Proof. LetV be irreducible and letf g∈ I(V). Put V1 =V ∩ V(f) andV2=V ∩ V(g). By Prop. 2.9, the intersection of affine varieties is an affine variety and thus V1 andV2 are also affine varieties. By Prop. 2.4, V(f g) ⊇ V(I(V)). Moreover, sinceV ⊆ V(I(V)), we have V = V ∩ V(f g). It follows by Prop. 2.11,V =V ∩(V(f)∪ V(g)) = (V ∩ V(f))∪(V ∩ V(g)) =V1∪V2. But V is irreducible and so V =V1 orV =V2. Without loss of generality, letV =V1. Then the polynomial f vanishes onV and thusf ∈ I(V). Hence, the idealI(V) is prime.
Conversely, suppose V is reducible. Then there are affine varieties V1 andV2 contained inV such that V = V1 ∪V2. By Prop. 2.16, we have I(V) ⊆ I(V1) and I(V) ⊆ I(V2). By the ideal-variety correspondence, we haveV1=V(I(V1)),V2=V(I(V2)), andV =V(I(V)). SinceV16=V andV26=V, it follows thatI(V1)6=I(V) andI(V2)6=I(V). Take polynomialsf ∈ I(V1)\I(V) andg∈ I(V2)\I(V).
Thenf g∈ I(V1∪V2) =I(V) and hence I(V) is not prime. ⊓⊔ Example 2.25.The idealsI1=hX, YiandI2=hZiare prime inR[X, Y, Z]. Thus the corresponding affine varieties, the (x, y)-planeV(I1) ={(x, y,0)|x, y∈R} and thez-axis V(I2) ={(0,0, z)|z∈R},
are irreducible. ♦
Proposition 2.26.Each affine variety V in Kn can be uniquely written (up to permutation) in the form
V =V1∪. . .∪Vm, whereV1, . . . , Vmare pairwise distinct irreducible affine varieties.
Proof. LetV be an affine variety that cannot be written as a finite union of irreducible affine varieties.
ThenV is reducible with V =V1∪V1′ such thatV16=V andV1′6=V. Furthermore, at least one ofV1
andV1′ cannot be described as a union of irreducible affine varieties. Assume that V1 is not a union of irreducible affine varieties. Then again write V1 =V2∪V2′, where V2 6=V1 and V2′ 6=V1. Continuing in this way, we obtain an infinite descending sequence of affine varietiesV ⊃V1 ⊃ V2 ⊃. . .. By the ideal-variety correspondence, the operatorI is one-to-one and thus gives rise to an ascending sequence of idealsI(V)⊂ I(V1)⊂ I(V2)⊂. . .. By the ascending chain condition 1.28, this chain must become stationary contradicting the infiniteness of the sequence of affine varieties.
Assume there are two such expressionsV =U1∪. . .∪Ur=W1∪. . .∪Ws. ConsiderU1=V∩U1= (W1∩U1)∪. . .∪(Ws∩U1). SinceU1 is irreducible, there is an indexj such that U1=Wj∩U1; that is,U1⊆Wj. Likewise, there is an indexksuch that W1⊆Uk. It follows thatU1 ⊆Uk, which implies by hypothesis thatj=k and soU1=Wj Continuing we see thatr=sand that one decomposition is
only a renumbering of the other. ⊓⊔
Example 2.27.The affine variety V = {(x, y, z) | xz = yz = 0, x, y, z ∈ R} in R3 is reducible, since V decomposes into the union of the (x, y)-plane V1 = {(x, y,0) | x, y ∈ R} and the z-axis V2={(0,0, z)|z∈R}. Both are irreducible affine varieties. ♦
2.5 Elimination Theory
We provide a straightforward method for solving systems of polynomial equations based on the elimi-nation orderings.
Let I be an ideal in K[X1, . . . , Xn] and letk ≥0 be an integer. The kth elimination ideal of I is given as
Ik =I∩K[Xk+1, . . . , Xn]. (2.5) Note that the 0-th elimination ideal isI0=I. Clearly,Ik is an ideal ofK[X1, . . . , Xn].
A monomial ordering > on K[X1, . . . , Xn] has the elimination property for X1, . . . , Xk if f ∈ K[X1, . . . , Xn] andlm>(f)∈K[Xk+1, . . . , Xn] impliesf ∈K[Xk+1, . . . , Xn]. That is, monomials which contain one ofX1, . . . , Xkare always larger than monomials which contain none ofX1, . . . , Xk. A mono-mial ordering >on K[X1, . . . , Xn] is an elimination ordering for X1, . . . , Xk if it has the elimination property forX1, . . . , Xk.
For instance, the lp ordering has the elimination property for any sequence X1, . . . , Xk, k ≥ 0.
Product orderings provide a large class of elimination orderings. For this, let>1be a monomial ordering onK[X1, . . . , Xm] and>2 be a monomial ordering on K[Y1, . . . , Yn]. Then the product ordering >on K[X1, . . . , Xm, Y1, . . . , Yn], denoted by (>1, >2), is defined by
XαYβ > XγYδ :⇐⇒ Xα>1Xγ∨(Xα=Xγ∧Yβ>2Yδ).
The product ordering>onK[X1, . . . , Xm, Y1, . . . , Yn] is an elimination ordering forX1, . . . , Xm. Example 2.28 (Singular). Product orderings can be specified by the ring definitions.
> ring r = 0, (w,x,y,z), (dp(2),Dp(2)); // mixed product ordering
> poly f = wx2z+w2x2yz+wx+yz2+y2z;
> f;
w2x2yz+wx2z+wx+y2z+yz2
♦ Theorem 2.29. (Elimination)Let I be an ideal of K[X1, . . . , Xn] and let k≥0 be an integer. If G is a Groebner basis ofI with respect to an elimination ordering>forX1, . . . , Xk, the k-th elimination idealIk of I has the Groebner basis
Gk=G∩K[Xk+1, . . . , Xn].
Proof. LetG={g1, . . . , gs}be a Groebner basis ofI. Assume that the firstr≤selements ofGlie in K[Xk+1, . . . , Xn].
Claim that Gk = {g1, . . . , gr} is a generating set of Ik. Indeed, by definition, Gk ⊆ Ik and so hg1, . . . , gri ⊆Ik. Conversely, letf ∈Ik. Then divide f into g1, . . . , gsgiving the remainderfG = 0. In view of the given elimination ordering, the leading terms of the (eliminated) polynomialsgr+1, . . . , gs
must involve at least one of the variablesX1, . . . , Xk and these terms are greater than any term inf. It follows that the division of f into g1, . . . , gs does not involvegr+1, . . . , gs and thereforef is of the form
f =h1g1+. . .+hrgr+ 0·gr+1+. . .+ 0·gs+ 0.
Hence,f ∈ hg1, . . . , grias required.
Claim thatGk ={g1, . . . , gr}is a Groebner basis ofIk. Indeed, divide the S-polynomialS(gi, gj)∈Ik
intoGk for each pairi6=j, 1≤i, j≤r. The previous paragraph shows that the remainderS(gi, gj)Gk is zero. Thus by the Buchberger S-criterion,Gk is a Groebner basis ofIk. ⊓⊔ Example 2.30 (Singular).
2.5 Elimination Theory 37
> ring r = 0, (w,x,y,z), lp;
> ideal i = w2,x4,y5,z3,wxyz;
> eliminate(i,z);
_[1]=w2 _[2]=x4 _[3]=y5
> eliminate(i,yz);
_[1]=w2 _[2]=x4
> eliminate(i,xyz);
_[1]=w2
♦ An ideal IofK[X1, . . . , Xn] generated byf1, f2, . . . , fs provides a system of polynomial equations
f1= 0, f2= 0, . . . , fs= 0.
Any point (a1, . . . , an)∈ V(I) is a solution of the system of equations, and any point (ak+1, . . . , an) in V(Ik) is apartial solutionof the system of equations. Each solution truncates to a partial solution, but not each partial solution extends to a solution. This is where the following Extension results comes into play. For this, note that each polynomial f in Ik−1 can be written as a polynomial inXk, whose coefficients are polynomials inXk+1, . . . , Xn,
f =cXkN + terms in whichXk has degree < N , (2.6) where 06=c∈K[Xk+1, . . . , Xn] is called the leading coefficient polynomial off.
Theorem 2.31. (Extension)LetKis an algebraically closed field. A partial solution(ak+1, . . . , an)in V(Ik)extends to a partial solution(ak, ak+1, . . . , an)inV(Ik−1)if the leading coefficient polynomials of the elements of the Groebner basis ofIk−1 with respect to an eliminiation ordering withX1> . . . > Xn
do not all vanish at(ak+1, . . . , an).
Note that the condition of the theorem is particularly fulfilled if the leading coefficient polynomials are constants inK. The Elimination theorem shows that a Groebner basisGof the given idealIwith respect to the lp ordering eliminates successively more and more variables. This gives the following strategy for finding all solutions of the system of equations: Start with the polynomials in Gwith the fewest variables, solve them, and then extend these partial solutions to solutions of the whole system adding one variable at a time.
Example 2.32 (Singular). Consider the system of equations X2+Y2+Z2 = 2,
X2+ 2Y2 = 3, XZ = 1.
Take the idealI=hX2+Y2+Z2−2, X2+ 2Y2−3, XZ−1iinC[X, Y, Z]. First, compute a Groebner basis ofIwith respect to an elimination ordering.
> ring r = 0, (x,y,z), lp; // lexicographical ordering
> ideal i = x2+y2+z2-2, x2+2y2-3, xz-1;
> ideal j = std(i); j;
j[1]=2z4-z2+1 j[2]=y2-z2-1 j[3]=x+2y2z-3z
The corresponding Groebner basis isG={2Z4−Z2+ 1, Y2−Z2−1, X+ 2Y2Z−3Z}. The second elimination idealI2=I∩C[Z] has the Groebner basisG2={2Z4−Z2+1}. The generating polynomial is irreducible which can be tested byMaple.
> with(PolynomialTools):
> factor( 2z^4-z^2-1+1 );
The zeros of this polynomial can be numerically found as follows.
> LIB "solve.lib";
> ring r2 = 0, (z), lp;
> ideal i2 = 2z4-z2+1;
> solve (i2,6);
[1]:
(-0.691776+i*0.478073) [2]:
(0.691776-i*0.478073) [3]:
(0.691776-i*0.478073) [4]:
(-0.691776+i*0.478073)
However, these roots arealgebraic numbersand can be symbolically established byMaple.
> with(PolynomialTools):
> solve( 2z^4-z^2-1+1, z );
This produces the four solutions
±1 2
q 1 +i√
7, 1 2
q 1−i√
7.
Note that each algebraic number has adegree which is the degree of its minimal polynomial over Q;
for instance, the above algebraic numbers have degree 4. By elimination, the first elimination ideal I1=I∩C[Y, Z] is generated by the polynomialsY2−Z2−1 and 2Z4−Z2+ 1. The leading coefficient polynomial ofY2−Z2−1∈C[Z][Y] is the leading term ofY2 which is a nonzero constant. Thus by extension, each partial solution inV(I2) extends to a solution inV(I1). There are eight such points. To find them, substitute a root of the generator 2Z4−Z2+ 1 for Z and solve the resulting equation for Y. For instance, theMaplecommand
> subs(Z=(1/2)*sqrt(1+I*sqrt(7)), G);
produces
2.5 Elimination Theory 39 We can check that the first expression is a zero by usingMaple
> evalf(%);
Finally, the leading coefficient polynomial ofX+ 2Z3−Z∈C[Y, Z][X] is the coefficient of the termX which is a nonzero constant. By extension, each partial solution inV(I1) can be extended to a point in V(I). For instance, for the above value ofZ we obtain
X = 1 This gives rise to the following solutions of the system of equations,
(1
All other solutions can be derived in the same way. ♦
Example 2.33 (Singular). Consider the system of equations
XY = 1, (2.7)
The leading coefficient polynomial of XZ−1 equalsZ. By extension, each partial solution (a, a) witha6= 0 extends to a solution (X, Y, Z) = (1/a, a, a) inV(I) and thus solves the system of equations.
Note that the partial solution (0,0) cannot be extended. ♦
The above examples is rather simple because the coordinates of the solutions can all be expressed in terms of roots of complex numbers. Unfortunately, general systems of polynomial equations are rarely this nice. For instance, it is known that there are no general formulae involving only the field operations inKand extraction of roots, forming so-called radicals, for solving single variable polynomial equations of degree 5 or higher. This is a famous result due to Evariste Galois (1811-1832). Thus if elimination leads to a one-variable equations of degree 5 or higher, we may not be able to give radical formulae for the roots.
Example 2.34 (Maple). Consider the system of equations X5+Y2+Z2 = 2, X2+ 2Y2 = 3, XZ = 1.
To solve these equations, we first compute a Groebner basis of the idealI=hX5+Y2+Z2−2, X2+ 2Y2−3, XZ−1iwith respect to thelpordering.
> with(Groebner):
> F2 := [x^5+y^2+z^2-2, x^2+2*y^2-3, x*z-1]:
> G2 := gbasis(F2, plex(x,y,z));
This gives the output
[2Z7−Z5−Z3+ 2,4Y2−2Z5+Z3+Z−6,2X+ 2Z6−Z4−Z2].
By elimination, the second elimination idealI2=I∩C[Z] is generated by the polynomial 2Z7−Z5− Z3+ 2. This generator is irreducible overQ. In this situation, we need to decide what kind of answer is required.
If we want a purely algebraic description of the solutions, then Maple can represent solutions of systems like this by thesolve command. Entering
> solve(convert(G2, set), {x,y,z});
gives the output X = 1
2 ·Root of(2 Z7− Z5− Z3+ 2)2+1
2 ·Root of(2Z7− Z5− Z3+ 2)4
−Root of(2Z7− Z5− Z3+ 2)6, Y = 1
2 ·Root of(−6 +Y′)5,
Y′= Root of(2 Z7− Z5− Z3+ 2) + Root of(2 Z7− Z5− Z3+ 2)3
−2·Root of(2 Z7− Z5− Z3+ 2)5+ Z7, Z= Root of(2 Z7− Z5− Z3+ 2).
Here Root of(2Z7− Z5− Z3+ 2) stands for any of the roots of the polynomial equation 2Z7− Z5− Z3+ 2 = 0 in the dummy variable Z.
On the other hand, in many practical situations where equations must be solved, knowing a numer-ical approximation to a real or complex solution is often more useful and perfectly acceptable provided