Gröbner基与代数方程求解：消元

翻斗花园爷爷
66 阅读4分钟

Gröbner Bases and Solving Algebraic Equations: Elimination

Gröbner基与代数方程求解：消元

The theory of Gröbner bases is very useful in explicitly solving systems of algebraic equations, and is the basis by which computer algebra programs attempt to solve systems of equations. Suppose S={f1,…,fm}S = \left\{ {{f}_{1},\ldots ,{f}_{m}}\right\} is a collection of polynomials in nn variables x1,…,xn{x}_{1},\ldots ,{x}_{n} and we are trying to find the solutions of the system of equations f1=0{f}_{1} = 0 , f2=0,…,fm=0{f}_{2} = 0,\ldots ,{f}_{m} = 0 (i.e.,the common set of zeros of the polynomials in SS ). If (a1,…,an)\left( {{a}_{1},\ldots ,{a}_{n}}\right) is any solution to this system,then every element ff of the ideal II generated by SS also satisfies f(a1,…,an)=0f\left( {{a}_{1},\ldots ,{a}_{n}}\right) = 0 . Furthermore,it is an easy exercise to see that if S′={g1,…,gs}{S}^{\prime } = \left\{ {{g}_{1},\ldots ,{g}_{s}}\right\} is any set of generators for the ideal II then the set of solutions to the system g1=0,…,gs=0{g}_{1} = 0,\ldots ,{g}_{s} = 0 is the same as the original solution set.

格罗布纳基理论在显式求解代数方程组中非常有用，它是计算机代数程序尝试解方程组的基础。假设 S={f1,…,fm}S = \left\{ {{f}_{1},\ldots ,{f}_{m}}\right\} 是一组 nn 变量的多项式集合 x1,…,xn{x}_{1},\ldots ,{x}_{n} ，我们试图找到方程组 f1=0{f}_{1} = 0 ，f2=0,…,fm=0{f}_{2} = 0,\ldots ,{f}_{m} = 0 （即 SS 中多项式的公共零点集）的解。如果 (a1,…,an)\left( {{a}_{1},\ldots ,{a}_{n}}\right) 是该方程组的任一解，那么由 SS 生成的理想 II 的每个元素 ff 也满足 f(a1,…,an)=0f\left( {{a}_{1},\ldots ,{a}_{n}}\right) = 0 。此外，很容易验证，如果 S′={g1,…,gs}{S}^{\prime } = \left\{ {{g}_{1},\ldots ,{g}_{s}}\right\} 是理想 II 的任一生成集合，那么方程组的解集 g1=0,…,gs=0{g}_{1} = 0,\ldots ,{g}_{s} = 0 与原始解集相同。

In the situation where f1,…,fm{f}_{1},\ldots ,{f}_{m} are linear polynomials,a solution to the system of equations can be obtained by successively eliminating the variables x1,x2,…{x}_{1},{x}_{2},\ldots by elementary means-using linear combinations of the original equations to eliminate the variable x1{x}_{1} ,then using these equations to eliminate x2{x}_{2} ,etc.,producing a system of equations that can be easily solved (this is “Gauss-Jordan elimination” in linear algebra, cf. the exercises in Section 11.2).

在 f1,…,fm{f}_{1},\ldots ,{f}_{m} 是线性多项式的情况下，可以通过逐个消除变量 x1,x2,…{x}_{1},{x}_{2},\ldots 来获得方程组的解，使用基本的手段——利用原始方程的线性组合来消除变量 x1{x}_{1} ，然后使用这些方程来消除 x2{x}_{2} 等等，从而产生一个可以轻松解决的方程组（这是线性代数中的“高斯-若尔当消元法”，参见第11.2节的练习）。

The situation for polynomial equations that are nonlinear is naturally more complicated, but the basic principle is the same. If there is a nonzero polynomial in the ideal II involving only one of the variables,say p(xn)p\left( {x}_{n}\right) ,then the last coordinate an{a}_{n} is a solution of p(xn)=0p\left( {x}_{n}\right) = 0 . If now there is a polynomial in II involving only xn−1{x}_{n - 1} and xn{x}_{n} ,say q(xn−1,xn)q\left( {{x}_{n - 1},{x}_{n}}\right) ,then the coordinate an−1{a}_{n - 1} would be a solution of q(xn−1,an)=0q\left( {{x}_{n - 1},{a}_{n}}\right) = 0 , etc. If we can successively find polynomials in II that eliminate the variables x1,x2,…{x}_{1},{x}_{2},\ldots then we will be able to determine all the solutions (a1,…,an)\left( {{a}_{1},\ldots ,{a}_{n}}\right) to our original system of equations explicitly.

对于非线性多项式方程的情况自然更为复杂，但基本原理是相同的。如果理想 II 中存在一个只涉及一个变量，比如说 p(xn)p\left( {x}_{n}\right) 的非零多项式，那么最后一个坐标 an{a}_{n} 就是 p(xn)=0p\left( {x}_{n}\right) = 0 的解。如果现在理想 II 中有一个只涉及 xn−1{x}_{n - 1} 和 xn{x}_{n} 的多项式，比如说 q(xn−1,xn)q\left( {{x}_{n - 1},{x}_{n}}\right) ，那么坐标 an−1{a}_{n - 1} 将是 q(xn−1,an)=0q\left( {{x}_{n - 1},{a}_{n}}\right) = 0 的解，等等。如果我们能够依次找到理想 II 中消去变量 x1,x2,…{x}_{1},{x}_{2},\ldots 的多项式，那么我们将能够明确地确定原方程组的所有解 (a1,…,an)\left( {{a}_{1},\ldots ,{a}_{n}}\right) 。

Finding equations that follow from the system of equations in SS ,i.e.,finding elements of the ideal II that do not involve some of the variables,is referred to as elimination theory. The polynomials in II that do not involve the variables x1,…,xi{x}_{1},\ldots ,{x}_{i} ,i.e., I∩F[xi+1,…,xn]I \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack ,is easily seen to be an ideal in F[xi+1,…,xn]F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack and is given a name.

寻找由 SS 的方程组得出的方程，即寻找不涉及某些变量的理想 II 中的元素，这个过程被称为消去理论。在 II 中不涉及变量 x1,…,xi{x}_{1},\ldots ,{x}_{i} 的多项式，即 I∩F[xi+1,…,xn]I \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack ，可以容易地看出是 F[xi+1,…,xn]F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack 的一个理想，并给它一个名称。

Definition. If II is an ideal in F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack then Ii=I∩F[xi+1,…,xn]{I}_{i} = I \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack is called the ith {i}^{\text{th }} elimination ideal of II with respect to the ordering x1>⋯>xn{x}_{1} > \cdots > {x}_{n} .

定义。如果 II 是 F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack 的一个理想，那么 Ii=I∩F[xi+1,…,xn]{I}_{i} = I \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack 被称为关于排序 x1>⋯>xn{x}_{1} > \cdots > {x}_{n} 的 II 的 ith {i}^{\text{th }} 消去理想。

The success of using elimination to solve a system of equations depends on being able to determine the elimination ideals (and, ultimately, on whether these elimination ideals are nonzero).

使用消去法解方程组成功与否取决于能否确定消去理想（最终取决于这些消去理想是否非零）。

The following fundamental proposition shows that if the lexicographic monomial ordering x1>⋯>xn{x}_{1} > \cdots > {x}_{n} is used to compute a Gröbner basis for II then the elements in the resulting basis not involving the variables x1,…,xi{x}_{1},\ldots ,{x}_{i} not only determine the ith {i}^{\text{th }} elimination ideal,but in fact give a Gröbner basis for the ith {i}^{\text{th }} elimination ideal of II .

以下基本命题表明，如果使用字典序单项式排序 x1>⋯>xn{x}_{1} > \cdots > {x}_{n} 来计算 II 的 Gröbner 基，那么结果基中不涉及变量 x1,…,xi{x}_{1},\ldots ,{x}_{i} 的元素不仅确定了 ith {i}^{\text{th }} 的消去理想，实际上还给出了 ith {i}^{\text{th }} 消去理想的 Gröbner 基。

Proposition 29. (Elimination) Suppose G={g1,…,gm}G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} is a Gröbner basis for the nonzero ideal II in F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack with respect to the lexicographic monomial ordering x1>⋯>xn{x}_{1} > \cdots > {x}_{n} . Then G∩F[xi+1,…,xn]G \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack is a Gröbner basis of the ith {i}^{\text{th }} elimination ideal Ii=I∩F[xi+1,…,xn]{I}_{i} = I \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack of II . In particular, I∩F[xi+1,…,xn]=0I \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack = 0 if and only if G∩F[xi+1,…,xn]=∅G \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack = \varnothing .

命题 29.（消去）假设 G={g1,…,gm}G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} 是关于字典序单项式排序 x1>⋯>xn{x}_{1} > \cdots > {x}_{n} 的非零理想 II 在 F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack 上的 Gröbner 基。那么 G∩F[xi+1,…,xn]G \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack 是 II 的 ith {i}^{\text{th }} 消去理想 Ii=I∩F[xi+1,…,xn]{I}_{i} = I \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack 的 Gröbner 基。特别地，当且仅当 I∩F[xi+1,…,xn]=0I \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack = 0 时，G∩F[xi+1,…,xn]=∅G \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack = \varnothing。

Proof: Denote Gi=G∩F[xi+1,…,xn]{G}_{i} = G \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack . Then Gi⊆Ii{G}_{i} \subseteq {I}_{i} ,so by Proposition 24,to see that Gi{G}_{i} is a Gröbner basis of Ii{I}_{i} it suffices to see that LT(Gi){LT}\left( {G}_{i}\right) ,the leading terms of the elements in Gi{G}_{i} ,generate LT(Ii){LT}\left( {I}_{i}\right) as an ideal in F[xi+1,…,xn]F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack . Certainly (LT(Gi))⊆LT(Ii)\left( {{LT}\left( {G}_{i}\right) }\right) \subseteq {LT}\left( {I}_{i}\right) as ideals in F[xi+1,…,xn]F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack . To show the reverse containment, let ff be any element in Ii{I}_{i} . Then f∈If \in I and since GG is a Gröbner basis for II we have

证明：记 Gi=G∩F[xi+1,…,xn]{G}_{i} = G \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack。那么 Gi⊆Ii{G}_{i} \subseteq {I}_{i}，因此根据命题 24，要证明 Gi{G}_{i} 是 Ii{I}_{i} 的 Gröbner 基，只需证明 LT(Gi){LT}\left( {G}_{i}\right)，即 Gi{G}_{i} 中元素的领先项，作为 F[xi+1,…,xn]F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack 中的理想生成 LT(Ii){LT}\left( {I}_{i}\right)。显然 (LT(Gi))⊆LT(Ii)\left( {{LT}\left( {G}_{i}\right) }\right) \subseteq {LT}\left( {I}_{i}\right) 作为 F[xi+1,…,xn]F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack 中的理想。为了证明反向包含，设 ff 是 Ii{I}_{i} 中的任意元素。那么 f∈If \in I，由于 GG 是 II 的 Gröbner 基，我们有

LT(f)=a1(x1,…,xn)LT(g1)+⋯+am(x1,…,xn)LT(gm){LT}\left( f\right) = {a}_{1}\left( {{x}_{1},\ldots ,{x}_{n}}\right) {LT}\left( {g}_{1}\right) + \cdots + {a}_{m}\left( {{x}_{1},\ldots ,{x}_{n}}\right) {LT}\left( {g}_{m}\right)

for some polynomials a1,…,am∈F[x1,…,xn]{a}_{1},\ldots ,{a}_{m} \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack . Writing each polynomial ai{a}_{i} as a sum of monomial terms we see that LT(f){LT}\left( f\right) is a sum of monomial terms of the form ax1s1…xnsnLT(gi)a{x}_{1}^{{s}_{1}}\ldots {x}_{n}^{{s}_{n}}{LT}\left( {g}_{i}\right) . Since LT(f){LT}\left( f\right) involves only the variables xi+1,…,xn{x}_{i + 1},\ldots ,{x}_{n} ,the sum of all such terms containing any of the variables x1,…,xi{x}_{1},\ldots ,{x}_{i} must be 0,so LT(f){LT}\left( f\right) is also the sum of those monomial terms only involving xi+1,…,xn{x}_{i + 1},\ldots ,{x}_{n} . It follows that LT(f){LT}\left( f\right) can be written as a F[xi+1,…,xn]F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack -linear combination of some monomial terms LT(gt){LT}\left( {g}_{t}\right) where LT(gt){LT}\left( {g}_{t}\right) does not involve the variables x1,…,xi{x}_{1},\ldots ,{x}_{i} . But by the choice of the ordering, if LT(gt){LT}\left( {g}_{t}\right) does not involve x1,…,xi{x}_{1},\ldots ,{x}_{i} ,then neither do any of the other terms in gt{g}_{t} , i.e., gt∈Gi{g}_{t} \in {G}_{i} . Hence LT(f){LT}\left( f\right) can be written as a F[xi+1,…,xn]F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack -linear combination of elements LT(Gi){LT}\left( {G}_{i}\right) ,completing the proof.

对于某些多项式 a1,…,am∈F[x1,…,xn]{a}_{1},\ldots ,{a}_{m} \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack ，将每个多项式 ai{a}_{i} 写作单项式的和，我们可以看到 LT(f){LT}\left( f\right) 是形如 ax1s1…xnsnLT(gi)a{x}_{1}^{{s}_{1}}\ldots {x}_{n}^{{s}_{n}}{LT}\left( {g}_{i}\right) 的单项式的和。由于 LT(f){LT}\left( f\right) 只涉及变量 xi+1,…,xn{x}_{i + 1},\ldots ,{x}_{n} ，所有包含这些变量的项的和必须是 0，因此 LT(f){LT}\left( f\right) 也仅是涉及 xi+1,…,xn{x}_{i + 1},\ldots ,{x}_{n} 的那些单项式的和。因此 LT(f){LT}\left( f\right) 可以写成某些单项式 LT(gt){LT}\left( {g}_{t}\right) 的 F[xi+1,…,xn]F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack -线性组合，其中 LT(gt){LT}\left( {g}_{t}\right) 不涉及变量 x1,…,xi{x}_{1},\ldots ,{x}_{i} 。但是根据排序的选择，如果 LT(gt){LT}\left( {g}_{t}\right) 不涉及 x1,…,xi{x}_{1},\ldots ,{x}_{i} ，那么 gt{g}_{t} 中的其他任何项也不涉及，即 gt∈Gi{g}_{t} \in {G}_{i} 。因此 LT(f){LT}\left( f\right) 可以写成元素 LT(Gi){LT}\left( {G}_{i}\right) 的 F[xi+1,…,xn]F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack -线性组合，从而完成了证明。

Note also that Gröbner bases can be used to eliminate any variables simply by using an appropriate monomial ordering.

还应注意，通过使用适当的单项式排序，可以利用 Gröbner 基来消去任何变量。

Examples

示例

(1) The ellipse 2x2+2xy+y2−2x−2y=02{x}^{2} + {2xy} + {y}^{2} - {2x} - {2y} = 0 intersects the circle x2+y2=1{x}^{2} + {y}^{2} = 1 in two points. To find them we compute a Gröbner basis for the ideal I=(2x2+2xy+y2−I = \left( {2{x}^{2} + {2xy} + {y}^{2} - }\right. 2x−2y,x2+y2−1)⊂R[x,y]{2x} - {2y},{x}^{2} + {y}^{2} - 1) \subset \mathbb{R}\left\lbrack {x,y}\right\rbrack using the lexicographic monomial order x>yx > y to eliminate xx ,obtaining g1=2x+y2+5y3−2{g}_{1} = {2x} + {y}^{2} + 5{y}^{3} - 2 and g2=5y4−4y3{g}_{2} = 5{y}^{4} - 4{y}^{3} . Hence 5y4=4y35{y}^{4} = 4{y}^{3} and y=0y = 0 or y=4/5y = 4/5 . Substituting these values into g1=0{g}_{1} = 0 and solving for xx we find the two intersection points are(1,0)and (−3/5,4/5)\left( {-3/5,4/5}\right) .

(1) 椭圆 2x2+2xy+y2−2x−2y=02{x}^{2} + {2xy} + {y}^{2} - {2x} - {2y} = 0 与圆 x2+y2=1{x}^{2} + {y}^{2} = 1 在两点相交。为了找到这两个点，我们计算理想 I=(2x2+2xy+y2−I = \left( {2{x}^{2} + {2xy} + {y}^{2} - }\right. 2x−2y,x2+y2−1)⊂R[x,y]{2x} - {2y},{x}^{2} + {y}^{2} - 1) \subset \mathbb{R}\left\lbrack {x,y}\right\rbrack 的 Gröbner 基，使用字典序单项式排序 x>yx > y 来消去 xx ，得到 g1=2x+y2+5y3−2{g}_{1} = {2x} + {y}^{2} + 5{y}^{3} - 2 和 g2=5y4−4y3{g}_{2} = 5{y}^{4} - 4{y}^{3} 。因此 5y4=4y35{y}^{4} = 4{y}^{3} 和 y=0y = 0 或者 y=4/5y = 4/5 。将这些值代入 g1=0{g}_{1} = 0 并求解 xx ，我们找到的两个交点是 (1,0) 和 (−3/5,4/5)\left( {-3/5,4/5}\right) 。

Instead using the lexicographic monomial order y>xy > x to eliminate yy results in the Gröbner basis {y2+x2−1,2yx−2y+x2−2x+1,5x3−7x2−x+3}\left\{ {{y}^{2} + {x}^{2} - 1,{2yx} - {2y} + {x}^{2} - {2x} + 1,5{x}^{3} - 7{x}^{2} - x + 3}\right\} . Then 5x3−7x2−x+3=(x−1)2(5x+3)5{x}^{3} - 7{x}^{2} - x + 3 = {\left( x - 1\right) }^{2}\left( {{5x} + 3}\right) shows that xx is 1 or −3/5- 3/5 and we obtain the same solutions as before, although with more effort.

使用字典序单项式排列 y>xy > x 来消去 yy 结果得到 Gröbner 基 {y2+x2−1,2yx−2y+x2−2x+1,5x3−7x2−x+3}\left\{ {{y}^{2} + {x}^{2} - 1,{2yx} - {2y} + {x}^{2} - {2x} + 1,5{x}^{3} - 7{x}^{2} - x + 3}\right\} 。然后 5x3−7x2−x+3=(x−1)2(5x+3)5{x}^{3} - 7{x}^{2} - x + 3 = {\left( x - 1\right) }^{2}\left( {{5x} + 3}\right) 显示 xx 是 1 或 −3/5- 3/5 ，我们得到了与之前相同的解，尽管付出了更多的努力。

(2)  \textbf{(2)}\; In the previous example the solutions could also have been found by elementary means. Consider now the solutions in C\mathbb{C} to the system of two equations

(2)  \textbf{(2)}\; 在之前的例子中，解也可以通过基本方法找到。现在考虑 C\mathbb{C} 中两个方程组的解

x3−2xy+y3=0 and x5−2x2y2+y5=0.{x}^{3} - {2xy} + {y}^{3} = 0\text{ and }{x}^{5} - 2{x}^{2}{y}^{2} + {y}^{5} = 0.

Computing a Gröbner basis for the ideal generated by f1=x3−2xy+y3{f}_{1} = {x}^{3} - {2xy} + {y}^{3} and f2=x5−2x2y2+y5{f}_{2} = {x}^{5} - 2{x}^{2}{y}^{2} + {y}^{5} with respect to the lexicographic monomial order x>yx > y we obtain the basis

对于由 f1=x3−2xy+y3{f}_{1} = {x}^{3} - {2xy} + {y}^{3} 和 f2=x5−2x2y2+y5{f}_{2} = {x}^{5} - 2{x}^{2}{y}^{2} + {y}^{5} 生成的理想，相对于字典序单项式排列 x>yx > y 计算其 Gröbner 基，我们得到基

g1=x3−2xy+y3{g}_{1} = {x}^{3} - {2xy} + {y}^{3}
g2=200xy2+193y9+158y8−45y7−456y6+50y5−100y4{g}_{2} = {200x}{y}^{2} + {193}{y}^{9} + {158}{y}^{8} - {45}{y}^{7} - {456}{y}^{6} + {50}{y}^{5} - {100}{y}^{4}
g3=y10−y8−2y7+2y6.{g}_{3} = {y}^{10} - {y}^{8} - 2{y}^{7} + 2{y}^{6}.

Any solution to our original equations would satisfy g1=g2=g3=0{g}_{1} = {g}_{2} = {g}_{3} = 0 . Since g3=y6(y−1)2(y2+2y+2){g}_{3} = {y}^{6}{\left( y - 1\right) }^{2}\left( {{y}^{2} + {2y} + 2}\right) ,we have y=0,y=1y = 0,y = 1 or y=−1±iy = - 1 \pm i . Since g1(x,0)=x3{g}_{1}\left( {x,0}\right) = {x}^{3} and g2(x,0)=0{g}_{2}\left( {x,0}\right) = 0 ,we see that(0,0)is the only solution with y=0y = 0 . Since g1(x,1)=x3−2x+1{g}_{1}\left( {x,1}\right) = {x}^{3} - {2x} + 1 and g2(x,1)=200(x−1){g}_{2}\left( {x,1}\right) = {200}\left( {x - 1}\right) have only x=1x = 1 as a common zero,the only solution with y=1y = 1 is(1,1). Finally,

我们原始方程的任何解都将满足 g1=g2=g3=0{g}_{1} = {g}_{2} = {g}_{3} = 0 。由于 g3=y6(y−1)2(y2+2y+2){g}_{3} = {y}^{6}{\left( y - 1\right) }^{2}\left( {{y}^{2} + {2y} + 2}\right) ，我们有 y=0,y=1y = 0,y = 1 或 y=−1±iy = - 1 \pm i 。由于 g1(x,0)=x3{g}_{1}\left( {x,0}\right) = {x}^{3} 和 g2(x,0)=0{g}_{2}\left( {x,0}\right) = 0 ，我们看到 (0,0) 是唯一满足 y=0y = 0 的解。由于 g1(x,1)=x3−2x+1{g}_{1}\left( {x,1}\right) = {x}^{3} - {2x} + 1 和 g2(x,1)=200(x−1){g}_{2}\left( {x,1}\right) = {200}\left( {x - 1}\right) 只有一个公共零点 x=1x = 1 ，所以唯一满足 y=1y = 1 的解是 (1,1)。最后，

g1(x,−1±i)=x3+(2∓2i)x+(2±2i){g}_{1}\left( {x, - 1 \pm i}\right) = {x}^{3} + \left( {2 \mp {2i}}\right) x + \left( {2 \pm {2i}}\right)
g2(x,−1±i)=−400i(x+1±i),{g}_{2}\left( {x, - 1 \pm i}\right) = - {400i}\left( {x + 1 \pm i}\right) ,

and a quick check shows the common zero x=−1∓ix = - 1 \mp i when y=−1±iy = - 1 \pm i ,respectively. Hence, there are precisely four solutions to the original pair of equations, namely

快速检查显示当 y=−1±iy = - 1 \pm i 时，公共零点 x=−1∓ix = - 1 \mp i 分别是。因此，原始方程组恰好有四个解，即

(x,y)=(0,0),  (1,1),  (−1+i,−1−i),   or   (−1−i,−1+i).\left( {x,y}\right) = \left( {0,0}\right) ,\;\left( {1,1}\right) ,\;\left( {-1 + i, - 1 - i}\right) ,\;\text{ or }\;\left( {-1 - i, - 1 + i}\right) .

(3) Consider the solutions in C\mathbb{C} to the system of equations

(3) 考虑 C\mathbb{C} 中方程组的解

x+y+z=1x + y + z = 1
x2+y2+z2=2{x}^{2} + {y}^{2} + {z}^{2} = 2
x3+y3+z3=3{x}^{3} + {y}^{3} + {z}^{3} = 3

The reduced Gröbner basis with respect to the lexicographic ordering x>y>zx > y > z is

相对于字典序排列的简化 Gröbner 基 x>y>zx > y > z 是

{x+y+z−1,  y2+yz−y+z2−z−(1/2),  z3−z2−(1/2)z−(1/6)}\{ x + y + z - 1,\;{y}^{2} + {yz} - y + {z}^{2} - z - \left( {1/2}\right) ,\;{z}^{3} - {z}^{2} - \left( {1/2}\right) z - \left( {1/6}\right) \}

and so zz is a root of the polynomial t3−t2−(1/2)t−(1/6){t}^{3} - {t}^{2} - \left( {1/2}\right) t - \left( {1/6}\right) (by symmetry,also xx and yy are roots of this same polynomial). For each of the three roots of this polynomial, there are two values of yy and one corresponding value of xx making the first two polynomials in the Gröbner basis equal to 0 . The resulting six solutions are quickly checked to be the three distinct roots of the polynomial t3−t2−(1/2)t−(1/6){t}^{3} - {t}^{2} - \left( {1/2}\right) t - \left( {1/6}\right) (which is irreducible over Q\mathbb{Q} ) in some order.

因此 zz 是多项式 t3−t2−(1/2)t−(1/6){t}^{3} - {t}^{2} - \left( {1/2}\right) t - \left( {1/6}\right) 的一个根（由于对称性，xx 和 yy 也是这个相同多项式的根）。对于这个多项式的每一个根，都有两个 yy 的值和一个相应的 xx 的值，使得格罗布纳基中的前两个多项式等于 0。得到的六个解很快就被验证为多项式 t3−t2−(1/2)t−(1/6){t}^{3} - {t}^{2} - \left( {1/2}\right) t - \left( {1/6}\right)（在 Q\mathbb{Q} 上不可约）的三个不同根（以某种顺序）。

As the previous examples show, the study of solutions to systems of polynomial equations f1=0,f2=0,…,fm=0{f}_{1} = 0,{f}_{2} = 0,\ldots ,{f}_{m} = 0 is intimately related to the study of the ideal I=(f1,f2,…,fm)I = \left( {{f}_{1},{f}_{2},\ldots ,{f}_{m}}\right) the polynomials generate in F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack . This fundamental connection is the starting point for the important and active branch of mathematics called “algebraic geometry”, introduced in Chapter 15, where additional applications of Gröbner bases are given.

如前例所示，研究多项式方程组 f1=0,f2=0,…,fm=0{f}_{1} = 0,{f}_{2} = 0,\ldots ,{f}_{m} = 0 的解与研究由多项式生成的理想 I=(f1,f2,…,fm)I = \left( {{f}_{1},{f}_{2},\ldots ,{f}_{m}}\right) 在 F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack 中的关系密切相关。这个基本联系是重要的、活跃的数学分支“代数几何”的起点，在第15章中引入了代数几何，并给出了格罗布纳基的额外应用。

We close this section by showing how to compute the basic set-theoretic operations of sums, products and intersections of ideals in polynomial rings. Suppose I=(f1,…,fs)I = \left( {{f}_{1},\ldots ,{f}_{s}}\right) and J=(h1,…,ht)J = \left( {{h}_{1},\ldots ,{h}_{t}}\right) are two ideals in F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack . Then I+J=(f1,…,fs,h1,…,ht)I + J = \left( {{f}_{1},\ldots ,{f}_{s},{h}_{1},\ldots ,{h}_{t}}\right) and IJ=(f1h1,…,fihj,…,fsht){IJ} = \left( {{f}_{1}{h}_{1},\ldots ,{f}_{i}{h}_{j},\ldots ,{f}_{s}{h}_{t}}\right) . The following proposition shows how to compute the intersection of any two ideals.

我们通过展示如何在多项式环中计算基本集合运算，如理想的和、积和交集来结束这一节。假设 I=(f1,…,fs)I = \left( {{f}_{1},\ldots ,{f}_{s}}\right) 和 J=(h1,…,ht)J = \left( {{h}_{1},\ldots ,{h}_{t}}\right) 是 F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack 中的两个理想。那么 I+J=(f1,…,fs,h1,…,ht)I + J = \left( {{f}_{1},\ldots ,{f}_{s},{h}_{1},\ldots ,{h}_{t}}\right) 和 IJ=(f1h1,…,fihj,…,fsht){IJ} = \left( {{f}_{1}{h}_{1},\ldots ,{f}_{i}{h}_{j},\ldots ,{f}_{s}{h}_{t}}\right)。以下命题展示了如何计算任意两个理想的交集。

Proposition 30. If II and JJ are any two ideals in F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack then tI+(1−t)J{tI} + \left( {1 - t}\right) J is an ideal in F[t,x1,…,xn]F\left\lbrack {t,{x}_{1},\ldots ,{x}_{n}}\right\rbrack and I∩J=(tI+(1−t)J)∩F[x1,…,xn]I \cap J = \left( {{tI} + \left( {1 - t}\right) J}\right) \cap F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack . In particular, I∩JI \cap J is the first elimination ideal of tI+(1−t)J{tI} + \left( {1 - t}\right) J with respect to the ordering t>x1>⋯>xnt > {x}_{1} > \cdots > {x}_{n} .

命题30。如果 II 和 JJ 是 F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack 中的任意两个理想，那么 tI+(1−t)J{tI} + \left( {1 - t}\right) J 是 F[t,x1,…,xn]F\left\lbrack {t,{x}_{1},\ldots ,{x}_{n}}\right\rbrack 中的一个理想，且 I∩J=(tI+(1−t)J)∩F[x1,…,xn]I \cap J = \left( {{tI} + \left( {1 - t}\right) J}\right) \cap F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack。特别是，I∩JI \cap J 是相对于顺序 t>x1>⋯>xnt > {x}_{1} > \cdots > {x}_{n} 的 tI+(1−t)J{tI} + \left( {1 - t}\right) J 的第一个消元理想。

Proof: First, tI{tI} and (1−t)J\left( {1 - t}\right) J are clearly ideals in F[x1,…,xn,t]F\left\lbrack {{x}_{1},\ldots ,{x}_{n},t}\right\rbrack ,so also their sum tI+(1−t)J{tI} + \left( {1 - t}\right) J is an ideal in F[x1,…,xn,t]F\left\lbrack {{x}_{1},\ldots ,{x}_{n},t}\right\rbrack . If f∈I∩Jf \in I \cap J ,then f=tf+(1−t)ff = {tf} + \left( {1 - t}\right) f shows I∩J⊆(tI+(1−t)J)∩F[x1,…,xn].  Conversely,suppose  f=tf1+(1−t)f2  is\begin{matrix} I \cap J \subseteq \left( {{tI} + \left( {1 - t}\right) J}\right) \cap F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack .\;\text{Conversely,suppose}\;f = t{f}_{1} + \left( {1 - t}\right) {f}_{2}\;\text{is} \end{matrix} an element of F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack ,where f1∈I{f}_{1} \in I and f2∈J{f}_{2} \in J . Then t(f1−f2)=f−f2∈t\left( {{f}_{1} - {f}_{2}}\right) = f - {f}_{2} \in F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack shows that f1−f2=0{f}_{1} - {f}_{2} = 0 and f=f2f = {f}_{2} ,so f=f1=f2∈I∩Jf = {f}_{1} = {f}_{2} \in I \cap J . Since I∩J=(tI+(1−t)J)∩F[x1,…,xn],I∩JI \cap J = \left( {{tI} + \left( {1 - t}\right) J}\right) \cap F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack ,I \cap J is the first elimination ideal of tI+(1−t)J{tI} + \left( {1 - t}\right) J with respect to the ordering t>x1>⋯>xnt > {x}_{1} > \cdots > {x}_{n} .

证明：首先，tI{tI} 和 (1−t)J\left( {1 - t}\right) J 显然是 F[x1,…,xn,t]F\left\lbrack {{x}_{1},\ldots ,{x}_{n},t}\right\rbrack 中的理想，因此它们的和 tI+(1−t)J{tI} + \left( {1 - t}\right) J 也是 F[x1,…,xn,t]F\left\lbrack {{x}_{1},\ldots ,{x}_{n},t}\right\rbrack 中的理想。如果 f∈I∩Jf \in I \cap J ，那么 f=tf+(1−t)ff = {tf} + \left( {1 - t}\right) f 显示 I∩J⊆(tI+(1−t)J)∩F[x1,…,xn].  Conversely,suppose  f=tf1+(1−t)f2  is\begin{matrix} I \cap J \subseteq \left( {{tI} + \left( {1 - t}\right) J}\right) \cap F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack .\;\text{Conversely,suppose}\;f = t{f}_{1} + \left( {1 - t}\right) {f}_{2}\;\text{is} \end{matrix} 是 F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack 的一个元素，其中 f1∈I{f}_{1} \in I 和 f2∈J{f}_{2} \in J 。那么 t(f1−f2)=f−f2∈t\left( {{f}_{1} - {f}_{2}}\right) = f - {f}_{2} \in F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack 显示 f1−f2=0{f}_{1} - {f}_{2} = 0 和 f=f2f = {f}_{2} ，所以 f=f1=f2∈I∩Jf = {f}_{1} = {f}_{2} \in I \cap J 。由于 I∩J=(tI+(1−t)J)∩F[x1,…,xn],I∩JI \cap J = \left( {{tI} + \left( {1 - t}\right) J}\right) \cap F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack ,I \cap J 是相对于排序 t>x1>⋯>xnt > {x}_{1} > \cdots > {x}_{n} 的 tI+(1−t)J{tI} + \left( {1 - t}\right) J 的第一个消元理想。

We have tI+(1−t)J=(tf1,…,tfs,(1−t)h1,…,(1−t)ht){tI} + \left( {1 - t}\right) J = \left( {t{f}_{1},\ldots ,t{f}_{s},\left( {1 - t}\right) {h}_{1},\ldots ,\left( {1 - t}\right) {h}_{t}}\right) if I=(f1,…,fs)I = \left( {{f}_{1},\ldots ,{f}_{s}}\right) and J=(h1,…,ht)J = \left( {{h}_{1},\ldots ,{h}_{t}}\right) . By Proposition 29,the elements not involving tt in a Gröbner basis for this ideal in F[t,x1,…,xn]F\left\lbrack {t,{x}_{1},\ldots ,{x}_{n}}\right\rbrack ,computed for the lexicographic monomial ordering t>x1>⋯>xnt > {x}_{1} > \cdots > {x}_{n} ,give a Gröbner basis for the ideal I∩JI \cap J in F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack .

如果 I=(f1,…,fs)I = \left( {{f}_{1},\ldots ,{f}_{s}}\right) 和 J=(h1,…,ht)J = \left( {{h}_{1},\ldots ,{h}_{t}}\right) ，那么我们有 tI+(1−t)J=(tf1,…,tfs,(1−t)h1,…,(1−t)ht){tI} + \left( {1 - t}\right) J = \left( {t{f}_{1},\ldots ,t{f}_{s},\left( {1 - t}\right) {h}_{1},\ldots ,\left( {1 - t}\right) {h}_{t}}\right) 。根据命题29，在这个理想在 F[t,x1,…,xn]F\left\lbrack {t,{x}_{1},\ldots ,{x}_{n}}\right\rbrack 中的Gröbner基中不涉及 tt 的元素，对于字典序单项式排序 t>x1>⋯>xnt > {x}_{1} > \cdots > {x}_{n} 计算得到的 I∩JI \cap J 理想在 F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack 中的Gröbner基。

Example

示例

Let I=(x,y)2=(x2,xy,y2)I = {\left( x,y\right) }^{2} = \left( {{x}^{2},{xy},{y}^{2}}\right) and let J=(x)J = \left( x\right) . For the lexicographic monomial ordering t>x>yt > x > y the reduced Gröbner basis for tI+(1−t)J{tI} + \left( {1 - t}\right) J in F[t,x,y]F\left\lbrack {t,x,y}\right\rbrack is {tx−x,ty2,x2,xy}\left\{ {{tx} - x,t{y}^{2},{x}^{2},{xy}}\right\} and so I∩J=(x2,xy)I \cap J = \left( {{x}^{2},{xy}}\right) .

设 I=(x,y)2=(x2,xy,y2)I = {\left( x,y\right) }^{2} = \left( {{x}^{2},{xy},{y}^{2}}\right) 并且让 J=(x)J = \left( x\right) 。对于字典序单项式排序 t>x>yt > x > y ，tI+(1−t)J{tI} + \left( {1 - t}\right) J 在 F[t,x,y]F\left\lbrack {t,x,y}\right\rbrack 中的简化Gröbner基是 {tx−x,ty2,x2,xy}\left\{ {{tx} - x,t{y}^{2},{x}^{2},{xy}}\right\} ，因此 I∩J=(x2,xy)I \cap J = \left( {{x}^{2},{xy}}\right) 。

EXERCISES

练习

  1. Suppose II is an ideal in F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack generated by a (possibly infinite) set S\mathcal{S} of polynomials. Prove that a finite subset of the polynomials in S\mathcal{S} suffice to generate II . [Use Theorem 21 to write I=(f1,…,fm)I = \left( {{f}_{1},\ldots ,{f}_{m}}\right) and then write each fi∈I{f}_{i} \in I using polynomials in S.]

  2. 假设 II 是由 F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack 中的一组（可能是无限的）多项式 S\mathcal{S} 生成的理想。证明 S\mathcal{S} 中多项式的一个有限子集足以生成 II 。[使用定理21将 I=(f1,…,fm)I = \left( {{f}_{1},\ldots ,{f}_{m}}\right) 写出来，然后用 S 中的多项式写出每个 fi∈I{f}_{i} \in I 。]

  3. Let ≥\geq be any monomial ordering.

  4. 设 ≥\geq 是任意的单项式排序。

(a) Prove that LT(fg)=LT(f)LT(g){LT}\left( {fg}\right) = {LT}\left( f\right) {LT}\left( g\right) and ∂(fg)=∂(f)+∂(g)\partial \left( {fg}\right) = \partial \left( f\right) + \partial \left( g\right) for any nonzero polynomials ff and gg .

(a) 证明对于任何非零多项式 ff 和 gg ，LT(fg)=LT(f)LT(g){LT}\left( {fg}\right) = {LT}\left( f\right) {LT}\left( g\right) 和 ∂(fg)=∂(f)+∂(g)\partial \left( {fg}\right) = \partial \left( f\right) + \partial \left( g\right) 成立。

(b) Prove that ∂(f+g)≤max⁡(∂(f),∂(g))\partial \left( {f + g}\right) \leq \max \left( {\partial \left( f\right) ,\partial \left( g\right) }\right) with equality if ∂(f)≠∂(g)\partial \left( f\right) \neq \partial \left( g\right) .

(b) 证明 ∂(f+g)≤max⁡(∂(f),∂(g))\partial \left( {f + g}\right) \leq \max \left( {\partial \left( f\right) ,\partial \left( g\right) }\right) 在 ∂(f)≠∂(g)\partial \left( f\right) \neq \partial \left( g\right) 时取等。

(c) Prove that m≥1m \geq 1 for every monomial mm .

(c) 证明 m≥1m \geq 1 对于每一个单项式 mm 成立。

(d) Prove that if m1{m}_{1} divides m2{m}_{2} then m2≥m1{m}_{2} \geq {m}_{1} . Deduce that the leading term of a polynomial does not divide any of its lower order terms.

(d) 证明如果 m1{m}_{1} 整除 m2{m}_{2} ，那么 m2≥m1{m}_{2} \geq {m}_{1} 。推导出多项式的首项不整除它的任何低阶项。

  1. Prove that if ≥\geq is any total or partial ordering on a nonempty set then the following are equivalent:

  2. 证明如果 ≥\geq 是非空集合上的任何全序或偏序，那么以下条件是等价的：

(i) Every nonempty subset contains a minimum element.

(i) 每个非空子集都包含一个最小元素。

(ii) There is no infinite strictly decreasing sequence a1>a2>a3>⋯{a}_{1} > {a}_{2} > {a}_{3} > \cdots (this is called the descending chain condition or D.C.C.).

(ii) 不存在无限严格递减序列 a1>a2>a3>⋯{a}_{1} > {a}_{2} > {a}_{3} > \cdots（这被称为降链条件或D.C.C.）。

Deduce that General Polynomial Division always terminates in finitely many steps.

推导出多项式除法总是在有限步骤内终止。

  1. Let ≥\geq be a monomial ordering,and for monomials m1,m2{m}_{1},{m}_{2} define m1≥gm2{m}_{1}{ \geq }_{g}{m}_{2} if either deg⁡m1>deg⁡m2\deg {m}_{1} > \deg {m}_{2} ,or deg⁡m1=deg⁡m2\deg {m}_{1} = \deg {m}_{2} and m1≥m2{m}_{1} \geq {m}_{2} .

  2. 设 ≥\geq 是单项式排序，对于单项式 m1,m2{m}_{1},{m}_{2} 定义 m1≥gm2{m}_{1}{ \geq }_{g}{m}_{2} 如果 deg⁡m1>deg⁡m2\deg {m}_{1} > \deg {m}_{2} ，或者 deg⁡m1=deg⁡m2\deg {m}_{1} = \deg {m}_{2} 且 m1≥m2{m}_{1} \geq {m}_{2} 。

(a) Prove that ≥g{ \geq }_{g} is also a monomial ordering. (The relation ≥g{ \geq }_{g} is called the grading of ≥\geq . An ordering in which the most important criterion for comparison is degree is sometimes called a graded or a degree ordering, so this exercise gives a method for constructing graded orderings.)

(a) 证明 ≥g{ \geq }_{g} 也是一个单项式排序。（关系 ≥g{ \geq }_{g} 被称为 ≥\geq 的分级。在比较中最重要的标准是次数的排序有时被称为分级排序或次数排序，因此这个练习提供了一种构造分级排序的方法。）

(b) The grading of the lexicographic ordering x1>⋯>xn{x}_{1} > \cdots > {x}_{n} is called the grlex monomial ordering. Show that x24>x12x2>x1x22>x22>x1{x}_{2}^{4} > {x}_{1}^{2}{x}_{2} > {x}_{1}{x}_{2}^{2} > {x}_{2}^{2} > {x}_{1} with respect to the grlex ordering and x12x2>x1x22>x1>x24>x22{x}_{1}^{2}{x}_{2} > {x}_{1}{x}_{2}^{2} > {x}_{1} > {x}_{2}^{4} > {x}_{2}^{2} with respect to the lexicographic ordering.

(b) 字典序排序 x1>⋯>xn{x}_{1} > \cdots > {x}_{n} 的分级被称为grlex单项式排序。证明在grlex排序下 x24>x12x2>x1x22>x22>x1{x}_{2}^{4} > {x}_{1}^{2}{x}_{2} > {x}_{1}{x}_{2}^{2} > {x}_{2}^{2} > {x}_{1} ，在字典序下 x12x2>x1x22>x1>x24>x22{x}_{1}^{2}{x}_{2} > {x}_{1}{x}_{2}^{2} > {x}_{1} > {x}_{2}^{4} > {x}_{2}^{2} 。

  1. The grevlex monomial ordering is defined by first choosing an ordering of the variables {x1,x2,…,xn}\left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\} ,then defining m1≥m2{m}_{1} \geq {m}_{2} for monomials m1,m2{m}_{1},{m}_{2} if either deg⁡m1>deg⁡m2\deg {m}_{1} > \deg {m}_{2} or deg⁡m1=deg⁡m2\deg {m}_{1} = \deg {m}_{2} and the first exponent of xn,xn−1,…,x1{x}_{n},{x}_{n - 1},\ldots ,{x}_{1} (in that order) where m1{m}_{1} and m2{m}_{2} differ is smaller in m1{m}_{1} .

  2. grevlex单项式排序通过首先选择变量的排序 {x1,x2,…,xn}\left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\} ，然后定义 m1≥m2{m}_{1} \geq {m}_{2} 对于单项式 m1,m2{m}_{1},{m}_{2} 如果 deg⁡m1>deg⁡m2\deg {m}_{1} > \deg {m}_{2} 或者 deg⁡m1=deg⁡m2\deg {m}_{1} = \deg {m}_{2} 且第一个指数 xn,xn−1,…,x1{x}_{n},{x}_{n - 1},\ldots ,{x}_{1} （按该顺序）在 m1{m}_{1} 和 m2{m}_{2} 不同时在 m1{m}_{1} 中较小。

(a) Prove that grevlex is a monomial ordering that satisfies x1>x2>⋯>xn{x}_{1} > {x}_{2} > \cdots > {x}_{n} .

(a) 证明 grevlex 是一个满足 x1>x2>⋯>xn{x}_{1} > {x}_{2} > \cdots > {x}_{n} 的单项式排序。

(b) Prove that the grevlex ordering on F[x1,x2]F\left\lbrack {{x}_{1},{x}_{2}}\right\rbrack with respect to {x1,x2}\left\{ {{x}_{1},{x}_{2}}\right\} is the graded lexicographic ordering with x1>x2{x}_{1} > {x}_{2} ,but that the grevlex ordering on F[x1,x2,x3]F\left\lbrack {{x}_{1},{x}_{2},{x}_{3}}\right\rbrack is not the grading of any lexicographic ordering.

(b) 证明在 F[x1,x2]F\left\lbrack {{x}_{1},{x}_{2}}\right\rbrack 上关于 {x1,x2}\left\{ {{x}_{1},{x}_{2}}\right\} 的 grevlex 排序是分级的词典序，且与 x1>x2{x}_{1} > {x}_{2} 相关，但在 F[x1,x2,x3]F\left\lbrack {{x}_{1},{x}_{2},{x}_{3}}\right\rbrack 上的 grevlex 排序不是任何词典排序的分级。

(c) Show that x1x22x3>x12x32>x22x32>x2x32>x1x2>x22>x1x3>x32>x1>x2{x}_{1}{x}_{2}^{2}{x}_{3} > {x}_{1}^{2}{x}_{3}^{2} > {x}_{2}^{2}{x}_{3}^{2} > {x}_{2}{x}_{3}^{2} > {x}_{1}{x}_{2} > {x}_{2}^{2} > {x}_{1}{x}_{3} > {x}_{3}^{2} > {x}_{1} > {x}_{2} for the grevlex monomial ordering with respect to {x1,x2,x3}\left\{ {{x}_{1},{x}_{2},{x}_{3}}\right\} .

(c) 显示 x1x22x3>x12x32>x22x32>x2x32>x1x2>x22>x1x3>x32>x1>x2{x}_{1}{x}_{2}^{2}{x}_{3} > {x}_{1}^{2}{x}_{3}^{2} > {x}_{2}^{2}{x}_{3}^{2} > {x}_{2}{x}_{3}^{2} > {x}_{1}{x}_{2} > {x}_{2}^{2} > {x}_{1}{x}_{3} > {x}_{3}^{2} > {x}_{1} > {x}_{2} 对于关于 {x1,x2,x3}\left\{ {{x}_{1},{x}_{2},{x}_{3}}\right\} 的 grevlex 单项式排序成立。

  1. Show that x3y>x3z2>x3z>x2y2z>x2y>xz2>y2z2>y2z{x}^{3}y > {x}^{3}{z}^{2} > {x}^{3}z > {x}^{2}{y}^{2}z > {x}^{2}y > x{z}^{2} > {y}^{2}{z}^{2} > {y}^{2}z with respect to the lexicographic monomial ordering x>y>zx > y > z . Show that for the corresponding grlex monomial ordering x3z2>x2y2z>x3y>x3z>y2z2>x2y>xz2>y2z{x}^{3}{z}^{2} > {x}^{2}{y}^{2}z > {x}^{3}y > {x}^{3}z > {y}^{2}{z}^{2} > {x}^{2}y > x{z}^{2} > {y}^{2}z ,and that x2y2z>x3z2>x3y>x3z>y2z2>x2y>y2z>xz2{x}^{2}{y}^{2}z > {x}^{3}{z}^{2} > {x}^{3}y > {x}^{3}z > {y}^{2}{z}^{2} > {x}^{2}y > {y}^{2}z > x{z}^{2} for the grevlex monomial ordering with respect to {x,y,z}\{ x,y,z\} .

  2. 证明 x3y>x3z2>x3z>x2y2z>x2y>xz2>y2z2>y2z{x}^{3}y > {x}^{3}{z}^{2} > {x}^{3}z > {x}^{2}{y}^{2}z > {x}^{2}y > x{z}^{2} > {y}^{2}{z}^{2} > {y}^{2}z 对于词典序单项式排序 x>y>zx > y > z 成立。证明对于相应的 grlex 单项式排序 x3z2>x2y2z>x3y>x3z>y2z2>x2y>xz2>y2z{x}^{3}{z}^{2} > {x}^{2}{y}^{2}z > {x}^{3}y > {x}^{3}z > {y}^{2}{z}^{2} > {x}^{2}y > x{z}^{2} > {y}^{2}z ，并且 x2y2z>x3z2>x3y>x3z>y2z2>x2y>y2z>xz2{x}^{2}{y}^{2}z > {x}^{3}{z}^{2} > {x}^{3}y > {x}^{3}z > {y}^{2}{z}^{2} > {x}^{2}y > {y}^{2}z > x{z}^{2} 对于关于 {x,y,z}\{ x,y,z\} 的 grevlex 单项式排序成立。

  3. Order the monomials x2z,x2y2z,xy2z,x3y,x3z2,x2,x2yz2,x2z2{x}^{2}z,{x}^{2}{y}^{2}z,x{y}^{2}z,{x}^{3}y,{x}^{3}{z}^{2},{x}^{2},{x}^{2}y{z}^{2},{x}^{2}{z}^{2} for the lexicographic monomial ordering x>y>zx > y > z ,for the corresponding grlex monomial order,and for the grevlex monomial ordering with respect to {x,y,z}\{ x,y,z\} .

  4. 对于词典序单项式排序 x2z,x2y2z,xy2z,x3y,x3z2,x2,x2yz2,x2z2{x}^{2}z,{x}^{2}{y}^{2}z,x{y}^{2}z,{x}^{3}y,{x}^{3}{z}^{2},{x}^{2},{x}^{2}y{z}^{2},{x}^{2}{z}^{2} ，相应的 grlex 单项式排序，以及关于 x>y>zx > y > z 的 grevlex 单项式排序，对单项式 {x,y,z}\{ x,y,z\} 进行排序。

  5. Show there are nn ! distinct lexicographic monomial orderings on F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack . Show similarly that there are nn ! distinct grlex and grevlex monomial orderings.

  6. 证明存在 nn ! 个不同的词典序单项式排序。类似地，证明存在 nn ! 个不同的 grlex 和 grevlex 单项式排序。

  7. It can be shown that any monomial ordering on F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack may be obtained as follows. For k≤nk \leq n let v1,v2,…,vk{v}_{1},{v}_{2},\ldots ,{v}_{k} be nonzero vectors in Euclidean nn -space, Rn{\mathbb{R}}^{n} ,that are pairwise orthogonal: vi⋅vj=0{v}_{i} \cdot {v}_{j} = 0 for all i≠ji \neq j ,where ⋅\cdot is the usual dot product,and suppose also that all the coordinates of v1{v}_{1} are nonnegative. Define an order, ≥\geq ,on monomials by m1>m2{m}_{1} > {m}_{2} if and only if for some t≤kt \leq k we have vi⋅∂(m1)=vi⋅∂(m2){v}_{i} \cdot \partial \left( {m}_{1}\right) = {v}_{i} \cdot \partial \left( {m}_{2}\right) for all i∈{1,2,…,t−1}i \in \{ 1,2,\ldots ,t - 1\} and vt⋅∂(m1)>vt⋅∂(m2){v}_{t} \cdot \partial \left( {m}_{1}\right) > {v}_{t} \cdot \partial \left( {m}_{2}\right) .

  8. 可以证明，任何在 F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack 上的单项式排序都可以按照以下方式得到。对于 k≤nk \leq n ，设 v1,v2,…,vk{v}_{1},{v}_{2},\ldots ,{v}_{k} 是欧几里得 nn -空间中的非零向量，Rn{\mathbb{R}}^{n} ，它们两两正交：vi⋅vj=0{v}_{i} \cdot {v}_{j} = 0 对于所有 i≠ji \neq j ，其中 ⋅\cdot 是通常的点积，并且假设 v1{v}_{1} 的所有坐标都是非负的。通过 ≥\geq 定义单项式上的一个排序，当且仅当对于某些 m1>m2{m}_{1} > {m}_{2} ，我们有 t≤kt \leq k 对于所有 vi⋅∂(m1)=vi⋅∂(m2){v}_{i} \cdot \partial \left( {m}_{1}\right) = {v}_{i} \cdot \partial \left( {m}_{2}\right) 和 i∈{1,2,…,t−1}i \in \{ 1,2,\ldots ,t - 1\} 成立 vt⋅∂(m1)>vt⋅∂(m2){v}_{t} \cdot \partial \left( {m}_{1}\right) > {v}_{t} \cdot \partial \left( {m}_{2}\right) 。

(a) Let k=nk = n and let vi=(0,…,0,1,0,…,0){v}_{i} = \left( {0,\ldots ,0,1,0,\ldots ,0}\right) with 1 in the ith {i}^{\text{th }} position. Show that ≥\geq defines the lexicographic order with x1>x2>⋯>xn{x}_{1} > {x}_{2} > \cdots > {x}_{n} .

(a) 设 k=nk = n 并且设 vi=(0,…,0,1,0,…,0){v}_{i} = \left( {0,\ldots ,0,1,0,\ldots ,0}\right) 在 ith {i}^{\text{th }} 位置上为1。证明 ≥\geq 定义了与 x1>x2>⋯>xn{x}_{1} > {x}_{2} > \cdots > {x}_{n} 的字典序。

(b) Let k=nk = n and define v1=(1,1,…,1){v}_{1} = \left( {1,1,\ldots ,1}\right) and vi=(1,1,…,1,−n+i−1,0,…,0){v}_{i} = \left( {1,1,\ldots ,1, - n + i - 1,0,\ldots ,0}\right) ,

(b) 设 k=nk = n 并定义 v1=(1,1,…,1){v}_{1} = \left( {1,1,\ldots ,1}\right) 和 vi=(1,1,…,1,−n+i−1,0,…,0){v}_{i} = \left( {1,1,\ldots ,1, - n + i - 1,0,\ldots ,0}\right)，

where there are i−2i - 2 trailing zeros, 2≤i≤n2 \leq i \leq n . Show that ≥\geq defines the grlex order with respect to {x1,…,xn}\left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} .

  1. Suppose II is a monomial ideal generated by monomials m1,…,mk{m}_{1},\ldots ,{m}_{k} . Prove that the polynomial f∈F[x1,…,xn]f \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack is in II if and only if every monomial term fi{f}_{i} of ff is a multiple of one of the mj{m}_{j} . [For polynomials a1,…,ak∈F[x1,…,xn]{a}_{1},\ldots ,{a}_{k} \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack expand the polynomial a1m1+⋯+akmk{a}_{1}{m}_{1} + \cdots + {a}_{k}{m}_{k} and note that every monomial term is a multiple of at least one of the mj]\left. {m}_{j}\right\rbrack .] Show that x2yz+3xy2{x}^{2}{yz} + {3x}{y}^{2} is an element of the ideal I=(xyz,y2)⊂F[x,y,z]I = \left( {{xyz},{y}^{2}}\right) \subset F\left\lbrack {x,y,z}\right\rbrack but is not an element of the ideal I′=(xz2,y2){I}^{\prime } = \left( {x{z}^{2},{y}^{2}}\right) .

  2. Fix a monomial ordering on R=F[x1,…,xn]R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack and suppose {g1,…,gm}\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} is a Gröbner basis for the ideal II in RR . Prove that h∈LT(I)h \in {LT}\left( I\right) if and only if hh is a sum of monomial terms each divisible by some LT(gi),1≤i≤m{LT}\left( {g}_{i}\right) ,1 \leq i \leq m . [Use the previous exercise.]

  3. Suppose II is a monomial ideal with monomial generators g1,…,gm{g}_{1},\ldots ,{g}_{m} . Use the previous exercise to prove directly that {g1,…,gm}\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} is a Gröbner basis for II .

  4. Suppose II is a monomial ideal with monomial generators g1,…,gm{g}_{1},\ldots ,{g}_{m} . Use Buchberger’s Criterion to prove that {g1,…,gm}\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} is a Gröbner basis for II .

  5. Suppose II is a monomial ideal in R=F[x1,…,xn]R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack and suppose {m1,…,mk}\left\{ {{m}_{1},\ldots ,{m}_{k}}\right\} is a minimal set of monomials generating II ,i.e.,each mi{m}_{i} is a monomial and no proper subset of {m1,…,mk}\left\{ {{m}_{1},\ldots ,{m}_{k}}\right\} generates II . Prove that the mi,1≤i≤k{m}_{i},1 \leq i \leq k are unique. [Use Exercise 10.]

  6. Fix a monomial ordering on R=F[x1,…,xn]R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack .

(a) Prove that {g1,…,gm}\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} is a minimal Gröbner basis for the ideal II in RR if and only if {LT(g1),…,LT(gm)}\left\{ {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right\} is a minimal generating set for LT(I){LT}\left( I\right) .

(b) Prove that the leading terms of a minimal Gröbner basis for II are uniquely determined and the number of elements in any two minimal Gröbner bases for II is the same. [Use (a) and the previous exercise.]

  1. Fix a monomial ordering on F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack and suppose G={g1,…,gm}G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} is a set of generators for the nonzero ideal II . Show that if S(gi,gj)≢0  mod⁡  GS\left( {{g}_{i},{g}_{j}}\right) ≢ 0{\;\operatorname{mod}\;G} then the ideal (LT(g1),...,LT(gm),LT(S(gi,gj)) is strictly larger than the ideal (LT(g1),...,LT(gm)).\left( {{LT}\left( {g}_{1}\right) ,...,{LT}\left( {g}_{m}\right) ,{LT}\left( {S\left( {{g}_{i},{g}_{j}}\right) }\right) \text{ is strictly larger than the ideal }\left( {{LT}\left( {g}_{1}\right) ,...,{LT}\left( {g}_{m}\right) }\right) .}\right. Conclude that the algorithm for computing a Gröbner basis described following Proposition 26 terminates after a finite number of steps. [Use Exercise 1.]

  2. Fix the lexicographic ordering x>yx > y on F[x,y]F\left\lbrack {x,y}\right\rbrack . Use Buchberger’s Criterion to show that {x2y−y2,x3−xy}\left\{ {{x}^{2}y - {y}^{2},{x}^{3} - {xy}}\right\} is a Gröbner basis for the ideal I=(x2y−y2,x3−xy)I = \left( {{x}^{2}y - {y}^{2},{x}^{3} - {xy}}\right) .

  3. Show {x−y3,y5−y6}\left\{ {x - {y}^{3},{y}^{5} - {y}^{6}}\right\} is the reduced Gröbner basis for the ideal I=(x−y3,−x2+xy2)I = \left( {x - {y}^{3}, - {x}^{2} + x{y}^{2}}\right) with respect to the lexicographic ordering defined by x>yx > y in F[x,y]F\left\lbrack {x,y}\right\rbrack .

  4. Fix the lexicographic ordering x>yx > y on F[x,y]F\left\lbrack {x,y}\right\rbrack .

(a) Show that {x3−y,x2y−y2,xy2−y2,y3−y2}\left\{ {{x}^{3} - y,{x}^{2}y - {y}^{2},x{y}^{2} - {y}^{2},{y}^{3} - {y}^{2}}\right\} is the reduced Gröbner basis for the ideal   I=(−x3+y,x2y−y2).\;I = \left( {-{x}^{3} + y,{x}^{2}y - {y}^{2}}\right) .

(b) Determine whether the polynomial f=x6−x5yf = {x}^{6} - {x}^{5}y is an element of the ideal II .

  1. Fix the lexicographic ordering x>y>zx > y > z on F[x,y,z]F\left\lbrack {x,y,z}\right\rbrack . Show that {x2+xy+z,xyz+\left\{ {{x}^{2} + {xy} + z,{xyz} + }\right. z2,xz2,z3}\left. {{z}^{2},x{z}^{2},{z}^{3}}\right\} is the reduced Gröbner basis for the ideal I=(x2+xy+z,xyz+z2)I = \left( {{x}^{2} + {xy} + z,{xyz} + {z}^{2}}\right) and in particular conclude that the leading term ideal LT(I){LT}\left( I\right) requires four generators.

  2. Fix the lexicographic ordering x>yx > y on F[x,y]F\left\lbrack {x,y}\right\rbrack . Use Buchberger’s Criterion to show that {x2y−y2,x3−xy}\left\{ {{x}^{2}y - {y}^{2},{x}^{3} - {xy}}\right\} is a Gröbner basis for the ideal I=(x2y−y2,x3−xy)I = \left( {{x}^{2}y - {y}^{2},{x}^{3} - {xy}}\right) .

  3. Let I=(x2−y,x2y−z)I = \left( {{x}^{2} - y,{x}^{2}y - z}\right) in F[x,y,z]F\left\lbrack {x,y,z}\right\rbrack .

(a) Show that {x2−y,y2−z}\left\{ {{x}^{2} - y,{y}^{2} - z}\right\} is the reduced Gröbner basis for II with respect to the lexicographic ordering defined by x>y>zx > y > z .

(b) Show that {x2−y,z−y2}\left\{ {{x}^{2} - y,z - {y}^{2}}\right\} is the reduced Gröbner basis for II with respect to the

lexicographic ordering defined by z>x>yz > x > y (note these are essentially the same polynomials as in (a)).

由 z>x>yz > x > y 定义的字典序（注意这些本质上与 (a) 中的多项式相同）。

(c) Show that {y−x2,z−x4}\left\{ {y - {x}^{2},z - {x}^{4}}\right\} is the reduced Gröbner basis for II with respect to the lexicographic ordering defined by z>y>xz > y > x .

(c) 证明 {y−x2,z−x4}\left\{ {y - {x}^{2},z - {x}^{4}}\right\} 是关于 II 的简化Gröbner基，相对于由 z>y>xz > y > x 定义的字典序。

  1. Show that the ideals I=(x2y+xy2−2y,x2+xy−x+y2−2y,xy2−x−y+y3)I = \left( {{x}^{2}y + x{y}^{2} - {2y},{x}^{2} + {xy} - x + {y}^{2} - {2y},x{y}^{2} - x - y + {y}^{3}}\right) and J=(x−y2,xy−y,x2−y)J = \left( {x - {y}^{2},{xy} - y,{x}^{2} - y}\right) in F[x,y]F\left\lbrack {x,y}\right\rbrack are equal.

  2. 证明在 F[x,y]F\left\lbrack {x,y}\right\rbrack 中的理想 I=(x2y+xy2−2y,x2+xy−x+y2−2y,xy2−x−y+y3)I = \left( {{x}^{2}y + x{y}^{2} - {2y},{x}^{2} + {xy} - x + {y}^{2} - {2y},x{y}^{2} - x - y + {y}^{3}}\right) 和 J=(x−y2,xy−y,x2−y)J = \left( {x - {y}^{2},{xy} - y,{x}^{2} - y}\right) 是相等的。

  3. Use reduced Gröbner bases to show that the ideal I=(x3−yz,yz+y)I = \left( {{x}^{3} - {yz},{yz} + y}\right) and the ideal J=(x3z+x3,x3+y)J = \left( {{x}^{3}z + {x}^{3},{x}^{3} + y}\right) in F[x,y,z]F\left\lbrack {x,y,z}\right\rbrack are equal.

  4. 使用简化Gröbner基证明在 F[x,y,z]F\left\lbrack {x,y,z}\right\rbrack 中的理想 I=(x3−yz,yz+y)I = \left( {{x}^{3} - {yz},{yz} + y}\right) 和理想 J=(x3z+x3,x3+y)J = \left( {{x}^{3}z + {x}^{3},{x}^{3} + y}\right) 是相等的。

  5. Show that the reduced Gröbner basis using the lexicographic ordering x>yx > y for the ideal I=(x2+xy2,x2−y3,y3−y2)I = \left( {{x}^{2} + x{y}^{2},{x}^{2} - {y}^{3},{y}^{3} - {y}^{2}}\right) is {x2−y2,y3−y2‾,xy2+y2}.\{ {x}^{2} - {y}^{2},{y}^{3} - \overline{{y}^{2}},x{y}^{2} + {y}^{2}\} .

  6. 证明使用字典序 x>yx > y 对理想 I=(x2+xy2,x2−y3,y3−y2)I = \left( {{x}^{2} + x{y}^{2},{x}^{2} - {y}^{3},{y}^{3} - {y}^{2}}\right) 进行简化Gröbner基是 {x2−y2,y3−y2‾,xy2+y2}.\{ {x}^{2} - {y}^{2},{y}^{3} - \overline{{y}^{2}},x{y}^{2} + {y}^{2}\} .。

  7. Show that the reduced Gröbner basis for the ideal I=(xy+y2,x2y+xy2+x2)I = \left( {{xy} + {y}^{2},{x}^{2}y + x{y}^{2} + {x}^{2}}\right) is {x2,xy+y2,y3}\left\{ {{x}^{2},{xy} + {y}^{2},{y}^{3}}\right\} with respect to the lexicographic ordering x>yx > y and is {y2+yx,x2}\left\{ {{y}^{2} + {yx},{x}^{2}}\right\} with respect to the lexicographic ordering y>xy > x .

  8. 证明对于理想 I=(xy+y2,x2y+xy2+x2)I = \left( {{xy} + {y}^{2},{x}^{2}y + x{y}^{2} + {x}^{2}}\right) 的简化Gröbner基相对于字典序 x>yx > y 是 {x2,xy+y2,y3}\left\{ {{x}^{2},{xy} + {y}^{2},{y}^{3}}\right\}，并且相对于字典序 y>xy > x 是 {y2+yx,x2}\left\{ {{y}^{2} + {yx},{x}^{2}}\right\}。

There are generally substantial differences in computational complexity when using different monomial orders. The grevlex monomial ordering often provides the most efficient computation and produces simpler polynomials.

使用不同的单项式顺序时，计算复杂性通常存在较大差异。grevlex单项式顺序通常提供最有效的计算，并产生更简单的多项式。

  1. Show that {x3−y3,x2+xy2+y4,x2y+xy3+y2}\left\{ {{x}^{3} - {y}^{3},{x}^{2} + x{y}^{2} + {y}^{4},{x}^{2}y + x{y}^{3} + {y}^{2}}\right\} is a reduced Gröbner basis for the ideal II in the example following Corollary 28 with respect to the grlex monomial ordering. (Note that while this gives three generators for II rather than two for the lexicographic ordering as in the example, the degrees are smaller.)

  2. 证明 {x3−y3,x2+xy2+y4,x2y+xy3+y2}\left\{ {{x}^{3} - {y}^{3},{x}^{2} + x{y}^{2} + {y}^{4},{x}^{2}y + x{y}^{3} + {y}^{2}}\right\} 是相对于grlex单项式顺序在例28之后的例子中对理想 II 的简化Gröbner基。（注意，虽然这为 II 提供了三个生成元，而不是字典序中的两个，但它们的次数较小。）

  3. Let I=(x4−y4+z3−1,x3+y2+z2−1)I = \left( {{x}^{4} - {y}^{4} + {z}^{3} - 1,{x}^{3} + {y}^{2} + {z}^{2} - 1}\right) . Show that there are five elements in a reduced Gröbner basis for II with respect to the lexicographic ordering with x>y>zx > y > z (the maximum degree among the five generators is 12 and the maximum number of monomial terms among the five generators is 35 ), that there are two elements for the lexicographic ordering y>z>xy > z > x (maximum degree is 6 and maximum number of terms is 8 ),and that {x3+y2+z2−1,xy2+xz2−x+y4−z3+1}\left\{ {{x}^{3} + {y}^{2} + {z}^{2} - 1,x{y}^{2} + x{z}^{2} - x + {y}^{4} - {z}^{3} + 1}\right\} is the reduced Gröbner basis for the grevlex monomial ordering.

  4. 设 I=(x4−y4+z3−1,x3+y2+z2−1)I = \left( {{x}^{4} - {y}^{4} + {z}^{3} - 1,{x}^{3} + {y}^{2} + {z}^{2} - 1}\right) 。证明在字典序下，关于 x>y>zx > y > z 的一个化简Gröbner基有五个元素（这五个生成元中的最大次数为12，且最多单项式项数为35），而在字典序 y>z>xy > z > x 下有两个元素（最大次数为6，最多项数为8），并且 {x3+y2+z2−1,xy2+xz2−x+y4−z3+1}\left\{ {{x}^{3} + {y}^{2} + {z}^{2} - 1,x{y}^{2} + x{z}^{2} - x + {y}^{4} - {z}^{3} + 1}\right\} 是关于grevlex单项式序的化简Gröbner基。

  5. Solve the system of equations x2−yz=3,y2−xz=4,z2−xy=5{x}^{2} - {yz} = 3,{y}^{2} - {xz} = 4,{z}^{2} - {xy} = 5 over C\mathbb{C} .

  6. 解方程组 x2−yz=3,y2−xz=4,z2−xy=5{x}^{2} - {yz} = 3,{y}^{2} - {xz} = 4,{z}^{2} - {xy} = 5 关于 C\mathbb{C} 。

  7. Find a Gröbner basis for the ideal I=(x2+xy+y2−1,x2+4y2−4)I = \left( {{x}^{2} + {xy} + {y}^{2} - 1,{x}^{2} + 4{y}^{2} - 4}\right) for the lexicographic ordering   x  >  y  \;x\; > \;y\; and use it to find the four points of intersection of the ellipse   x2  +  xy  +  y2=1\;{x}^{2}\; + \;{xy}\; + \;{y}^{2} = 1 with the ellipse x2+4y2=4{x}^{2} + 4{y}^{2} = 4 in R2{\mathbb{R}}^{2} .

  8. 对于理想 I=(x2+xy+y2−1,x2+4y2−4)I = \left( {{x}^{2} + {xy} + {y}^{2} - 1,{x}^{2} + 4{y}^{2} - 4}\right) 求其在字典序   x  >  y  \;x\; > \;y\; 下的Gröbner基，并利用它找到椭圆   x2  +  xy  +  y2=1\;{x}^{2}\; + \;{xy}\; + \;{y}^{2} = 1 与椭圆 x2+4y2=4{x}^{2} + 4{y}^{2} = 4 在 R2{\mathbb{R}}^{2} 中的四个交点。

  9. Use Gröbner bases to find all six solutions to the system of equations 2x3+2x2y2+3y3=02{x}^{3} + 2{x}^{2}{y}^{2} + 3{y}^{3} = 0 and 3x5+2x3y3+2y5=03{x}^{5} + 2{x}^{3}{y}^{3} + 2{y}^{5} = 0 over C\mathbb{C} .

  10. 使用Gröbner基找到方程组 2x3+2x2y2+3y3=02{x}^{3} + 2{x}^{2}{y}^{2} + 3{y}^{3} = 0 和 3x5+2x3y3+2y5=03{x}^{5} + 2{x}^{3}{y}^{3} + 2{y}^{5} = 0 关于 C\mathbb{C} 的所有六个解。

  11. Use Gröbner bases to show that (x,z)∩(y2,x−yz)=(xy,x−yz)\left( {x,z}\right) \cap \left( {{y}^{2},x - {yz}}\right) = \left( {{xy},x - {yz}}\right) in F[x,y,z]F\left\lbrack {x,y,z}\right\rbrack .

  12. 使用Gröbner基证明 (x,z)∩(y2,x−yz)=(xy,x−yz)\left( {x,z}\right) \cap \left( {{y}^{2},x - {yz}}\right) = \left( {{xy},x - {yz}}\right) 在 F[x,y,z]F\left\lbrack {x,y,z}\right\rbrack 中。

  13. Use Gröbner bases to compute the intersection of the ideals (x3y−xy2+1,x2y2−y3−1)\left( {{x}^{3}y - x{y}^{2} + 1,{x}^{2}{y}^{2} - {y}^{3} - 1}\right) and (x2−y2,x3+y3)\left( {{x}^{2} - {y}^{2},{x}^{3} + {y}^{3}}\right) in F[x,y]F\left\lbrack {x,y}\right\rbrack .

  14. 使用Gröbner基计算理想 (x3y−xy2+1,x2y2−y3−1)\left( {{x}^{3}y - x{y}^{2} + 1,{x}^{2}{y}^{2} - {y}^{3} - 1}\right) 和 (x2−y2,x3+y3)\left( {{x}^{2} - {y}^{2},{x}^{3} + {y}^{3}}\right) 在 F[x,y]F\left\lbrack {x,y}\right\rbrack 中的交集。

The following four exercises deal with the ideal quotient of two ideals II and JJ in a ring RR . Definition. The ideal quotient (I:J)\left( {I : J}\right) of two ideals I,JI,J in a ring RR is the ideal

以下四个练习涉及环 RR 中两个理想 II 和 JJ 的理想商。定义。环 RR 中两个理想 I,JI,J 的理想商 (I:J)\left( {I : J}\right) 是理想

(I:J)={r∈R∣rJ∈I}.\left( {I : J}\right) = \{ r \in R \mid {rJ} \in I\} .

  1. (a) Suppose RR is an integral domain, 0≠f∈R0 \neq f \in R and II is an ideal in RR . Show that if {g1,…,gs}\left\{ {{g}_{1},\ldots ,{g}_{s}}\right\} are generators for the ideal I∩(f)I \cap \left( f\right) ,then {g1/f,…,gs/f}\left\{ {{g}_{1}/f,\ldots ,{g}_{s}/f}\right\} are generators for the ideal quotient (I:(f))\left( {I : \left( f\right) }\right) .

  2. (a) 假设 RR 是一个整环，0≠f∈R0 \neq f \in R 并且 II 是 RR 中的一个理想。证明如果 {g1,…,gs}\left\{ {{g}_{1},\ldots ,{g}_{s}}\right\} 是理想 I∩(f)I \cap \left( f\right) 的生成元，那么 {g1/f,…,gs/f}\left\{ {{g}_{1}/f,\ldots ,{g}_{s}/f}\right\} 是理想商 (I:(f))\left( {I : \left( f\right) }\right) 的生成元。

(b) If II is an ideal in the commutative ring RR and f1,…,fs∈R{f}_{1},\ldots ,{f}_{s} \in R ,show that the ideal quotient (I:(f1,…fs))\left( {I : \left( {{f}_{1},\ldots {f}_{s}}\right) }\right) is the ideal ∩i=1s(I:(fi)){ \cap }_{i = 1}^{s}\left( {I : \left( {f}_{i}\right) }\right) .

(b) 如果 II 是交换环 RR 中的理想且 f1,…,fs∈R{f}_{1},\ldots ,{f}_{s} \in R ，证明理想商 (I:(f1,…fs))\left( {I : \left( {{f}_{1},\ldots {f}_{s}}\right) }\right) 是理想 ∩i=1s(I:(fi)){ \cap }_{i = 1}^{s}\left( {I : \left( {f}_{i}\right) }\right) 。

  1. If I=(x2y+z3,x+y3−z,2y4z−yz2−z3)I = \left( {{x}^{2}y + {z}^{3},x + {y}^{3} - z,2{y}^{4}z - y{z}^{2} - {z}^{3}}\right) and J=(x2y5,x3z4,y3z7)J = \left( {{x}^{2}{y}^{5},{x}^{3}{z}^{4},{y}^{3}{z}^{7}}\right) in Q[x,y,z]\mathbb{Q}\left\lbrack {x,y,z}\right\rbrack show (I:J)\left( {I : J}\right) is the ideal (z2,y+z,x−z)\left( {{z}^{2},y + z,x - z}\right) . [Use the previous exercise and Proposition 30.]

  2. 如果 I=(x2y+z3,x+y3−z,2y4z−yz2−z3)I = \left( {{x}^{2}y + {z}^{3},x + {y}^{3} - z,2{y}^{4}z - y{z}^{2} - {z}^{3}}\right) 和 J=(x2y5,x3z4,y3z7)J = \left( {{x}^{2}{y}^{5},{x}^{3}{z}^{4},{y}^{3}{z}^{7}}\right) 在 Q[x,y,z]\mathbb{Q}\left\lbrack {x,y,z}\right\rbrack 中，证明 (I:J)\left( {I : J}\right) 是理想 (z2,y+z,x−z)\left( {{z}^{2},y + z,x - z}\right) 。[使用前一个练习和命题30。]

  3. Suppose that KK is an ideal in RR ,that II is an ideal containing KK ,and JJ is any ideal. If Iˉ\bar{I} and Jˉ\bar{J} denote the images of II and JJ in the quotient ring R/KR/K ,show that (I:J)‾=(Iˉ:Jˉ)\overline{\left( I : J\right) } = \left( {\bar{I} : \bar{J}}\right) where (I:J)‾\overline{\left( I : J\right) } is the image in R/KR/K of the ideal quotient (I:J)\left( {I : J}\right) .

  4. 假设 KK 是 RR 中的理想，II 是包含 KK 的理想，JJ 是任意理想。如果 Iˉ\bar{I} 和 Jˉ\bar{J} 分别表示 II 和 JJ 在商环 R/KR/K 中的像，证明 (I:J)‾=(Iˉ:Jˉ)\overline{\left( I : J\right) } = \left( {\bar{I} : \bar{J}}\right) ，其中 (I:J)‾\overline{\left( I : J\right) } 是理想商 (I:J)\left( {I : J}\right) 在 R/KR/K 中的像。

  5. Let KK be the ideal (y5−z4)\left( {{y}^{5} - {z}^{4}}\right) in R=Q[y,z]R = \mathbb{Q}\left\lbrack {y,z}\right\rbrack . For each of the following pairs of ideals II and JJ ,use the previous two exercises together with Proposition 30 to verify the ideal quotients (Iˉ:Jˉ)\left( {\bar{I} : \bar{J}}\right) in the ring R/KR/K :

  6. 设 KK 是 R=Q[y,z]R = \mathbb{Q}\left\lbrack {y,z}\right\rbrack 中的理想 (y5−z4)\left( {{y}^{5} - {z}^{4}}\right) 。对于以下每一对理想 II 和 JJ ，使用前两个练习以及命题30来验证环 R/KR/K 中的理想商 (Iˉ:Jˉ)\left( {\bar{I} : \bar{J}}\right) ：

i. I=(y3,y5−z4),J=(z),(Iˉ:Jˉ)=(yˉ3,zˉ3)I = \left( {{y}^{3},{y}^{5} - {z}^{4}}\right) ,J = \left( z\right) ,\left( {\bar{I} : \bar{J}}\right) = \left( {{\bar{y}}^{3},{\bar{z}}^{3}}\right) .

i. I=(y3,y5−z4),J=(z),(Iˉ:Jˉ)=(yˉ3,zˉ3)I = \left( {{y}^{3},{y}^{5} - {z}^{4}}\right) ,J = \left( z\right) ,\left( {\bar{I} : \bar{J}}\right) = \left( {{\bar{y}}^{3},{\bar{z}}^{3}}\right) 。

ii. I=(y3,z,y5−z4),J=(y),(Iˉ:Jˉ)=(yˉ2,zˉ)I = \left( {{y}^{3},z,{y}^{5} - {z}^{4}}\right) ,J = \left( y\right) ,\left( {\bar{I} : \bar{J}}\right) = \left( {{\bar{y}}^{2},\bar{z}}\right) .

ii. I=(y3,z,y5−z4),J=(y),(Iˉ:Jˉ)=(yˉ2,zˉ)I = \left( {{y}^{3},z,{y}^{5} - {z}^{4}}\right) ,J = \left( y\right) ,\left( {\bar{I} : \bar{J}}\right) = \left( {{\bar{y}}^{2},\bar{z}}\right) 。

iii. I=(y,y3,z,y5−z4),J=(1),(Iˉ:Jˉ)=(yˉ,zˉ)I = \left( {y,{y}^{3},z,{y}^{5} - {z}^{4}}\right) ,J = \left( 1\right) ,\left( {\bar{I} : \bar{J}}\right) = \left( {\bar{y},\bar{z}}\right) .

iii. I=(y,y3,z,y5−z4),J=(1),(Iˉ:Jˉ)=(yˉ,zˉ)I = \left( {y,{y}^{3},z,{y}^{5} - {z}^{4}}\right) ,J = \left( 1\right) ,\left( {\bar{I} : \bar{J}}\right) = \left( {\bar{y},\bar{z}}\right) 。

Exercises 38 to 44 develop some additional elementary properties of monomial ideals in F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack . It follows from Hilbert’s Basis Theorem that ideals are finitely generated, however one need not assume this in these exercises - the arguments are the same for finitely or infinitely generated ideals. These exercises may be used to give an independent proof of Hilbert’s Basis Theorem (Exercise 44). In these exercises, MM and NN are monomial ideals with monomial generators {mi∣i∈I}\left\{ {{m}_{i} \mid i \in I}\right\} and {nj∣j∈J}\left\{ {{n}_{j} \mid j \in J}\right\} for some index sets II and JJ respectively.

练习38至44研究了 F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack 中单项式理想的一些额外基本性质。根据希尔伯特基定理，理想是有限生成的，但在这些练习中无需假设这一点 - 对于有限生成或无限生成的理想，论证都是相同的。这些练习可以用来给出希尔伯特基定理的独立证明（练习44）。在这些练习中，MM 和 NN 是单项式理想，其单项式生成元分别为 {mi∣i∈I}\left\{ {{m}_{i} \mid i \in I}\right\} 和 {nj∣j∈J}\left\{ {{n}_{j} \mid j \in J}\right\} ，对应于某些指标集 II 和 JJ 。

  1. Prove that the sum and product of two monomial ideals is a monomial ideal by showing that M+N=(mi,nj∣i∈I,j∈J)M + N = \left( {{m}_{i},{n}_{j} \mid i \in I,j \in J}\right) ,and MN=(minj∣i∈I,j∈J){MN} = \left( {{m}_{i}{n}_{j} \mid i \in I,j \in J}\right) .

  2. 证明两个单项式理想的和与积是单项式理想，通过展示 M+N=(mi,nj∣i∈I,j∈J)M + N = \left( {{m}_{i},{n}_{j} \mid i \in I,j \in J}\right) ，以及 MN=(minj∣i∈I,j∈J){MN} = \left( {{m}_{i}{n}_{j} \mid i \in I,j \in J}\right) 。

  3. Show that if {Ms∣s∈S}\left\{ {{M}_{s} \mid s \in S}\right\} is any nonempty collection of monomial ideals that is totally ordered under inclusion then ∪s∈SMs{ \cup }_{s \in S}{M}_{s} is a monomial ideal. (In particular,the union of any increasing sequence of monomial ideals is a monomial ideal, cf. Exercise 19, Section 7.3.)

  4. 证明如果 {Ms∣s∈S}\left\{ {{M}_{s} \mid s \in S}\right\} 是任何非空的单项式理想集合，并且在包含关系下完全有序，那么 ∪s∈SMs{ \cup }_{s \in S}{M}_{s} 是单项式理想。（特别地，任何单调递增的单项式理想序列的并是单项式理想，参见练习19，第7.3节。）

  5. Prove that the intersection of two monomial ideals is a monomial ideal by showing that M∩N=(ei,j∣i∈I,j∈J)M \cap N = \left( {{e}_{i,j} \mid i \in I,j \in J}\right) ,where ei,j{e}_{i,j} is the least common multiple of mi{m}_{i} and nj{n}_{j} . [Use Exercise 10.]

  6. 证明两个单项式理想的交集是单项式理想，通过展示 M∩N=(ei,j∣i∈I,j∈J)M \cap N = \left( {{e}_{i,j} \mid i \in I,j \in J}\right) ，其中 ei,j{e}_{i,j} 是 mi{m}_{i} 和 nj{n}_{j} 的最小公倍数。[使用练习10。]

  7. Prove that for any monomial nn ,the ideal quotient (M:(n))\left( {M : \left( n\right) }\right) is (mi/di∣i∈I)\left( {{m}_{i}/{d}_{i} \mid i \in I}\right) ,where di{d}_{i} is the greatest common divisor of mi{m}_{i} and nn (cf. Exercise 34). Show that if NN is finitely generated,then the ideal quotient (M:N)\left( {M : N}\right) of two monomial ideals is a monomial ideal.

  8. 证明对于任何单项式 nn ，理想商 (M:(n))\left( {M : \left( n\right) }\right) 是 (mi/di∣i∈I)\left( {{m}_{i}/{d}_{i} \mid i \in I}\right) ，其中 di{d}_{i} 是 mi{m}_{i} 和 nn 的最大公约数（参见练习34）。证明如果 NN 是有限生成的，那么两个单项式理想的理想商 (M:N)\left( {M : N}\right) 是单项式理想。

  9. (a) Show that MM is a monomial prime ideal if and only if M=(S)M = \left( S\right) for some subset of SS of {x1,x2,…,xn}\left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\} . (In particular,there are only finitely many monomial prime ideals, and each is finitely generated.)

  10. (a) 证明 MM 是单项式素理想当且仅当 M=(S)M = \left( S\right) 对于某个 SS 的子集 {x1,x2,…,xn}\left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\} 成立。（特别地，只有有限多个单项式素理想，每个都是有限生成的。）

(b) Show that (x1,…,xn)\left( {{x}_{1},\ldots ,{x}_{n}}\right) is the only monomial maximal ideal.

(b) 证明 (x1,…,xn)\left( {{x}_{1},\ldots ,{x}_{n}}\right) 是唯一的单项式极大理想。

  1. (Dickson's Lemma - a special case of Hilbert's Basis Theorem) Prove that every monomial ideal in F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack is finitely generated as follows.

  2. （迪克森引理 - 希尔伯特基定理的特殊情况）证明 F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack 中的每个单项式理想都是有限生成的，如下所示。

Let S={N∣N\mathcal{S} = \{ N \mid N is a monomial ideal that is not finitely generated }\} ,and assume by way of contradiction S≠∅\mathcal{S} \neq \varnothing .

设 S={N∣N\mathcal{S} = \{ N \mid N 是一个不是有限生成的单项式理想 }\} ，并且假设反证 S≠∅\mathcal{S} \neq \varnothing 。

(a) Show that S\mathcal{S} contains a maximal element MM . [Use Exercise 30 and Zorn’s Lemma.]

(a) 证明 S\mathcal{S} 包含一个极大元素 MM 。[使用练习30和Zorn引理。]

(b) Show that there are monomials x,yx,y not in MM with xy∈M{xy} \in M . [Use Exercise 33(a).]

(b) 证明存在单项式 x,yx,y 不在 MM 中，且满足 xy∈M{xy} \in M 。[使用练习33(a)。]

(c) For xx as in (b),show that MM contains a finitely generated monomial ideal M0{M}_{0} such that M0+(x)=M+(x){M}_{0} + \left( x\right) = M + \left( x\right) and M=M0+(x)(M:(x))M = {M}_{0} + \left( x\right) \left( {M : \left( x\right) }\right) ,where (M:(x))\left( {M : \left( x\right) }\right) is the (monomial) ideal defined in Exercise 32,and (x)(M:(x))\left( x\right) \left( {M : \left( x\right) }\right) is the product of these two ideals. Deduce that MM is finitely generated,a contradiction which proves S=∅.\mathcal{S} = \varnothing . [Use the maximality of MM and previous exercises.]

(c) 对于 (b) 中的 xx，证明 MM 包含一个有限生成的单项式理想 M0{M}_{0}，使得 M0+(x)=M+(x){M}_{0} + \left( x\right) = M + \left( x\right) 和 M=M0+(x)(M:(x))M = {M}_{0} + \left( x\right) \left( {M : \left( x\right) }\right)，其中 (M:(x))\left( {M : \left( x\right) }\right) 是练习32中定义的（单项式）理想，(x)(M:(x))\left( x\right) \left( {M : \left( x\right) }\right) 是这两个理想的乘积。推导出 MM 是有限生成的，这与 MM 的极大性以及之前的练习相矛盾，从而证明了 S=∅.\mathcal{S} = \varnothing . [使用 MM 的极大性和之前的练习。]

  1. If II is a nonzero ideal in F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack ,use Dickson’s Lemma to prove that LT(I){LT}\left( I\right) is finitely generated. Conclude that II has a Gröbner basis and deduce Hilbert’s Basis Theorem. [cf. Proposition 24.]

  2. 如果 II 是 F[x1,…,xn]F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack 中的一个非零理想，使用Dickson引理证明 LT(I){LT}\left( I\right) 是有限生成的。得出 II 有一个Gröbner基，并推导出希尔伯特基定理。[参见命题24。]

  3. (n-colorings of graphs) A finite graph G\mathcal{G} of size NN is a set of vertices i∈{1,2,…,N}i \in \{ 1,2,\ldots ,N\} and a collection of edges(i,j)connecting vertex ii with vertex jj . An nn -coloring of G\mathcal{G} is an assignment of one of nn colors to each vertex in such a way that vertices connected by an edge have distinct colors. Let FF be any field containing at least nn elements. If we introduce a variable xi{x}_{i} for each vertex ii and represent the nn colors by choosing a set SS of nn distinct elements from FF ,then an nn -coloring of G\mathcal{G} is equivalent to assigning a value xi=αi{x}_{i} = {\alpha }_{i} for each i=1,2,…,Ni = 1,2,\ldots ,N where αi∈S{\alpha }_{i} \in S and αi≠αj{\alpha }_{i} \neq {\alpha }_{j} if(i,j)is an edge in G\mathcal{G} . If f(x)=∏α∈S(x−α)f\left( x\right) = \mathop{\prod }\limits_{{\alpha \in S}}\left( {x - \alpha }\right) is the polynomial in F[x]F\left\lbrack x\right\rbrack of degree nn whose roots are the elements in SS ,then xi=αi{x}_{i} = {\alpha }_{i} for some αi∈S{\alpha }_{i} \in S is equivalent to the statement that xi{x}_{i} is a solution to the equation f(xi)=0f\left( {x}_{i}\right) = 0 . The statement αi≠αj{\alpha }_{i} \neq {\alpha }_{j} is then the statement that f(xi)=f(xj)f\left( {x}_{i}\right) = f\left( {x}_{j}\right) but xi≠xj{x}_{i} \neq {x}_{j} ,so xi{x}_{i} and xj{x}_{j} satisfy the equation g(xi,xj)=0g\left( {{x}_{i},{x}_{j}}\right) = 0 ,where g(xi,xj)g\left( {{x}_{i},{x}_{j}}\right) is the polynomial (f(xi)−f(xj))/(xi−xj)\left( {f\left( {x}_{i}\right) - f\left( {x}_{j}\right) }\right) /\left( {{x}_{i} - {x}_{j}}\right) in F[xi,xj]F\left\lbrack {{x}_{i},{x}_{j}}\right\rbrack . It follows that finding an nn -coloring of G\mathcal{G} is equivalent to solving the system of equations

  4. （图的 n-着色）一个有限图 G\mathcal{G} 的大小为 NN ，它是一组顶点 i∈{1,2,…,N}i \in \{ 1,2,\ldots ,N\} 和一个连接顶点 ii 与顶点 jj 的边(i,j)的集合。一个 nn -着色是对 G\mathcal{G} 中的每个顶点分配一种 nn 种颜色之一，使得通过边连接的顶点具有不同的颜色。设 FF 为包含至少 nn 个元素的任意域。如果我们为每个顶点 ii 引入一个变量 xi{x}_{i} ，并通过从 FF 中选择一组 nn 个不同的元素作为 SS 来表示 nn 种颜色，那么 G\mathcal{G} 的一个 nn -着色等价于为每个 i=1,2,…,Ni = 1,2,\ldots ,N 分配一个值 xi=αi{x}_{i} = {\alpha }_{i} ，其中 αi∈S{\alpha }_{i} \in S 且如果(i,j)是 G\mathcal{G} 中的一条边。如果 f(x)=∏α∈S(x−α)f\left( x\right) = \mathop{\prod }\limits_{{\alpha \in S}}\left( {x - \alpha }\right) 是 F[x]F\left\lbrack x\right\rbrack 中的一个多项式，其次数为 nn ，其根是 SS 中的元素，那么对于某个 αi∈S{\alpha }_{i} \in S ，xi=αi{x}_{i} = {\alpha }_{i} 等价于 xi{x}_{i} 是方程 f(xi)=0f\left( {x}_{i}\right) = 0 的解。那么 αi≠αj{\alpha }_{i} \neq {\alpha }_{j} 的陈述是 f(xi)=f(xj)f\left( {x}_{i}\right) = f\left( {x}_{j}\right) 但 xi≠xj{x}_{i} \neq {x}_{j} ，因此 xi{x}_{i} 和 xj{x}_{j} 满足方程 g(xi,xj)=0g\left( {{x}_{i},{x}_{j}}\right) = 0 ，其中 g(xi,xj)g\left( {{x}_{i},{x}_{j}}\right) 是 F[xi,xj]F\left\lbrack {{x}_{i},{x}_{j}}\right\rbrack 中的多项式 (f(xi)−f(xj))/(xi−xj)\left( {f\left( {x}_{i}\right) - f\left( {x}_{j}\right) }\right) /\left( {{x}_{i} - {x}_{j}}\right) 。因此，找到一个 nn -着色等价于解方程组。

{f(xi)=0, for i=1,2,…,N.g(xi,xj)=0, for all edges (i,j) in ζ\left\{ \begin{array}{ll} f\left( {x}_{i}\right) = 0, & \text{ for }i = 1,2,\ldots ,N. \\ g\left( {{x}_{i},{x}_{j}}\right) = 0, & \text{ for all edges }\left( {i,j}\right) \text{ in }\zeta \end{array}\right.

(note also we may use any polynomial gg satisfying αi≠αj{\alpha }_{i} \neq {\alpha }_{j} if g(αi,αj)=0g\left( {{\alpha }_{i},{\alpha }_{j}}\right) = 0 ). It follows by “Hilbert’s Nullstellensatz” (cf. Corollary 33 in Section 15.3) that this system of equations has a solution,hence G\mathcal{G} has an nn -coloring,unless the ideal II in F[x1,x2,…,xN]F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{N}}\right\rbrack generated by the polynomials f(xi)f\left( {x}_{i}\right) for i=1,2,…,Ni = 1,2,\ldots ,N ,together with the polynomials g(xi,xj)g\left( {{x}_{i},{x}_{j}}\right) for all the edges(i,j)in the graph G\mathcal{G} ,is not a proper ideal. This in turn is equivalent to the statement that the reduced Gröbner basis for II (with respect to any monomial ordering) is simply [1]. Further,when an nn -coloring does exist,solving this system of equations as in the examples following Proposition 29 provides an explicit coloring for G\mathcal{G} .

（注意我们还可能使用任何满足 gg 的多项式 αi≠αj{\alpha }_{i} \neq {\alpha }_{j} ，如果 g(αi,αj)=0g\left( {{\alpha }_{i},{\alpha }_{j}}\right) = 0 ）。根据“希尔伯特零点定理”（参见第15.3节的推论33），这个方程组有解，因此 G\mathcal{G} 有一个 nn -着色，除非由多项式 f(xi)f\left( {x}_{i}\right) 生成的主理想 II 在 F[x1,x2,…,xN]F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{N}}\right\rbrack 中，以及图中所有边 (i,j) 的多项式 g(xi,xj)g\left( {{x}_{i},{x}_{j}}\right) ，不是一个适当理想。这反过来等价于这样一个陈述：对于 II 的简化Groebner基（对于任何单项式排序）仅仅是 [1]。进一步，当一个 nn -着色确实存在时，按照命题29后的例子解决这个方程组，为 G\mathcal{G} 提供了一个显式的着色。

There are many possible choices of field FF and set SS . For example,use any field FF containing a set SS of distinct nth {n}^{\text{th }} roots of unity,in which case f(x)=xn−1f\left( x\right) = {x}^{n} - 1 and we may take g(xi,xj)=(xin−xjn)/(xi−xj)=xin−1+xin−2xj+⋯+xixjn−2+xjn−1g\left( {{x}_{i},{x}_{j}}\right) = \left( {{x}_{i}^{n} - {x}_{j}^{n}}\right) /\left( {{x}_{i} - {x}_{j}}\right) = {x}_{i}^{n - 1} + {x}_{i}^{n - 2}{x}_{j} + \cdots + {x}_{i}{x}_{j}^{n - 2} + {x}_{j}^{n - 1} ,or use any subset SS of F=FpF = {\mathbb{F}}_{p} with a prime p≥np \geq n (in the special case n=pn = p ,then,by Fermat’s Little Theorem,we have f(x)=xp−xf\left( x\right) = {x}^{p} - x and g(xi,xj)=(xi−xj)p−1−1g\left( {{x}_{i},{x}_{j}}\right) = {\left( {x}_{i} - {x}_{j}\right) }^{p - 1} - 1 ).

字段 FF 和集合 SS 有许多可能的选择。例如，使用包含一组 nth {n}^{\text{th }} 个不同单位根的集合 SS 的任何字段 FF ，在这种情况下 f(x)=xn−1f\left( x\right) = {x}^{n} - 1 ，我们可以取 g(xi,xj)=(xin−xjn)/(xi−xj)=xin−1+xin−2xj+⋯+xixjn−2+xjn−1g\left( {{x}_{i},{x}_{j}}\right) = \left( {{x}_{i}^{n} - {x}_{j}^{n}}\right) /\left( {{x}_{i} - {x}_{j}}\right) = {x}_{i}^{n - 1} + {x}_{i}^{n - 2}{x}_{j} + \cdots + {x}_{i}{x}_{j}^{n - 2} + {x}_{j}^{n - 1} ，或者使用 F=FpF = {\mathbb{F}}_{p} 的任何子集 SS ，其具有素数 p≥np \geq n（在特殊情况 n=pn = p 中，根据费马小定理，我们有 f(x)=xp−xf\left( x\right) = {x}^{p} - x 和 g(xi,xj)=(xi−xj)p−1−1g\left( {{x}_{i},{x}_{j}}\right) = {\left( {x}_{i} - {x}_{j}\right) }^{p - 1} - 1 ）。

(a) Consider a possible 3-coloring of the graph G\mathcal{G} with eight vertices and 14 edges (1,3), (1,4),\left( {1,4}\right) , (1,5),\left( {1,5}\right) , (2,4),\left( {2,4}\right) , (2,7),\left( {2,7}\right) , (2,8),\left( {2,8}\right) , (3,4),\left( {3,4}\right) , (3,6),\left( {3,6}\right) , (3,8),\left( {3,8}\right) , (4,5),\left( {4,5}\right) , (5,6),\left( {5,6}\right) , (6,7),\left( {6,7}\right) , (6,8),\left( {6,8}\right) , (7,8).\left( {7,8}\right) . Take F=F3F = {\mathbb{F}}_{3} with ‘colors’ 0,1,2∈F30,1,2 \in {\mathbb{F}}_{3} and suppose vertex1is colored by0.In this case f(x)=x(x−1)(x−2)=x3−x∈F3[x]f\left( x\right) = x\left( {x - 1}\right) \left( {x - 2}\right) = {x}^{3} - x \in {\mathbb{F}}_{3}\left\lbrack x\right\rbrack and g(xi,xj)=xi2+xixj+xj2−1g\left( {{x}_{i},{x}_{j}}\right) = {x}_{i}^{2} + {x}_{i}{x}_{j} + {x}_{j}^{2} - 1 . If II is the ideal generated by x1,xi3−xi,2≤i≤8{x}_{1},{x}_{i}^{3} - {x}_{i},2 \leq i \leq 8 and g(xi,xj)g\left( {{x}_{i},{x}_{j}}\right) for the edges(i,j)in G\mathcal{G} ,show that the reduced Gröbner basis for II with respect to the lexicographic monomial ordering x1>x2>⋯>x8{x}_{1} > {x}_{2} > \cdots > {x}_{8} is {x1,x2,x3+x8,x4+2x8,x5+x8,x6,x7+x8,x82+2}\left\{ {{x}_{1},{x}_{2},{x}_{3} + {x}_{8},{x}_{4} + 2{x}_{8},{x}_{5} + {x}_{8},{x}_{6},{x}_{7} + {x}_{8},{x}_{8}^{2} + 2}\right\} . Deduce that G\mathcal{G} has two distinct 3-colorings,determined by the coloring of vertex 8 (which must be colored by a nonzero element in F3{\mathbb{F}}_{3} ),and exhibit the colorings of G\mathcal{G} .

(a) 考虑图 G\mathcal{G} 的一个可能的三着色，该图有八个顶点和14条边 (1,3), (1,4),\left( {1,4}\right) , (1,5),\left( {1,5}\right) , (2,4),\left( {2,4}\right) , (2,7),\left( {2,7}\right) , (2,8),\left( {2,8}\right) , (3,4),\left( {3,4}\right) , (3,6),\left( {3,6}\right) , (3,8),\left( {3,8}\right) , (4,5),\left( {4,5}\right) , (5,6),\left( {5,6}\right) , (6,7),\left( {6,7}\right) , (6,8),\left( {6,8}\right) , (7,8).\left( {7,8}\right) . 取 F=F3F = {\mathbb{F}}_{3} 中的‘颜色’ 0,1,2∈F30,1,2 \in {\mathbb{F}}_{3} 并假设顶点1被染成0色。在这种情况下 f(x)=x(x−1)(x−2)=x3−x∈F3[x]f\left( x\right) = x\left( {x - 1}\right) \left( {x - 2}\right) = {x}^{3} - x \in {\mathbb{F}}_{3}\left\lbrack x\right\rbrack 和 g(xi,xj)=xi2+xixj+xj2−1g\left( {{x}_{i},{x}_{j}}\right) = {x}_{i}^{2} + {x}_{i}{x}_{j} + {x}_{j}^{2} - 1 。如果 II 是由 x1,xi3−xi,2≤i≤8{x}_{1},{x}_{i}^{3} - {x}_{i},2 \leq i \leq 8 和 g(xi,xj)g\left( {{x}_{i},{x}_{j}}\right) 生成的理想，适用于 G\mathcal{G} 中的边(i,j)，则证明相对于字典序单项式排序 x1>x2>⋯>x8{x}_{1} > {x}_{2} > \cdots > {x}_{8} 的 II 的简化Gröbner基是 {x1,x2,x3+x8,x4+2x8,x5+x8,x6,x7+x8,x82+2}\left\{ {{x}_{1},{x}_{2},{x}_{3} + {x}_{8},{x}_{4} + 2{x}_{8},{x}_{5} + {x}_{8},{x}_{6},{x}_{7} + {x}_{8},{x}_{8}^{2} + 2}\right\} 。推断出 G\mathcal{G} 有两种不同的三着色，由顶点8的着色决定（必须用 F3{\mathbb{F}}_{3} 中的非零元素着色），并展示 G\mathcal{G} 的着色。

Show that if the edge(3,7)is added to G\mathcal{G} then the graph cannot be 3-colored.

证明如果将边(3,7)添加到 G\mathcal{G} 中，则该图不能被三着色。

(b) Take F=F5F = {\mathbb{F}}_{5} with four ’colors’ 1,2,3,4∈F51,2,3,4 \in {\mathbb{F}}_{5} ,so f(x)=x4−1f\left( x\right) = {x}^{4} - 1 and we may use g(xi,xj)=xi3+xi2xj+xixj2+xj3g\left( {{x}_{i},{x}_{j}}\right) = {x}_{i}^{3} + {x}_{i}^{2}{x}_{j} + {x}_{i}{x}_{j}^{2} + {x}_{j}^{3} . Show that the graph G\mathcal{G} with five vertices having 9 edges (1,3),\left( {1,3}\right) , (1,4),\left( {1,4}\right) , (1,5),\left( {1,5}\right) , (2,3),\left( {2,3}\right) , (2,4),\left( {2,4}\right) , (2,5),\left( {2,5}\right) , (3,4),\left( {3,4}\right) , (3,5),\left( {3,5}\right) , (4,5)(the “complete graph on five vertices” with one edge removed) can be 4-colored but cannot be 3-colored.

(b) 取 F=F5F = {\mathbb{F}}_{5} 有四种‘颜色’ 1,2,3,4∈F51,2,3,4 \in {\mathbb{F}}_{5} ，因此 f(x)=x4−1f\left( x\right) = {x}^{4} - 1 我们可以使用 g(xi,xj)=xi3+xi2xj+xixj2+xj3g\left( {{x}_{i},{x}_{j}}\right) = {x}_{i}^{3} + {x}_{i}^{2}{x}_{j} + {x}_{i}{x}_{j}^{2} + {x}_{j}^{3} 。证明具有五个顶点和9条边的图 G\mathcal{G} （五个顶点的‘完全图’去掉一条边）可以被四着色，但不能被三着色。

(c) Use Gröbner bases to show that the graph G\mathcal{G} with nine vertices and 22 edges (1,4),(1,6)\left( {1,4}\right) ,\left( {1,6}\right) , (1,7),\left( {1,7}\right) , (1,8),\left( {1,8}\right) , (2,3),\left( {2,3}\right) , (2,4),\left( {2,4}\right) , (2,6),\left( {2,6}\right) , (2,7),\left( {2,7}\right) , (3,5),\left( {3,5}\right) , (3,7),\left( {3,7}\right) , (3,9),\left( {3,9}\right) , (4,5),\left( {4,5}\right) , (4,6),\left( {4,6}\right) , (4,7),\left( {4,7}\right) , (4,9),\left( {4,9}\right) , (5,6),  (5,7),  (5,8),  (5,9),  (6,7),  (6,9),  (7,8)  \left( {5,6}\right) ,\;\left( {5,7}\right) ,\;\left( {5,8}\right) ,\;\left( {5,9}\right) ,\;\left( {6,7}\right) ,\;\left( {6,9}\right) ,\;\left( {7,8}\right) \; has precisely four 4-colorings up to a permutation of the colors (so a total of 96 total 4-colorings). Show that if the edge (1,5) is added then G\mathcal{G} cannot be 4-colored.

(c) 使用Gröbner基来证明具有九个顶点和22条边 G\mathcal{G} 的图 (1,4),(1,6)\left( {1,4}\right) ,\left( {1,6}\right) (1,7),\left( {1,7}\right) , (1,8),\left( {1,8}\right) , (2,3),\left( {2,3}\right) , (2,4),\left( {2,4}\right) , (2,6),\left( {2,6}\right) , (2,7),\left( {2,7}\right) , (3,5),\left( {3,5}\right) , (3,7),\left( {3,7}\right) , (3,9),\left( {3,9}\right) , (4,5),\left( {4,5}\right) , (4,6),\left( {4,6}\right) , (4,7),\left( {4,7}\right) , (4,9),\left( {4,9}\right) , (5,6),  (5,7),  (5,8),  (5,9),  (6,7),  (6,9),  (7,8)  \left( {5,6}\right) ,\;\left( {5,7}\right) ,\;\left( {5,8}\right) ,\;\left( {5,9}\right) ,\;\left( {6,7}\right) ,\;\left( {6,9}\right) ,\;\left( {7,8}\right) \; 精确地有四种4-着色方法（即考虑到颜色排列的总共96种4-着色方法）。证明如果添加边(1,5)，那么 G\mathcal{G} 无法进行4-着色。

avatar
文章
阅读
粉丝