Proposition 4.6. (Division with remainder) Let \(f, f_1,\ldots , f_s \in k[X_1,\ldots ,X_n]\) be non-zero polynomials. Then there exist polynomials \(h_1,\ldots , h_s, r \in k[X_1,\ldots ,X_n]\) such that \[ f = h_1 f_1 + h_2 f_2 + \ldots + h_s f_s + r \] with the following properties:
Proof. We argue by induction on \(LM(f)\): by Lemma 4.5 there is no infinite strictly decreasing sequence of monomials, so we may assume that the statement holds for every non-zero polynomial whose leading monomial is smaller than \(LM(f)\).
Suppose first that there exists an \(i\) such that \(LT(f_i) | LT(f)\), and let \(i \in \{1,\ldots ,s\}\) be the smallest such index. Let \[ \tilde {f} = f - \frac {LT(f)}{LT(f_i)} f_i. \] If \(\tilde {f}=0\), we stop, setting \(h_i = LT(f)/LT(f_i)\), \(h_j = 0\) for \(j \neq i\) and \(r=0\). Otherwise, we have \(LT(\tilde {f}) < LT(f)\), so by induction we can write \[ \tilde {f} = \tilde {h}_1 f_1 + \ldots + \tilde {h}_s f_s + r \] with the required conditions. Then \[ f = h_1 f_1 + \ldots + h_s f_s + r \] where \(h_j = \tilde {h}_j\) for \(j \neq i\) and \(h_i = \tilde {h}_i + \frac {LT(f)}{LT(f_i)}\). The conditions on \(r\) are inherited from \(\tilde {f}\), and \(LT(h_j f_j) \leq LT(f)\) for all \(j\) because \(LT(\tilde {h}_j f_j) \leq LT(\tilde {f}) < LT(f)\) and \(LT\big (\frac {LT(f)}{LT(f_i)} f_i\big ) = LT(f)\).
Suppose now that no \(LT(f_i)\) divides \(LT(f)\), and let \(\tilde {f} = f - LT(f)\). If \(\tilde {f}=0\), we stop, setting all \(h_i=0\) and \(r=f\). Otherwise, by induction we can write \(\tilde {f} = h_1 f_1 + \ldots + h_s f_s + \tilde {r}\) with the required conditions, and we set \[ f = h_1 f_1 + \ldots + h_s f_s + r \] where \(r= \tilde {r} + LT(f)\). The term \(LT(f)\) is divisible by no \(LT(f_i)\) by assumption, and \(LT(r) = LT(f)\) since every term of \(\tilde {r}\) is at most \(LT(\tilde {r}) \leq LT(\tilde {f}) < LT(f)\). □
Example. Let \(f = XY^2-X\), \(f_1 = XY+1\), \(f_2 = Y^2-1\), and use the lex order with \(X > Y\). The leading terms are \(XY^2, XY, Y^2\) respectively. Since \(XY | XY^2\), we can write \(\tilde {f} = f - Yf_1 = -X -Y\). Neither \(X\) nor \(Y\) is divisible by \(XY\) or \(Y^2\), hence \[ f = Y f_1 + 0 f_2 + (-X -Y). \] Another possibility is to use that \(Y^2 | XY^2\); then \(\tilde {f} = f - Xf_2 = 0\), so we get \[ f = 0 f_1 + X f_2 + 0. \] Hence the remainder \(r\) is not unique: it depends on the order in which \(f_1,f_2,\ldots , f_s\) are listed.
We can implement the division with remainder algorithm in Macaulay2 as follows.
div = (f,G) -> ( R := ring f; p := f; r := 0_R; s := #G; Q := new MutableHashTable; for j from 0 to s-1 do Q#j = 0_R; while p != 0 do ( i := 0; while i < s and leadTerm(p) % leadTerm(G#i) != 0 do i = i+1; if i < s then ( Q#i = Q#i + (leadTerm(p) // leadTerm(G#i)); p = p - (leadTerm(p) // leadTerm(G#i)*G#i)) else ( r = r + leadTerm(p); p = p - leadTerm(p))); L := apply(s, j -> Q#j); return (r,L));
Using the above division function, you can do division in Macaulay2 as follows.
R = QQ[X,Y, MonomialOrder => Lex]
f= X*Y^2 - X
f1 = X*Y +1
f2 = Y^2 - 1
G = {f1,f2}
div(f,G)
G = {f2,f1}
div(f,G)
Definition 4.7. (Buchberger) Let \(I \subset k[X_1,\ldots ,X_n]\) be a non-zero ideal. The initial ideal of \(I\) is the monomial ideal \[ \ini (I) \coloneqq ( LT(f) : 0 \neq f \in I ). \] A finite subset \(G \subset I \setminus \{0\}\) is called a Gröbner basis of \(I\) (with respect to the fixed monomial ordering) if the leading terms of its elements generate the initial ideal, that is, \(\ini (I) = (LT(g) : g \in G)\).
\(\blacktriangleright \) Every non-zero ideal \(I\) has a Gröbner basis. Indeed, \(\ini (I)\) is finitely generated by the Hilbert basis theorem, and each of its finitely many generators is a combination of finitely many of the \(LT(f)\) with \(0 \neq f \in I\). Hence \(\ini (I) = (LT(g_1),\ldots ,LT(g_s))\) for some \(g_1,\ldots ,g_s \in I \setminus \{0\}\).
\(\blacktriangleright \) If \(I = (f_1,f_2,\ldots , f_s)\), then \(\ini (I)\) need not equal \((LT(f_1), \ldots , LT(f_s))\).
\(\sphericalangle \) Take \(I = (f_1,f_2)\) with \(f_1= X^3-2XY\) and \(f_2 = X^2Y - 2Y^2 + X\) and use \(\leq _{grlex}\). Show that \((LT(f_1), LT(f_2))\) is strictly contained in \(\ini (I)\).
If you want to check your calculations, you can ask Macaulay2 for help as follows:
R = QQ[x,y, MonomialOrder=> GLex]; I = ideal (x^3-2*x*y, x^2*y-2*y^2+x); inI = ideal leadTerm I
Note that the initial ideal depends on the monomial order you choose.
R = QQ[x,y, MonomialOrder=> Lex]; I = ideal (x^3-2*x*y, x^2*y-2*y^2+x); inI = ideal leadTerm I
Theorem 4.8. (Buchberger) Let \(G=\{g_1,\ldots , g_s\}\) be a Gröbner basis of a non-zero ideal \(I \subset k[X_1,\ldots , X_n]\).
Proof. (1) Suppose \(f \in I\) and \(r \neq 0\). Then \(r = f- \sum _{i=1}^s h_i g_i \in I\), and since \(G\) is a Gröbner basis, \[ LT(r) \in \ini (I) = (LT(g_1), LT(g_2),\ldots , LT(g_s)). \] By Lemma 4.4, there exists \(i\) such that \(LT(g_i) | LT(r)\). This contradicts the properties of the division with remainder. Hence \(r=0\). Conversely, if \(r = 0\) then \(f = \sum _{i=1}^s h_i g_i \in (G) \subset I\). It also follows that \(I= (G)\), since every \(f \in I\) is of the form \(\sum _{i=1}^s h_i g_i\).
Suppose \(f = \sum _{i=1}^s h_i g_i + r = \sum _{i=1}^s k_i g_i + \tilde {r}\) are two expressions obtained by division with remainder, with the \(g_i\) listed in different orders. Then \(\sum _{i=1}^s (h_i - k_i) g_i = \tilde {r} - r \in I\). If \(\tilde {r} \neq r\), then \(LT(\tilde {r} - r) \in \ini (I) = (LT(g):g \in G)\), hence there exists \(i\) such that \(LT(g_i) | LT(\tilde {r} - r)\). But every term of \(\tilde {r} - r\) is, up to a scalar, a term of \(r\) or of \(\tilde {r}\), and none of these is divisible by any \(LT(g_i)\). This is a contradiction, hence \(r= \tilde {r}\).
(2) Let \(f \in k[X_1,\ldots ,X_n]\) be non-zero. After division with remainder, none of the terms of \(r\) is divisible by any \(LT(g_i)\), so \(r\) is a \(k\)-linear combination of monomials \(X^\alpha \) with \(\alpha \in C(I)\). Since \(f + I = r + I\), the quotient \(k[X_1,\ldots , X_n]/I\) is spanned by \(\{X^\alpha + I : \alpha \in C(I) \}\).
Suppose that \(\sum _{i=1}^m a_i X^{\alpha _i} \in I\) for distinct \(\alpha _1,\ldots ,\alpha _m \in C(I)\) and \(a_i \in k\). If this polynomial is non-zero, dividing it by \(g_1,\ldots , g_s\) gives \[ \sum _{i=1}^m a_i X^{\alpha _i} = 0 \cdot g_1 + 0 \cdot g_2 + \ldots + 0 \cdot g_s + r \] with \(r = \sum _{i=1}^m a_i X^{\alpha _i}\), since no term of it is divisible by any \(LT(g_i)\). By part (1), \(r=0\), a contradiction. Hence all \(a_i = 0\), and the \(X^\alpha +I\) for \(\alpha \in C(I)\) are linearly independent. □
\(\blacktriangleright \) Although \(C(I)\) is defined using \(G\), it depends only on \(I\) and the monomial ordering: by Lemma 4.4, \(\alpha \in C(I)\) if and only if \(X^\alpha \notin \ini (I)\). The monomials \(X^\alpha \) with \(\alpha \in C(I)\) are called the standard monomials of \(I\).
Corollary 4.9. The variety \(\mathcal {V}(I) \subset k^n\) defined by \(I\) has finitely many points if \(C(I)\) is finite.
Proof. By the theorem, if \(C(I)\) is finite then the ring \(k[X_1,\ldots ,X_n]/I\) is finite-dimensional as a \(k\)-vector space. Now fix \(i\) and consider the monomials \(1,X_i,X_i^2, \ldots \). Their images in \(k[X_1,\ldots , X_n]/I\) are linearly dependent, so there exist \(m\) and \(c_0, c_1,\ldots , c_m \in k\), not all zero, such that \[ c_0 + c_1 X_i + c_2 X_i^2 + \ldots + c_m X_i^m \in I. \] This polynomial vanishes on \(\mathcal {V}(I)\), but a non-zero polynomial in one variable has at most finitely many roots. It follows that the \(i\)-th co-ordinates of the points of \(\mathcal {V}(I)\) take only finitely many values. The same is true for every \(i\), hence \(\mathcal {V}(I)\) is finite. □
Theorem 4.8 gives an algorithm for solving the ideal membership problem provided that we know a Gröbner basis for the ideal.
Problem: How to construct a Gröbner basis?
Definition 4.10. Let \(f,g \in k[X_1,\ldots , X_n]\) be non-zero polynomials. Let \(X^\alpha = LM(f)\) with \(\alpha = (\alpha _1,\ldots , \alpha _n)\) and \(X^\beta = LM(g)\) with \(\beta =(\beta _1,\ldots ,\beta _n)\). Define \(\gamma =(\gamma _1,\ldots , \gamma _n)\) with \(\gamma _i = \mathrm {max} \{ \alpha _i,\beta _i \}\) for all \(i\), i.e. \(X^\gamma = \mathrm {lcm}(LM(f),LM(g))\). The \(S\)-polynomial of \(f\) and \(g\) is defined to be \[ S(f,g) = \frac {X^\gamma }{LT(f)} f - \frac {X^\gamma }{LT(g)} g \]
Example. Let \(f = 4X^2 Z - 7 Y^2\) and \(g = XYZ^2 + 3XZ^4\). Using the \(\leq _{lex}\) order with \(X > Y > Z\), we see that \(\mathrm {lcm}(LM(f),LM(g)) =X^2 Y Z^2\), and \[ S(f,g) = \frac {X^2 Y Z^2}{4 X^2 Z} f - \frac {X^2 Y Z^2}{X Y Z^2} g = (1/4) YZ f - X g = -3X^2 Z^4 - (7/4) Y^3 Z \]
Here is how you may do this in Macaulay2:
R = QQ[x,y,z, MonomialOrder => Lex]; f = 4*x^2*z-7*y^2; g = x*y*z^2 + 3*x*z^4; Spoly = (f,g) -> ( m := lcm(leadMonomial f, leadMonomial g); (m // leadTerm f) * f - (m // leadTerm g) * g); Spoly(f,g)
\(\blacktriangleright \) If \(S(f,g) \neq 0\), then \(LM(S(f,g)) < \mathrm {lcm}(LM(f), LM(g))\). In this sense, we can say that the \(S\)-polynomial is designed to produce cancellation of leading terms.
The name \(S\)-polynomial is short for “syzygy polynomial”. You can blame Sylvester if you don’t know how to pronounce the word syzygy. In general, syzygy is a kind of linear relation. Here it is a linear relation between polynomials in several variables with polynomial coefficients. In astronomy, a syzygy is an alignment of three celestial bodies, such as the Sun, the Earth and the Moon during an eclipse.