$ ,we can write $f = \mathop{\sum }\limits_{{i = 1}}^{m}{h}_{i}{g}_{i}$ for some polynomials ${h}_{1},\ldots ,{h}_{m}$ . Such a representation is not unique. Among all such representations choose one for which the largest multidegree of any summand (i.e., $\mathop{\max }\limits_{{i = 1,\ldots ,m}}\partial \left( {{h}_{i}{g}_{i}}\right)$ ) is minimal,say $\alpha$ . It is clear that the multidegree of $f$ is no worse than the largest multidegree of all the summands ${h}_{i}{g}_{i}$ ,so $\partial \left( f\right) \leq \alpha$ . Write

假设现在 $S\left( {{g}_{i},{g}_{j}}\right) \equiv 0{\;\operatorname{mod}\;G}$ 对于 $1 \leq i < j \leq m$ 并且取任意元素 $f \in I$ 。为了证明 $G$ 是一个Gröbner基，我们需要看到 $\left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right)$ 包含 ${LT}\left( f\right)$ 。由于 $f \in I$ ，我们可以写出 $f = \mathop{\sum }\limits_{{i = 1}}^{m}{h}_{i}{g}_{i}$ 对于某些多项式 ${h}_{1},\ldots ,{h}_{m}$ 。这样的表示不是唯一的。在所有这样的表示中选择一个使得任意项的最大多重度（即 $\mathop{\max }\limits_{{i = 1,\ldots ,m}}\partial \left( {{h}_{i}{g}_{i}}\right)$ ）最小，比如说 $\alpha$ 。显然 $f$ 的多重度不比所有项的最大多重度差 ${h}_{i}{g}_{i}$ ，因此 $\partial \left( f\right) \leq \alpha$ 。写下

Suppose that $\partial \left( f\right) < \alpha$ . Then since the multidegree of the second two sums is also strictly smaller than $\alpha$ it follows that the multidegree of the first sum is strictly smaller than $\alpha$ . If ${a}_{i} \in F$ denotes the constant coefficient of the monomial term ${LT}\left( {h}_{i}\right)$ then ${LT}\left( {h}_{i}\right) = {a}_{i}{h}_{i}^{\prime }$ where ${h}_{i}^{\prime }$ is a monomial. We can apply Lemma 25 to $\sum {a}_{i}\left( {{h}_{i}^{\prime }{g}_{i}}\right)$ to write the first sum above as $\sum {b}_{i}S\left( {{h}_{i - 1}^{\prime }{g}_{i - 1},{h}_{i}^{\prime }{g}_{i}}\right)$ with $\partial \left( {{h}_{i - 1}^{\prime }{g}_{i - 1}}\right) = \partial \left( {{h}_{i}^{\prime }{g}_{i}}\right) = \alpha .$ Let ${\beta }_{i - 1,i}$ be the multidegree of the monic least common multiple of ${LT}\left( {g}_{i - 1}\right)$ and ${LT}\left( {g}_{i}\right)$ . Then an easy computation shows that $S\left( {{h}_{i - 1}^{\prime }{g}_{i - 1},{h}_{i}^{\prime }{g}_{i}}\right)$ is just $S\left( {{g}_{i - 1},{g}_{i}}\right)$ multiplied by the monomial of multidegree $\alpha - {\beta }_{i - 1,i}$ . The polynomial $S\left( {{g}_{i - 1},{g}_{i}}\right)$ has multidegree less than ${\beta }_{i - 1,i}$ and,by assumption, $S\left( {{g}_{i - 1},{g}_{i}}\right) \equiv 0{\;\operatorname{mod}\;G}$ . This means that after general polynomial division of $S\left( {{g}_{i - 1},{g}_{i}}\right)$ by ${g}_{1},\ldots ,{g}_{m}$ ,each $S\left( {{g}_{i - 1},{g}_{i}}\right)$ can be written as a sum $\sum {q}_{j}{g}_{j}$ with $\partial \left( {{q}_{j}{g}_{j}}\right) < {\beta }_{i - 1,i}$ . It follows that each $S\left( {{h}_{i - 1}^{\prime }{g}_{i - 1},{h}_{i}^{\prime }{g}_{i}}\right)$ is a sum $\sum {q}_{i}^{\prime }{g}_{j}$ with $\partial \left( {{q}_{i}^{\prime }{g}_{j}}\right) < \alpha$ . But then all the sums on the right hand side of equation (2) can be written as a sum of terms of the form ${p}_{i}{g}_{i}$ with polynomials ${p}_{i}$ satisfying $\partial \left( {{p}_{i}{g}_{i}}\right) < \alpha$ . This contradicts the minimality of $\alpha$ and shows that in fact $\partial \left( f\right) = \alpha$ ,i.e.,the leading term of $f$ has multidegree $\alpha$ .

假设 $\partial \left( f\right) < \alpha$ 。那么由于后两个和的多重度也严格小于 $\alpha$ ，因此第一个和的多重度也严格小于 $\alpha$ 。如果 ${a}_{i} \in F$ 表示单项式项 ${LT}\left( {h}_{i}\right)$ 的常数系数，那么 ${LT}\left( {h}_{i}\right) = {a}_{i}{h}_{i}^{\prime }$ ，其中 ${h}_{i}^{\prime }$ 是一个单项式。我们可以将引理25应用于 $\sum {a}_{i}\left( {{h}_{i}^{\prime }{g}_{i}}\right)$ ，将上述的第一个和写成 $\sum {b}_{i}S\left( {{h}_{i - 1}^{\prime }{g}_{i - 1},{h}_{i}^{\prime }{g}_{i}}\right)$ ，其中 $\partial \left( {{h}_{i - 1}^{\prime }{g}_{i - 1}}\right) = \partial \left( {{h}_{i}^{\prime }{g}_{i}}\right) = \alpha .$ 。设 ${\beta }_{i - 1,i}$ 为 ${LT}\left( {g}_{i - 1}\right)$ 和 ${LT}\left( {g}_{i}\right)$ 的首一最小公倍数的多重度。那么一个简单的计算表明 $S\left( {{h}_{i - 1}^{\prime }{g}_{i - 1},{h}_{i}^{\prime }{g}_{i}}\right)$ 只不过是 $S\left( {{g}_{i - 1},{g}_{i}}\right)$ 乘以多重度为 $\alpha - {\beta }_{i - 1,i}$ 的单项式。多项式 $S\left( {{g}_{i - 1},{g}_{i}}\right)$ 的多重度小于 ${\beta }_{i - 1,i}$ ，根据假设，$S\left( {{g}_{i - 1},{g}_{i}}\right) \equiv 0{\;\operatorname{mod}\;G}$ 。这意味着在将 $S\left( {{g}_{i - 1},{g}_{i}}\right)$ 进行一般多项式除法除以 ${g}_{1},\ldots ,{g}_{m}$ 后，每个 $S\left( {{g}_{i - 1},{g}_{i}}\right)$ 都可以写成和 $\sum {q}_{j}{g}_{j}$ ，其中 $\partial \left( {{q}_{j}{g}_{j}}\right) < {\beta }_{i - 1,i}$ 。因此，每个 $S\left( {{h}_{i - 1}^{\prime }{g}_{i - 1},{h}_{i}^{\prime }{g}_{i}}\right)$ 都是和 $\sum {q}_{i}^{\prime }{g}_{j}$ ，其中 $\partial \left( {{q}_{i}^{\prime }{g}_{j}}\right) < \alpha$ 。但是，方程（2）右边的所有和都可以写成形式为 ${p}_{i}{g}_{i}$ 的项的和，其中多项式 ${p}_{i}$ 满足 $\partial \left( {{p}_{i}{g}_{i}}\right) < \alpha$ 。这与 $\alpha$ 的极小性相矛盾，并表明实际上 $\partial \left( f\right) = \alpha$ ，即 $f$ 的首项具有多重度 $\alpha$ 。

If we now take the terms in equation (2) of multidegree $\alpha$ we see that

如果我们现在取方程（2）中多重度为 $\alpha$ 的项，我们会看到

so indeed ${LT}\left( f\right) \in \left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right)$ . It follows that $G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ is a Gröbner basis.

因此确实 ${LT}\left( f\right) \in \left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right)$ 。这意味着 $G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ 是一个Gröbner基。

Buchberger’s Algorithm

Buchberger算法

Buchberger's Criterion can be used to provide an algorithm to find a Gröbner basis for an ideal $I$ ,as follows. If $I = \left( {{g}_{1},\ldots ,{g}_{m}}\right)$ and each $S\left( {{g}_{i},{g}_{j}}\right)$ leaves a remainder of 0 when divided by $G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ using general polynomial division then $G$ is a Gröbner basis. Otherwise $S\left( {{g}_{i},{g}_{j}}\right)$ has a nonzero remainder $r$ . Increase $G$ by appending the polynomial ${g}_{m + 1} = r : {G}^{\prime } = \left\{ {{g}_{1},\ldots ,{g}_{m},{g}_{m + 1}}\right\}$ and begin again (note that this is again a set of generators for $I$ since ${g}_{m + 1} \in I$ ). It is not hard to check that this procedure terminates after a finite number of steps in a generating set $G$ that satisfies Buchberger’s Criterion,hence is a Gröbner basis for $I$ (cf. Exercise 16). Note that once an $S\left( {{g}_{i},{g}_{j}}\right)$ yields a remainder of 0 after division by the polynomials in $G$ it also yields a remainder of 0 when additional polynomials are appended to $G$ .

Buchberger准则可以用来提供一个算法，用于找到一个理想 $I$ 的Gröbner基，如下所示。如果 $I = \left( {{g}_{1},\ldots ,{g}_{m}}\right)$ 和每个 $S\left( {{g}_{i},{g}_{j}}\right)$ 在使用一般多项式除法除以 $G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ 时余数为0，那么 $G$ 是一个Gröbner基。否则 $S\left( {{g}_{i},{g}_{j}}\right)$ 有一个非零余数 $r$ 。将多项式 ${g}_{m + 1} = r : {G}^{\prime } = \left\{ {{g}_{1},\ldots ,{g}_{m},{g}_{m + 1}}\right\}$ 添加到 $G$ 中并重新开始（注意，这仍然是 $I$ 的一组生成元，因为 ${g}_{m + 1} \in I$ ）。不难验证，在满足Buchberger准则的生成集 $G$ 中，这个过程在有限步骤后此过程终止，因此它是 $I$ 的一个Gröbner基（参见练习16）。注意，一旦某个 $S\left( {{g}_{i},{g}_{j}}\right)$ 在除以 $G$ 中的多项式后余数为0，那么在向 $G$ 添加额外的多项式后，它同样余数为0。

If $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ is a Gröbner basis for the ideal $I$ and ${LT}\left( {g}_{j}\right)$ is divisible by ${LT}\left( {g}_{i}\right)$ for some $j \neq i$ ,then ${LT}\left( {g}_{j}\right)$ is not needed as a generator for ${LT}\left( I\right)$ . By Proposition 24 we may therefore delete ${g}_{j}$ and still retain a Gröbner basis for $I$ . We may also assume without loss that the leading term of each ${g}_{i}$ is monic. A Gröbner basis $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ for $I$ where each ${LT}\left( {g}_{i}\right)$ is monic and where ${LT}\left( {g}_{j}\right)$ is not divisible by ${LT}\left( {g}_{i}\right)$ for $\begin{matrix} i \neq j\text{ is called a minimal }G \\ r\ddot{o}b\text{ ner basis. While a minimal }G \\ r\ddot{o}b\text{ ner basis is not unique,} \end{matrix}$ the number of elements and their leading terms are unique (cf. Exercise 15).

如果 $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ 是理想 $I$ 的Gröbner基，且 ${LT}\left( {g}_{j}\right)$ 可以被某个 $j \neq i$ 的 ${LT}\left( {g}_{i}\right)$ 整除，那么 ${LT}\left( {g}_{j}\right)$ 不需要作为 ${LT}\left( I\right)$ 的生成元。根据命题24，我们可以删除 ${g}_{j}$ ，仍然保留 $I$ 的Gröbner基。我们也可以不失一般性地假设每个 ${g}_{i}$ 的首项是单项式。对于每个 ${LT}\left( {g}_{i}\right)$ 是单项式且 ${LT}\left( {g}_{j}\right)$ 不能被 ${LT}\left( {g}_{i}\right)$ 整除的 $I$ 的Gröbner基 $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ ，元素的数量及其首项是唯一的（参见练习15）。

Examples

示例

(1) Choose the lexicographic ordering $x > y$ on $F\left\lbrack {x,y}\right\rbrack$ and consider the ideal $I$ generated by ${f}_{1} = {x}^{3}y - x{y}^{2} + 1$ and ${f}_{2} = {x}^{2}{y}^{2} - {y}^{3} - 1$ as in Example 1 at the beginning of this section. To test whether $G = \left\{ {{f}_{1},{f}_{2}}\right\}$ is a Gröbner basis we compute $S\left( {{f}_{1},{f}_{2}}\right) =$ $y{f}_{1} - x{f}_{2} = x + y$ ,which is its own remainder when divided by $\left\{ {{f}_{1},{f}_{2}}\right\}$ ,so $G$ is not a Gröbner basis for $I$ . Set ${f}_{3} = x + y$ ,and increase the generating set: ${G}^{\prime } = \left\{ {{f}_{1},{f}_{2},{f}_{3}}\right\}$ . Now $S\left( {{f}_{1},{f}_{2}}\right) \equiv 0{\;\operatorname{mod}\;{G}^{\prime }}$ ,and a brief computation yields

(1) Choose the lexicographic ordering $x > y$ on $F\left\lbrack {x,y}\right\rbrack$ and consider the ideal $I$ generated by ${f}_{1} = {x}^{3}y - x{y}^{2} + 1$ and ${f}_{2} = {x}^{2}{y}^{2} - {y}^{3} - 1$ as in Example 1 at the beginning of this section. To test whether $G = \left\{ {{f}_{1},{f}_{2}}\right\}$ is a Gröbner basis we compute $S\left( {{f}_{1},{f}_{2}}\right) =$ $y{f}_{1} - x{f}_{2} = x + y$ ,which is its own remainder when divided by $\left\{ {{f}_{1},{f}_{2}}\right\}$ ,so $G$ is not a Gröbner basis for $I$ . Set ${f}_{3} = x + y$ ,and increase the generating set: ${G}^{\prime } = \left\{ {{f}_{1},{f}_{2},{f}_{3}}\right\}$ . Now $S\left( {{f}_{1},{f}_{2}}\right) \equiv 0{\;\operatorname{mod}\;{G}^{\prime }}$ ,and a brief computation yields

Let ${f}_{4} = {y}^{4} - {y}^{3} - 1$ and increase the generating set to ${G}^{\prime \prime } = \left\{ {{f}_{1},{f}_{2},{f}_{3},{f}_{4}}\right\}$ . The previous 0 remainder is still 0,and now $S\left( {{f}_{2},{f}_{3}}\right) \equiv 0{\;\operatorname{mod}\;{G}^{\prime \prime }}$ by the choice of ${f}_{4}$ . Some additional computation yields

Let ${f}_{4} = {y}^{4} - {y}^{3} - 1$ and increase the generating set to ${G}^{\prime \prime } = \left\{ {{f}_{1},{f}_{2},{f}_{3},{f}_{4}}\right\}$ . The previous 0 remainder is still 0,and now $S\left( {{f}_{2},{f}_{3}}\right) \equiv 0{\;\operatorname{mod}\;{G}^{\prime \prime }}$ by the choice of ${f}_{4}$ . Some additional computation yields

and so $\left\{ {{x}^{3}y - x{y}^{2} + 1,{x}^{2}{y}^{2} - {y}^{3} - 1,x + y,{y}^{4} - {y}^{3} - 1}\right\}$ is a Gröbner basis for I. In particular, ${LT}\left( I\right)$ is generated by the leading terms of these four polynomials, so ${LT}\left( I\right) = \left( {{x}^{3}y,{x}^{2}{y}^{2},x,{y}^{4}}\right) = \left( {x,{y}^{4}}\right)$ ,as previously mentioned. Then $x + y$ and ${y}^{4} - {y}^{3} - 1$ in $I$ have leading terms generating ${LT}\left( I\right)$ ,so by Proposition 24, $\left\{ {x + y,{y}^{4} - {y}^{3} - 1}\right\}$ gives a minimal Gröbner basis for $I$ :

and so $\left\{ {{x}^{3}y - x{y}^{2} + 1,{x}^{2}{y}^{2} - {y}^{3} - 1,x + y,{y}^{4} - {y}^{3} - 1}\right\}$ is a Gröbner basis for I. In particular, ${LT}\left( I\right)$ is generated by the leading terms of these four polynomials, so ${LT}\left( I\right) = \left( {{x}^{3}y,{x}^{2}{y}^{2},x,{y}^{4}}\right) = \left( {x,{y}^{4}}\right)$ ,as previously mentioned. Then $x + y$ and ${y}^{4} - {y}^{3} - 1$ in $I$ have leading terms generating ${LT}\left( I\right)$ ,so by Proposition 24, $\left\{ {x + y,{y}^{4} - {y}^{3} - 1}\right\}$ gives a minimal Gröbner basis for $I$ :

This description of $I$ is much simpler than $I = \left( {{x}^{3}y - x{y}^{2} + 1,{x}^{2}{y}^{2} - {y}^{3} - 1}\right)$ .

This description of $I$ is much simpler than $I = \left( {{x}^{3}y - x{y}^{2} + 1,{x}^{2}{y}^{2} - {y}^{3} - 1}\right)$ .

(2) Choose the lexicographic ordering $y > x$ on $F\left\lbrack {x,y}\right\rbrack$ and consider the ideal $I$ in the previous example. In this case, $S\left( {{f}_{1},{f}_{2}}\right)$ produces a remainder of ${f}_{3} = - x - y$ ; then $S\left( {{f}_{1},{f}_{3}}\right)$ produces a remainder of ${f}_{4} = - {x}^{4} - {x}^{3} + 1$ ,and then all remainders are 0 with respect to the Gröbner basis $\left\{ {\frac{{x}^{3}y - x}{{y}^{2} + 1},\frac{{x}^{2}{y}^{2} - {y}^{3} - 1}{, - x - y, - {x}^{4} - {x}^{3} + {x}^{3}}}\right.$ 1). Here ${LT}\left( I\right) = \left( {-x{y}^{2}, - {y}^{3}, - y, - {x}^{4}}\right) = \left( {y,{x}^{4}}\right)$ ,as previously mentioned,and $\left\{ {x + y,{x}^{4} + {x}^{3} - 1}\right\}$ gives a minimal Gröbner basis for $I$ with respect to this ordering:

(2) 在 $y > x$ 上选择字典序 $F\left\lbrack {x,y}\right\rbrack$ 并考虑前一个示例中的理想 $I$ 。在这种情况下，$S\left( {{f}_{1},{f}_{2}}\right)$ 产生余数 ${f}_{3} = - x - y$ ；然后 $S\left( {{f}_{1},{f}_{3}}\right)$ 产生余数 ${f}_{4} = - {x}^{4} - {x}^{3} + 1$ ，之后所有余数相对于 Gröbner 基 $\left\{ {\frac{{x}^{3}y - x}{{y}^{2} + 1},\frac{{x}^{2}{y}^{2} - {y}^{3} - 1}{, - x - y, - {x}^{4} - {x}^{3} + {x}^{3}}}\right.$ 1) 都是 0 。在这里 ${LT}\left( I\right) = \left( {-x{y}^{2}, - {y}^{3}, - y, - {x}^{4}}\right) = \left( {y,{x}^{4}}\right)$ ，如前所述，并且 $\left\{ {x + y,{x}^{4} + {x}^{3} - 1}\right\}$ 对于此排序给出了 $I$ 的最小 Gröbner 基：

a different simpler description of $I$ .

对 $I$ 的另一种更简单的描述。

In Example 1 above it is easy to check that $\left\{ {x + {y}^{4} - {y}^{3} + y - 1,\;{y}^{4} - {y}^{3} - 1}\right\}$ is again a minimal Gröbner basis for $I$ (this is just $\left\{ {{f}_{3} + {f}_{4},{f}_{4}}\right\}$ ),so even with a fixed monomial ordering on $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ a minimal Gröbner basis for an ideal $I$ is not unique. We can obtain an important uniqueness property by strengthening the condition on divisibility by the leading terms of the basis.

在上面的示例 1 中，很容易验证 $\left\{ {x + {y}^{4} - {y}^{3} + y - 1,\;{y}^{4} - {y}^{3} - 1}\right\}$ 再次是 $I$ 的最小 Gröbner 基（这仅仅是 $\left\{ {{f}_{3} + {f}_{4},{f}_{4}}\right\}$ ），所以即使在 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 上有一个固定的单项式排序，一个理想 $I$ 的最小 Gröbner 基也不是唯一的。我们可以通过加强基的首项的可除性条件来获得一个重要的唯一性性质。

Definition. Fix a monomial ordering on $R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ . A Gröbner basis $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ for the nonzero ideal $I$ in $R$ is called a reduced Gröbner basis if

定义。在 $R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 上固定一个单项式排序。对于 $R$ 中的非零理想 $I$ 的 Gröbner 基 $\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ 被称为简化 Gröbner 基，如果

(a) each ${g}_{i}$ has monic leading term,i.e., ${LT}\left( {g}_{i}\right)$ is monic, $i = 1,\ldots ,m$ ,and

(a) 每个 ${g}_{i}$ 的首项是单项式，即 ${LT}\left( {g}_{i}\right)$ 是单项式，$i = 1,\ldots ,m$ ，并且

(b) no term in ${g}_{j}$ is divisible by ${LT}\left( {g}_{i}\right)$ for $j \neq i$ .

(b) 在 ${g}_{j}$ 中没有任何项可以被 ${LT}\left( {g}_{i}\right)$ 整除，对于 $j \neq i$ 。

Note that a reduced Gröbner basis is, in particular, a minimal Gröbner basis. If $G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ is a minimal Gröbner basis for $I$ ,then the leading term ${LT}\left( {g}_{j}\right)$ is not divisible by ${LT}\left( {g}_{i}\right)$ for any $i \neq j$ . As a result,if we use polynomial division to divide ${g}_{j}$ by the other polynomials in $G$ we obtain a remainder ${g}_{j}^{\prime }$ in the ideal $I$ with the same leading term as ${g}_{j}$ (the remainder ${g}_{j}^{\prime }$ does not depend on the order of the polynomials used in the division by (2) of Theorem 23). By Proposition 24,replacing ${g}_{j}$ by ${g}_{j}^{\prime }$ in $G$ again gives a minimal Gröbner basis for $I$ ,and in this basis no term of ${g}_{i}^{\prime }$ is divisible by ${LT}\left( {g}_{i}\right)$ for any $i \neq j$ . Replacing each element in $G$ by its remainder after division by the other elements in $G$ therefore results in a reduced Gröbner basis for $I$ . The importance of reduced Gröbner bases is that they are unique (for a given monomial ordering), as the next result shows.

注意，一个简化的Groebner基尤其是一个最小Groebner基。如果 $G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ 是 $I$ 的最小Groebner基，那么首项 ${LT}\left( {g}_{j}\right)$ 不能被 ${LT}\left( {g}_{i}\right)$ 整除，对于任何 $i \neq j$ 都成立。因此，如果我们使用多项式除法将 ${g}_{j}$ 除以 $G$ 中的其他多项式，我们将在理想 $I$ 中得到一个余数 ${g}_{j}^{\prime }$ ，其首项与 ${g}_{j}$ 相同（余数 ${g}_{j}^{\prime }$ 不依赖于定理23中除法（2）中所用多项式的顺序）。根据命题24，在 $G$ 中将 ${g}_{j}$ 替换为 ${g}_{j}^{\prime }$ 再次给出 $I$ 的最小Groebner基，在这个基中，${g}_{i}^{\prime }$ 的任何项都不被 ${LT}\left( {g}_{i}\right)$ 整除，对于任何 $i \neq j$ 都成立。因此，将 $G$ 中的每个元素替换为它除以 $G$ 中其他元素后的余数，将得到 $I$ 的简化Groebner基。简化Groebner基的重要性在于它们是唯一的（对于给定的单项式排序），如下一个结果所示。

Theorem 27. Fix a monomial ordering on $R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ . Then there is a unique reduced Gröbner basis for every nonzero ideal $I$ in $R$ .

定理27。在 $R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 上固定一个单项式排序。那么对于 $R$ 中的每个非零理想 $I$ ，都存在唯一的简化Groebner基。

Proof: By Exercise 15, two reduced bases have the same number of elements and the same leading terms since reduced bases are also minimal bases. If $G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ and ${G}^{\prime } = \{ {g}_{1}^{\prime },\ldots ,{g}_{m}^{\prime }\}$ are two reduced bases for the same nonzero ideal $I$ ,then after a possible rearrangement we may assume ${LT}\left( {g}_{i}\right) = {LT}\left( {g}_{i}^{\prime }\right) = {h}_{i}$ for $i = 1,\ldots ,m$ . For any fixed $i$ ,consider the polynomial ${f}_{i} = {g}_{i} - {g}_{i}^{\prime }$ . If ${f}_{i}$ is nonzero,then since ${f}_{i} \in I$ ,its leading term must be divisible by some ${h}_{j}$ . By definition of a reduced basis, ${h}_{j}$ for $j \neq i$ does not divide any of the terms in either ${g}_{i}$ or ${g}_{i}^{\prime }$ ,hence does not divide ${LT}\left( {f}_{i}\right)$ . But ${h}_{i}$ also does not divide ${LT}\left( {f}_{i}\right)$ since all the terms in ${f}_{i}$ have strictly smaller multidegree. This forces ${f}_{i} = 0$ ,i.e., ${g}_{i} = {g}_{i}^{\prime }$ for every $i$ ,so $G = {G}^{\prime }$ .

证明：由练习15知，两个化简基有相同数量的元素和相同的首项，因为化简基也是最小基。如果 $G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\}$ 和 ${G}^{\prime } = \{ {g}_{1}^{\prime },\ldots ,{g}_{m}^{\prime }\}$ 是同一个非零理想 $I$ 的两个化简基，那么在可能的重新排列后，我们可以假设 ${LT}\left( {g}_{i}\right) = {LT}\left( {g}_{i}^{\prime }\right) = {h}_{i}$ 对于 $i = 1,\ldots ,m$ 。对于任意固定的 $i$ ，考虑多项式 ${f}_{i} = {g}_{i} - {g}_{i}^{\prime }$ 。如果 ${f}_{i}$ 非零，那么由于 ${f}_{i} \in I$ ，其首项必定能被某个 ${h}_{j}$ 整除。根据化简基的定义， ${h}_{j}$ 对于 $j \neq i$ 不整除 ${g}_{i}$ 或 ${g}_{i}^{\prime }$ 中的任何项，因此也不整除 ${LT}\left( {f}_{i}\right)$ 。但是 ${h}_{i}$ 也不整除 ${LT}\left( {f}_{i}\right)$ ，因为 ${f}_{i}$ 中的所有项的多元次数都严格较小。这迫使 ${f}_{i} = 0$ ，即 ${g}_{i} = {g}_{i}^{\prime }$ 对于每个 $i$ ，因此 $G = {G}^{\prime }$ 。

One application of the uniqueness of the reduced Gröbner basis is a computational method to determine when two ideals in a polynomial ring are equal.

化简Gröbner基的唯一性一个应用是计算确定多项环中的两个理想是否相等的方法。

Corollary 28. Let $I$ and $J$ be two ideals in $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ . Then $I = J$ if and only if $I$ and $J$ have the same reduced Gröbner basis with respect to any fixed monomial ordering on $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ .

推论28。设 $I$ 和 $J$ 是 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 中的两个理想。那么 $I = J$ 当且仅当 $I$ 和 $J$ 对于 $F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack$ 上任何固定的单项式排序都有相同的化简Gröbner基。

Examples

示例

(1) Consider the ideal $I = \left( {{h}_{1},{h}_{2},{h}_{3}}\right)$ with ${h}_{1} = {x}^{2} + x{y}^{5} + {y}^{4},{h}_{2} = x{y}^{6} - x{y}^{3} + {y}^{5} - {y}^{2}$ , and ${h}_{3} = x{y}^{5} - x{y}^{2}$ in $F\left\lbrack {x,y}\right\rbrack$ . Using the lexicographic ordering $x > y$ we find $S\left( {{h}_{1},{h}_{2}}\right) \equiv S\left( {{h}_{1},{h}_{3}}\right) \equiv 0{\;\operatorname{mod}\;\{ }{h}_{1},{h}_{2},{h}_{3}\}$ and $S\left( {{h}_{2},{h}_{3}}\right) \equiv {y}^{5}$ - ${y}^{2}{\;\operatorname{mod}\;\{ }{h}_{1},{h}_{2},{h}_{3}\} .$ Setting ${h}_{4} = {y}^{5} - {y}^{2}$ we find $S\left( {{h}_{i},{h}_{j}}\right) \equiv 0$ mod $\left\{ {{h}_{1},{h}_{2},{h}_{3},{h}_{4}}\right\}$ for $1 \leq i < j \leq 4$ ,

（1）考虑理想 $I = \left( {{h}_{1},{h}_{2},{h}_{3}}\right)$ ，其中 ${h}_{1} = {x}^{2} + x{y}^{5} + {y}^{4},{h}_{2} = x{y}^{6} - x{y}^{3} + {y}^{5} - {y}^{2}$ ，和 ${h}_{3} = x{y}^{5} - x{y}^{2}$ 在 $F\left\lbrack {x,y}\right\rbrack$ 中。使用字典序 $x > y$ 我们发现 $S\left( {{h}_{1},{h}_{2}}\right) \equiv S\left( {{h}_{1},{h}_{3}}\right) \equiv 0{\;\operatorname{mod}\;\{ }{h}_{1},{h}_{2},{h}_{3}\}$ 和 $S\left( {{h}_{2},{h}_{3}}\right) \equiv {y}^{5}$ - ${y}^{2}{\;\operatorname{mod}\;\{ }{h}_{1},{h}_{2},{h}_{3}\} .$ 。设置 ${h}_{4} = {y}^{5} - {y}^{2}$ 我们发现 $S\left( {{h}_{i},{h}_{j}}\right) \equiv 0$ 模 $\left\{ {{h}_{1},{h}_{2},{h}_{3},{h}_{4}}\right\}$ 对于 $1 \leq i < j \leq 4$ ，

is a Gröbner basis for $I$ . The leading terms of this basis are ${x}^{2},x{y}^{6},x{y}^{5},{y}^{5}$ . Since ${y}^{5}$ divides both $x{y}^{6}$ and $x{y}^{5}$ ,we may remove the second and third generators to obtain a minimal Gröbner basis $\left\{ {{x}^{2} + x{y}^{5} + {y}^{4},{y}^{5} - {y}^{2}}\right\}$ for $I$ . The second term in the first generator is divisible by the leading term ${y}^{5}$ of the second generator,so this is not a reduced Gröbner basis. Replacing ${x}^{2} + x{y}^{5} + {y}^{4}$ by its remainder ${x}^{2} + x{y}^{2} + {y}^{4}$ after division by the other polynomials in the basis (which in this case is only the polynomial $\left. {{y}^{5} - {y}^{2}}\right)$ ,we are left with the reduced Gröbner basis $\left\{ {{x}^{2} + x{y}^{2} + {y}^{4},{y}^{5} - {y}^{2}}\right\}$ for $I$ .

是 $I$ 的一个Gröbner基。该基的首项是 ${x}^{2},x{y}^{6},x{y}^{5},{y}^{5}$ 。由于 ${y}^{5}$ 同时整除 $x{y}^{6}$ 和 $x{y}^{5}$ ，我们可以移除第二个和第三个生成元，以得到一个 $I$ 的最小Gröbner基 $\left\{ {{x}^{2} + x{y}^{5} + {y}^{4},{y}^{5} - {y}^{2}}\right\}$ 。第一个生成元的第二项可以被第二个生成元的首项 ${y}^{5}$ 整除，因此这不是一个简化的Gröbner基。将 ${x}^{2} + x{y}^{5} + {y}^{4}$ 替换为它除以基中其他多项式（在这种情况下只有一个多项式 $\left. {{y}^{5} - {y}^{2}}\right)$ ）后的余数 ${x}^{2} + x{y}^{2} + {y}^{4}$ ，我们得到 $I$ 的简化Gröbner基 $\left\{ {{x}^{2} + x{y}^{2} + {y}^{4},{y}^{5} - {y}^{2}}\right\}$ 。

2. Consider the ideal $J = \left( {{h}_{1},{h}_{2},{h}_{3}}\right)$ with ${h}_{1} = x{y}^{3} + {y}^{3} + 1,{h}_{2} = {x}^{3}y - {x}^{3} + 1$ ,and ${h}_{3} = x + y$ in $F\left\lbrack {x,y}\right\rbrack$ . Using the lexicographic monomial ordering $x > y$ we find $S\left( {{h}_{1},{h}_{2}}\right) \equiv 0{\;\operatorname{mod}\;\left\{ {{h}_{1},{h}_{2},{h}_{3}}\right\} }$ and $S\left( {{h}_{1},{h}_{3}}\right) \equiv {y}^{4} - {y}^{3} - 1{\;\operatorname{mod}\;\left\{ {{h}_{1},{h}_{2},{h}_{3}}\right\} }$ . Setting ${h}_{4} = {y}^{4} - {y}^{3} - 1$ we find $S\left( {{h}_{i},{h}_{j}}\right) \equiv 0{\;\operatorname{mod}\;\left\{ {{h}_{1},{h}_{2},{h}_{3},{h}_{4}}\right\} }$ for $1 \leq i < j \leq 4$ ,so

3. 考虑理想 $J = \left( {{h}_{1},{h}_{2},{h}_{3}}\right)$ ，其中 ${h}_{1} = x{y}^{3} + {y}^{3} + 1,{h}_{2} = {x}^{3}y - {x}^{3} + 1$ 和 ${h}_{3} = x + y$ 在 $F\left\lbrack {x,y}\right\rbrack$ 中。使用字典序单项式排序 $x > y$ ，我们找到 $S\left( {{h}_{1},{h}_{2}}\right) \equiv 0{\;\operatorname{mod}\;\left\{ {{h}_{1},{h}_{2},{h}_{3}}\right\} }$ 和 $S\left( {{h}_{1},{h}_{3}}\right) \equiv {y}^{4} - {y}^{3} - 1{\;\operatorname{mod}\;\left\{ {{h}_{1},{h}_{2},{h}_{3}}\right\} }$ 。设 ${h}_{4} = {y}^{4} - {y}^{3} - 1$ ，我们发现 $S\left( {{h}_{i},{h}_{j}}\right) \equiv 0{\;\operatorname{mod}\;\left\{ {{h}_{1},{h}_{2},{h}_{3},{h}_{4}}\right\} }$ 对于 $1 \leq i < j \leq 4$ ，因此

is a Gröbner basis for $J$ . The leading terms of this basis are $x{y}^{3},{x}^{3}y,x$ ,and ${y}^{4}$ ,so $\left\{ {x + y,{y}^{4} - {y}^{3} - 1}\right\}$ is a minimal Gröbner basis for $J$ . In this case none of the terms in ${y}^{4} - {y}^{3} - 1$ are divisible by the leading term of $x + y$ and none of the terms in $x + y$ are divisible by the leading term in ${y}^{4} - {y}^{3} - 1$ ,so $\left\{ {x + y,{y}^{4} - {y}^{3} - 1}\right\}$ is the reduced Gröbner basis for $J$ . This is the basis for the ideal $I$ in Example 1 following Proposition 26, so these two ideals are equal:

是 $J$ 的一个Gröbner基。该基的首项是 $x{y}^{3},{x}^{3}y,x$ 和 ${y}^{4}$ ，因此 $\left\{ {x + y,{y}^{4} - {y}^{3} - 1}\right\}$ 是 $J$ 的最小Gröbner基。在这种情况下，${y}^{4} - {y}^{3} - 1$ 中的任何项都不被 $x + y$ 的首项整除，且 $x + y$ 中的任何项都不被 ${y}^{4} - {y}^{3} - 1$ 的首项整除，因此 $\left\{ {x + y,{y}^{4} - {y}^{3} - 1}\right\}$ 是 $J$ 的简化Gröbner基。这是命题26后例1中理想 $I$ 的基，因此这两个理想是相等的：

(and both are equal to the ideal $\left( {x + y,{y}^{4} - {y}^{3} - 1}\right)$ ).

（且这两个都等于理想 $\left( {x + y,{y}^{4} - {y}^{3} - 1}\right)$ ）。