# Gröbner Bases and their Applications

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1 Gröbner Bases and their Applications Kaitlyn Moran July 30, Introduction We know from the Hilbert Basis Theorem that any ideal in a polynomial ring over a field is finitely generated [3]. However, there remains question as to the best generators to choose to describe the ideal. Are there generators for a polynomial ideal I that make it easy to see if a given polynomial f belongs to I? For instance, does 2x 2 z 2 +2xyz 2 +2xz 3 +z 3 1 belong to I = (x + y + z, xy + xz + yz, xyz 1)? Deciding if a polynomial is in an ideal is called the Ideal Membership Problem. In polynomial rings of one variable, we use long division of polynomials to solve this problem. There is a corresponding algorithm for K[x 1,..., x n ], but because there are multiple variables and multiple divisors, the remainder of the division is not unique. Hence a remainder of 0 is a sufficient condition, but not a necessary condition, to determine ideal membership. However, if we choose the correct divisors, then the remainder is unique regardless of the order of the divisors. These divisors are called a Gröbner basis. In order to define Gröbner bases, we must first discuss monomial orderings. 2 Monomial Orderings Let K be a field. In K[x], we write polynomials in a canonical way, with the term of highest degree first and each subsequent term of lesser degree than the preceding one. This assists in the long division of polynomials, for we divide the term of highest degree of the dividend by the term of highest degree of the divisor, and if it is not divisible, then we are done. In order to generalize this division algorithm, we must have a corresponding ordering on monomials in K[x 1,..., x n ], called a monomial ordering. Definition 1. A monomial ordering is an order relation < on the set of all monomials of K[x 1,..., x n ] such that For any monomials m, n, then exactly one of the following is true:m < n, n < m, or m = n 1

2 If m 1 < m 2 and m 2 < m 3, then m 1 < m 3 For any monomial m 1, 1 < m If m 1 < m 2, then nm 1 < nm 2 for any monomial n To discuss the four common examples of monomial orderings, it is helpful to first define a way of discussing the degree of a multivariable monomial, called multidegree. Definition 2. The multidegree of a monomial m = x i 1 1 x i 2 2 x in n is defined to be mdeg(m) = i i n. Now we consider some examples of monomial orderings. Example 1. Four common monomial orderings are Lex, Deglex, Revlex, and Degrevlex. In Lex, the lexicographic ordering, m 1 = x i 1 1 x in n < x j 1 1 x jn n = m 2 if i 1 = j 1,..., i k 1 = j k 1, i k < j k for some k. That is, m 1 < m 2 if the first variable with different exponents has a lower degree in m 1 than in m 2. Notice this is alphabetical ordering, like words in a dictionary. In Deglex, m 1 = x i 1 1 x in n < x j 1 1 x jn n = m 2 if mdeg(m 1 ) < mdeg(m 2 ) or if mdeg(m 1 ) = mdeg(m 2 ) and m 1 < m 2 with respect to Lex. In Revlex, m 1 = x i 1 1 x in n < x j 1 1 x jn n = m 2 if i n = j n,..., i k+1 = j k+1, i k > j k for some k. That is, m 1 < m 2 if the last variable with different exponents has a higher degree in m 1 than in m 2. In Degrevlex, m 1 = x i 1 1 x in n < x j 1 1 x jn n = m 2 if mdeg(m 1 ) < mdeg(m 2 ) or if mdeg(m 1 ) = mdeg(m 2 ) and m 1 < m 2 with respect to Revlex. Each monomial ordering depends on the ordering of the variables. For instance, Lex with the ordering x > y > z is a different ordering than Lex with y > x > z. In the above examples, we order x 1 > x 2 > > x n, but this is arbitrary. Hence for each of the four common monomial orderings, we have n! possible orderings. Once a monomial ordering is chosen, then the terms of a polynomial can be ordered in an unambiguous way, where each term is less than the preceding term with respect to the chosen ordering. For example, let f = 4xy 2 z + 4z 2 5x 3 + 7x 2 z 2, with x > y > z. Each monomial ordering gives a different reordering of the terms of f. With respect to Lex, f = 5x 3 + 7x 2 z 2 + 4xy 2 z + 4z 2. With respect to Deglex, f = 7x 2 z 2 + 4xy 2 z 5x 3 + 4z 2. 2

4 Now, because f 1 is listed first, we consider lt f and lt f 1. Since lt f 1 divides lt f, we write the quotient q in a 1 and then subtract q f 1 from f. a 1 : a 2 : xy + 1 y + 1 y xy2 + 1 xy 2 + y y + 1 Now we repeat the process on the new polynomial p. Since lt f 1 does not divide lt p but lt f 2 does, we write the quotient in a 2 and then multiply and subtract. a 1 : a 2 : xy + 1 y + 1 y 1 xy2 + 1 xy 2 + y y + 1 y 1 2 Because neither lt f 1 nor lt f 2 divides the new polynomial, this is our remainder. Hence, xy } {{ } f = ( y }{{} )(xy + 1) + ( } {{ } }{{} 1 )(y + 1) + } {{ } }{{} 2. a 1 a 2 f 2 r The previous example ends with a nice remainder. f 1 However, it is possible to have a remainder with a leading term that is not divisible by any of the leading terms of the divisors, but one of the leading terms of the divisors divides a different term of the remainder. In this case, the division can continue by moving the leading term to a "remainder" column. Example 3. Let f = x 2 y + xy 2 + y 2, f 1 = xy 1, and f 2 = y 2 1. We will use Lex ordering 4

5 with x > y. Performing long division, we see a 1 : x + y a 2 : 1 r xy 1 y 2 1 x2 y + xy 2 + y 2 x 2 y x xy 2 + x + y 2 xy 2 y x + y 2 + y y 2 + y x y 2 1 y x + y 0 x + y + 1 Hence, xy 2 + xy 2 + y 2 } {{ } f = (x + y } {{ } )(xy 1) + ( } {{ } }{{} 1 )(y 2 1) + x + y + 1. } {{ } } {{ } a 1 f 1 a 2 f 2 r This example is a fairly complete illustration of the division algorithm. For a statement of the general algorithm, the reader may consult [1], page 62. This division algorithm proves the following proposition. Proposition 4 (Division Algorithm). Fix a monomial ordering. Let F = (f 1,..., f m ) be an ordered m-tuple of polynomials in K[x 1,..., x n ]. Then for every f K[x 1,..., x n ], f can be written as f = a 1 f a m f m + r for some a 1,..., a n, r K[x 1,..., x n ], where no term of r is divisible by lt(f 1 ),..., lt(f m ). We call r the remainder of f on division by F. 3.2 Gröbner Bases and Some Properties Now, we know the division algorithm gives a remainder. However, unlike the familiar division algorithm in one variable, the remainder from this division algorithm is not unique. For example, let I = (g 1, g 2 ) = (x 2, xy y 2 ) and let f = x 2 y. Now, f I since f is a nonzero multiple of one of the generators of I, and so by dividing first by g 1, the remainder of f is 0. 5

6 However, by dividing first by g 2, we see f = (x + y)(xy y 2 ) + y 3. This gives the remainder of f is y 3, but y 3 0. Hence the remainder is not unique. However, if we choose the right generators, the remainder is unique, and so it is easy to solve the Ideal Membership Problem. These generators are a Gröbner basis. Definition 5. Let I be an ideal in K[x 1,..., x n ]. A set g 1,..., g m, where g i I 1 i m, is a Gröbner basis for I if (lm(g 1 ),..., lm(g m )) = lm(i). Proposition 6. If g 1,..., g m is a Gröbner basis for an ideal I, then (g 1,..., g m ) = I. Proof. Clearly, (g 1,..., g m ) I, since g i I for all 1 i m. Let f I. Then we can divide f by {g 1,..., g m }, so f can be written as f = a 1 g a m g m + r where no term in r is divisible by lm(g i ) for any i = 1,..., m. We must show that r = 0. Notice that r = f a 1 g a m g m I. If r 0, lm(r) lm(i) = (lm(g 1 ),..., lm(g m )). Thus, because lm(r) is a monomial, lm(r) must be divisible by some lm(g i ). This is a contradiction, so r = 0 and therefore f (g 1,..., g m ). Proposition 7. Every ideal I K[x 1,..., x n ] other than (0) has a Gröbner basis. Proof. lm(i) is generated by the monomials lm(g), where g I \ {0}. Because lm(i) K[x 1,..., x n ], lm(i) is finitely generated, so there exist g 1,..., g m such that lm(i) = (lm(g 1 ),..., lm(g m )). By Proposition 6, {g 1,..., g m } generate I. Theorem 8. Let G = (g 1,..., g m ) be a Gröbner basis for an ideal I. Let f K[x 1,..., x n ]. Then there exists a unique r K[x 1,..., x n ] such that (i) no term of r divides lm(g i ) for any i, and (ii) there exists a g I such that f = g + r. Proof. The division algorithm for K[x 1,..., x n ] gives f = a 1 g a m g m + r, which satisfies (i). Define g = a 1 g a m g m to satisfy (ii). To prove uniqueness, let f = g +r 1 = h+r 2 as in (ii). Then r 1 r 2 = h g I. Hence, if r 1 r 2, then lm(r 1 r 2 ) lm(i), so lm(r 1 r 2 ) is divisible by some lm(g i ), since lm(i) is a monomial ideal. However, this contradicts the assumption that no term of r 1 or r 2 divides lm(g i ) for any i. Hence r 1 r 2 = 0. Therefore, r is unique. 6

7 The r in Theorem 8 denotes the remainder of division of f by G, and the uniqueness of r guarantees that r is the remainder regardless of the order of the elements of G in the division algorithm. This gives us a necessary and sufficient condition for membership to I, thereby solving the Ideal Membership Problem. Corollary 9. Let G be a Gröbner basis for an ideal I, and let f K[x 1,..., x n ]. We write f = g + r where g I as above. Then f I if and only if r = 0. Proof. If r = 0, then clearly f I. If f I, then f = f + 0 satisfies the two conditions of Theorem 8. By uniqueness of r, this means 0 is the remainder of f on division by G. 3.3 A Criterion for a Gröbner Basis Now that we have seen uses for Gröbner bases, how do we determine if a given generating set f 1,..., f k is a Gröbner basis? As discussed previously, not every generating set is a Gröbner basis, because sometimes there are combinations of the generators that have cancellations of leading terms, leaving only smaller terms. These smaller monomials now appear in lm(i), but are not accounted for in the ideal generated by the leading monomials of the generators of I. For instance, let I = (xy 2 y 3, xy) and let the ordering be Revlex. Then y 3 = (xy 2 y 3 ) y(xy) I, so y 3 lm(i). But y 3 / (lm(xy 2 y 3 ), lm(xy)) = (xy 2, xy) = (xy). To study this, we look at special combinations of the generators, called syzygies and S- polynomials. Definition 10. Let R = K[x 1,..., x n ], R k a free module, f R and F = (f 1,..., f k ) R k. A syzygy of F is an element s = (s 1,..., s k ) R k such that s 1 f s k f k = 0. A representation of f with respect to F is an element H = (h 1,..., h k ) such that h 1 f h k f k = f. Note that if H and H are representations of f, then H H is a syzygy of F. Similarly, if H is a representation and S is a syzygy of F, then H + S is a representation of f with respect to F. Proposition 11. Suppose m 1,..., m k are monomials and S = (s 1,..., s k ) is a syzygy of (m 1,..., m k ). Let e i denotes the ith basis element of the free module R k, i.e., e i = (0,..., 0, 1, 0..., 0). Then S is a linear combination of the syzygies where 1 i j k. s ij = lcm(m i, m j ) m i e i lcm(m i, m j ) m j e j, 7

8 Proof. Since s 1 m s k m k = 0, the terms s i m i must all cancel, so we can assume each s j is of the form c j n j, where n j is a monomial such that n j m j = x i 1 1 x in n =: m for some i 1,..., i n Z + for all j. Thus, c 1 + +c k = 0. The solutions to this are linear combinations of e i e j, 1 i j k. Hence, (s 1,..., s k ) are linear combinations of which is a multiple of s ij. m m i e i m m j e j, Definition 12. Let f 1, f 2 be monic elements of R, and let < be a monomial ordering on R. Then the S-polynomial of (f 1, f 2 ) on < is S(f 1, f 2 ) = lcm(lm(f 1), lm(f 2 )) lm(f 1 ) f 1 lcm(lm(f 1), lm(f 2 )) f 2. lm(f 2 ) Notice that S-polynomials are designed for cancellation of terms. Theorem 13 (Criterion for a Gröbner basis). Suppose I is a polynomial ideal, G = {f 1,..., f m } is a basis for I, and S i,j = S(f i, f j ) are the S-polynomials. Then G is a Gröbner basis of I if and only if for all pairs i j, the remainder on division of S i,j by G is zero. Proof. : If G is a Gröbner basis, then all S i,j I. Hence by Corollary 9, the remainder on division of S i,j is 0. : Assume the remainder of S i,j is 0, and let f I. We need to show that lm(f) (lm(f 1 ),..., lm(f m )). Because {f 1,..., f m } generate I, we can write f = a 1 f a m f m for some a 1,..., a m R. Now, either max{lm(a i f i )} > lm(f) or max{lm(a i f i )} = lm(f). In the latter case, then lm(f) = lm(a i ) lm(f i ) for some i, so lm(f) (lm(f 1 ),..., lm(f m )), so assume the former is true. Now, A = (a 1,..., a m ) is a representation of f. We will construct a syzygy Z such that A = A + Z is another representation of f with max{lm(a if i )} < max{lm(a i f i )}. Because there is no strictly decreasing sequence of monomials, this process will eventually terminate, and then we will be in the case where max{lm(a i f i )} = lm(f). Let J = {j : lm(a j f j ) = max{lm(a i f i )}}. We know the leading terms in i J a if i cancel in order for lm(f) < max{lm(a i f i )}. Let S A = (s 1,..., s m ) where s i = lt(a i ) if i J and s i = 0 otherwise. Then S A is a syzygy for (lm(f 1 ),..., lm(f m )). By Proposition 11, S A is a linear combination of s ij s, so S A = 1 i j m t ijs ij. By definition, the s ij represents S ij with respect to F = (f 1,..., f m ), so s 1 f s m f m = t ij S ij. Now, since the remainder of S ij is 0, S ij is a linear combination of f i s, so S ij = m t=1 h ij,tf t. Now, H ij = (h ij,1,..., h ij,m ) 8

9 is a representation of S ij with respect to F, so s ij H ij is a syzygy of F. Let Z = t ij (s ij H ij ), 1 i j m which is also a syzygy of F. Hence, A Z = A = (a 1,..., a m) is a representation of f with respect to F. By construction, max{lm(a i f i )} > max{lm(a if i )}. Example 4. We will show that G = {x 2, xy y 2, y 3 } is a Gröbner basis for I = (x 2, xy y 2 ) with respect to Lex. Notice y 3 = ( x y)(xy y 2 ) + y(x 2 ). Hence y 3 I, so G generates I. Now, S 1,2 = lcm(x2, xy) x 2 (x 2 ) lcm(x2, xy) (xy y 2 ) = x 2 y x 2 y + xy 2 = xy 2 xy S 2,3 = xy3 xy (xy y2 ) xy3 y 3 y3 = y 4 S 1,3 = x2 y 3 x 2 x2 x2 y 3 y 3 y3 = 0 Clearly, both S 2,3 and S 1,3 give a remainder of 0 when divided by G. Because xy 2 = y(xy y 2 ) + y 3, then the remainder of S 1,2 is also 0. Therefore, by Theorem 13, G is a Gröbner basis. Example 5. Let I = (x + y + z, xy + xz + yz, xyz 1), G = {x + y + z, y 2 + yz + z 2, z 3 1}, and I G = (G). We will show that G is a Gröbner basis for I with respect to Lex. Notice xy + xz + yz = (y + z)(x + y + z) (y 2 + yz + z 2 ) Hence I = I G. Now, S 1,2 = lcm(x, y2 ) x xyz 1 = yz(x + y + z) z(y 2 + yz + z 2 ) + (z 3 1) y 2 + yz + z 2 = (y + z)(x + y + z) (xy + xz + yz) z 3 1 = z 2 (x + y + z) z(xy + xz + yz) + (xyz 1) (x + y + z) lcm(x, y2 ) (y 2 + yz + z 2 ) = xyz xz 2 + y 3 + y 2 z y 2 = yz z 2 (x + y + z) + (y + z)(y 2 + yz + z 2 ) 9

10 S 1,3 = z 3 (x + y + z) x(z 3 1) = x + yz 3 + z 4 = x + y + z + (y + z)(z 3 1) S 2,3 = z 3 (y 2 + yz + z 2 ) y 2 (z 3 1) = y 2 + yz 4 + z 5 = y 2 + yz + z 2 + (yz + z 2 )(z 3 1) Therefore, each S ij is divisible by G, so G is a Gröbner basis. 3.4 Buchberger s Algorithm By Proposition 7, we know that every ideal has a Gröbner basis. Buchberger s algorithm provides instructions to construct a Gröbner basis from any given generating set. Given an ideal I = (f 1,..., f m ), calculate all of the S-polynomials. Divide the S- polynomials by {f 1,..., f m }.. If the remainder is nonzero, extend (f 1,..., f m ) with the remainder. Calculate all of the S-polynomials which have not previously been calculated, and repeat the process until all of the remainders of the S-polynomials are zero. This process will terminate in a finite number of steps. For each nonzero remainder, the leading monomial of the remainder is not in (lm(f 1 ),..., lm(f m )), so extending the generating set of I also extends the monomial ideal generated by the leading monomials of the generating set of I. Because the ideal of leading monomials of the generating set is a subset of lm(i) and lm(i) has a finite generating set, the process will terminate. Additionally, this algorithm does give a Gröbner basis by Theorem 13. Example 6. Let I = (x 2 y 1, xy 2 x) = (g 1, g 2 ), and let the monomial ordering be Deglex. Then S 1,2 = y(x 2 y 1) x(xy 2 x) = y+x 2. Since lm( y+x 2 ) = x 2 is not divisible by either lm(g 1 ) or lm(g 2 ), g 3 = x 2 y. Continuing the algorithm, S 1,3 = (x 2 y 1) y(x 2 y) = 1+y 2, so lm(s 1,3 ) = y 2. Hence g 4 = y 2 1. Further, S 2,3 = x(xy 2 x) y 2 (x 2 y) = y 4 x 2 = (y 2 + 1)(y 2 1) (x 2 y) y + 1 so the remainder of S 2,3 divided by G is g 5 = y + 1. Now, S 1,4 = y(x 2 y 1) x 2 (y 2 1) = x 2 y = g 3 S 2,4 = (xy 2 x) x(y 2 1) = 0 S 3,4 = y 2 (x 2 y) x 2 (y 2 1) = y 3 + x 2 = yg 4 + g 3 Hence each of these are divisible by {g 1, g 2, g 3, g 4 }. The reader may verify that for every i = 1, 2, 3, 4, S i,5 is divisible by {g 1, g 2, g 3, g 4, g 5 }. Therefore, {g 1, g 2, g 3, g 4, g 5 } is a Gröbner basis for I. 10

11 4 Applications of Gröbner Bases As aforementioned, Gröbner bases provide a solution to the Ideal Membership Problem. If f is a polynomial and I is an ideal, then we can determine if f I by finding a Gröbner basis G for I and calculating the unique remainder of f divided by G. As in Corollary 9, this remainder is 0 if and only if f I. Example 7. Let I = (x+y +z, xy +xz +yz, xyz 1) and f = 2x 2 z 2 +2xyz 2 +2xz 3 +z 3 1. As proven in Example 5, a Gröbner basis for I is G = {x + y + z, y 2 + yz + z 2, z 3 1}. Performing long division, we see that f = 2xz 2 (x + y + z) + (z 3 1). Hence f I. Example 8. Let I = (xz y 2, x 3 z 2 ) and f = xy 5z 2 + x. With respect to Deglex, a Gröbner basis for I is G = {xz y 2, x 3 z 2, x 2 y 2 z 3, xy 4 z 4, y 6 z 5 }. Now, lm(g) = (xz, x 3, x 2 y 2, xy 4, y 6 ). Clearly, lm(f) = xy / lm(g), so therefore f / I. Another application of Gröbner bases lies in algebraic geometry. Smith [3] shows that the projective closure of an affine variety is the vanishing of the homogenization of the ideal. However, the homogenization of the ideal is not always equal to the ideal generated by the homogenization of the generators. Yet if the generating set is a Gröbner basis, these are equal. Proposition 14. Let I be an ideal in K[x 1,..., x n ], and let h(i) be its homogenization in K[x 1,..., x n, y]. Suppose G = {g 1,..., g m } is a Gröbner basis for I with respect to a graded monomial order, i.e., a monomial ordering that depends on multidegree of polynomials. Then h(g) = {h(g 1 ),..., h(g m )} is the Gröbner basis for h(i). Proof. Because h(g) h(i), it suffices to show that lm(h(i)) = (lm(h(g 1 )),..., lm(h(g m ))), or if f h(i) is homogeneous, then lm(f) (lm(h(g 1 )),..., lm(h(g m ))). Because f is homogeneous, f = y p h(g) for some h I and some p Z +. Now, since the monomial ordering is graded, for any h h(i), lm(h(h)) = lm(h). Hence, lm(f) = y p lm(h(g)) = y p lm(g). So lm(f) (lm((g 1 )),..., lm((g m ))) = (lm(h(g 1 )),..., lm(h(g m ))). 4.1 Elimination Theory A third application of Gröbner bases lies in elimination theory. Elimination theory gives a way to solve systems of polynomial equations by eliminating some of the variables from 11

12 some equations, and then back-solving. For example, if our system of polynomials is x 2 + y + z = 1 x + y 2 + z = 1 x + y + z 2 = 1 then we can consider the ideal I = (x 2 + y + z 1, x + y 2 + z 1, x + y 2 + z 1). A Gröbner basis for I is: g 1 = x + y + z 2 1 g 2 = y 2 y z 2 + z g 3 = 2yz 2 + z 4 z 2 g 4 = z 6 4z 4 + 4z 3 z 2 = z 2 (z 1) 2 (z 2 + 2z 1) Notice that g 4 is only in terms of z, so we can obtain the possible values of z from this equation. We see that z can be 0, 1, or 1 ± 2. Since both g 3 and g 4 are in terms of only y and z, we can substitute for z and obtain the possible values for y. From here, we can solve for the possible values of x. This system of equations has 5 solutions: (1, 0, 0), (0, 1, 0), (0, 0, 1), ( 1 + 2, 1 + 2, 1 + 2), ( 1 2, 1 2, 1 2). In solving this system of equations, the process can be divided into two parts. First, we eliminate variables, called the Elimination Step, and then we extend our solutions by backsolving, called the Extension Step. First, we study the Elimination Step. Note that observing that g 4 is only in terms of z can also be stated as g 4 I C[z]. Generalizing this leads to a definition. Definition 15. Let I = (f 1,..., f m ) K[x 1,..., x n ]. The kth elimination ideal I k is the ideal of K[x k+1,..., x n ] defined as I k = I K[x k+1,..., x n ] Solving the Elimination Step means finding polynomials in I that are not in terms of x 1,..., x k, which amounts to finding the kth elimination ideal. Gröbner bases makes this easy. 12

13 Theorem 16 (The Elimination Theorem). Let I be an ideal and G a Gröbner basis with respect to Lex, where x 1 > x 2 >... > x n. Then, for every 0 k n, the set G k = G K[x k+1,..., x n ] is a Gröbner basis for I k. Proof. Fix k. Since G k I k, so it suffices to show that (lm(g k )) = lm(i k ). Clearly, (lm(g k )) lm(i k ). Let f I k. We want to show that lm(f) is divisible by lm(g) for some g G k. Now, f I, so lm(f) divisible by lm(g) for some g G. Since f I k, this means that lm(g) involves only x k+1,..., x n. Since the ordering is Lex with x 1 > x 2 >... > x n, this means all terms of g only involve x k+1,..., x n, so g K[x k+1,..., x n ]. Hence g G k. Now we may turn our attention to solving the Extension Step. First, we define a partial solution. Definition 17. Let I be a systems of equations in K[x 1,..., x n ]. A solution (a k+1,..., a n ) V(I k ) is a partial solution of the original equations. When we solve the equation in one variable, we find a partial solution. To extend a partial solution, we want to find a coordinate a k such that (a k, a k+1,..., a n ) V(I k 1 ). If I k 1 = (g 1,..., g m ), then we want solutions x k = a k to the equations g 1 (x k, a k+1,..., a n ) = = g m (x k, a k+1,..., a n ) = 0. These are polynomials in one variable, so the possible a k is just the roots of the greatest common divisor of the m polynomials. However, they might not have a common root, so not all partial solutions extend to complete solutions. Example 9. Consider the following system of equations: { xy = 1 xz = 1 Clearly, I 1 = (y z). So, the partial solutions are of the form (a, a), since these are precisely the elements in I 1. Hence the complete solutions are (1/a, a, a), because x = 1/y = 1/x, if a 0. If a = 0, the system is not consistent, so this does not admit a complete solution. We extend a partial solution one coordinate at a time until we have completed it, so we only need to study extensions by one coordinate. For now, we shall restrict to the case where k = 1. 13

14 Theorem 18. Let I = (f 1,..., f m ) C[x 1,..., x n ]. For each 0 i m, write f i = g i (x 2,..., x n )x N i 1 + terms of degree < N i. Suppose (a 2,..., a n ) is a partial solution. If (a 2,..., a n ) / V(g 1,..., g m ), then there exists a 1 C such that (a 1,..., a n ) V(I). The proof of this uses resultants, so it will not be covered here. 1 This theorem means that the Extension Step fails only when the leading coefficients of the polynomials vanish simultaneously. To return to Example 9, we see that the leading coefficients of x are y and z in each polynomial. Thus V(g 1, g 2 ) = V(y, z), which contains only the point (0, 0), the lone point which cannot be extended. An easy corollary to this theorem occurs when the leading coefficients are constant. Corollary 19. Assume that for some i, f i = cx N 1 +terms of degree < N. Then if (a 2,..., a n ) V(I 1 ), then there exists a 1 such that (a 1, a 2,..., a n ) V(I). Proof. Follows immediately from c 0 V(g 1,..., g m ) = The Geometry of Elimination Now, we will study a geometric interpretation of elimination theory. Elimination corresponds to projecting a variety onto a lower dimensional subspace. Definition 20. Let V = V(f 1,..., f m ) C n. We define the projection map π k : C n C n k (a 1,..., a n ) (a k+1,..., a n ) Notice π k (V ) C n k. We can relate this projection map to the kth elimination ideal. Proposition 21. π k (V ) V(I k ). Proof. Let f I k and (a 1,..., a n ) V. Then f vanishes at (a 1,..., a n ), but f involves only x k+1,..., x n, so f(a k+1,..., a n ) = f(π k (a 1,..., a n )) = 0. Hence f vanishes at all points of π k (V ). 1 For a version of the proof, the reader can consult [1], page

15 21. Now, we have a precise definition for π k (V ), using our new knowledge from Proposition π k (V ) = {(a k+1,..., a n ) V(I k ) : a 1,..., a k C with (a 1,..., a k, a k+1,..., a n ) V }. So π k (V ) is exactly the partial solutions which extend to complete solutions. For instance, if we return to Example 9, we see that π k (V ) = {(a, a) C 2 : a 0}. Although π 1 (V ) is not necessarily an affine variety, the Closure Theorem provides a strong statement about the relationship between π k (V ) and V(I k ). Theorem 22 (The Closure Theorem). V(I k ) is the smallest affine variety containing π k (V ) C n k. Proof. We have shown that V(I(S)) is the smallest affine variety containing S, so we must show that V(I k ) = V(I(π k (V ))). By the lemma, we know π k (V ) V(I k ). Since V(I(π k (V ))) is the smallest variety containing π k (V ), this imples that V(I(π k (V ))) V(I k ). Suppose f I(π k (V )), so f(a k+1,..., a n ) = 0 for all (a k+1,..., a n ) π k (V ). Then f(a 1,..., a k, a k+1,..., a n ) = 0 for all (a 1,..., a n ) V, since f only involves x k+1,..., x n. By Nullstellensatz, f N I for some integer N. But since f does not depend on x 1,..., x k, neither does f N, so f I k. So, f I k, hence I(π k (V )) I k. Thus, V(I k ) = V( I k ) V(I(π k (V ))). This shows that the projection map projects a variety into a very specific lower-dimensional variety, which is in fact the Zariski closure of the projection. 15

16 References [1] Cox, David, John Little, and Donal O Shea. Ideals, Varieties, and Algorithms. Springer: New York, [2] Fröberg, Ralf. An Introduction to Gröbner Bases. John Wiley and Sons: Chichester, [3] Smith, Karen. An Invitation to Algebraic Geometry. Springer: New York,

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