Lecture 1. (i,j) N 2 kx i y j, and this makes k[x, y]


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1 Lecture 1 1. Polynomial Rings, Gröbner Bases Definition 1.1. Let R be a ring, G an abelian semigroup, and R = i G R i a direct sum decomposition of abelian groups. R is graded (Ggraded) if R i R j R i+j for all i, j G. Similarly, let M = i G M i be an Rmodule. If R i M j M i+j for all i, j G then M is a graded Rmodule. M i is called the i th graded homogeneous component of M, and elements of M i are called homogeneous elements of degree i or i th forms. Example 1.2. Let k be a ring. Consider k[x] = kx n. Then k[x] is therefore N n N graded. Example 1.3. Let k be a ring. Consider k[x, y] = N 2 graded. (i,j) N 2 kx i y j, and this makes k[x, y] Remark 1.4. R 0 is a subring of R, and R 0 R as a direct summand (show this!). Also, each R i is a R 0 module because R 0 R i R i. The same is true for M i, since R 0 M i M i. Definition 1.5. Let M, N be graded Rmodules, and φ : M N where φ is Rlinear. Then φ is graded of degree d (sometimes called homogeneous if d = 0) if φ(m i ) N i+d for all i G. This gives the category of Rgraded modules where the objects are graded Rmodules and morphisms are graded homomorphisms of Rmodules. According to the definition, each x M, where M is an Rgraded module, can be written as a finite sum x = x i, where each x i 0, and x i M i. This is a unique representation and each x i has degree i G. By convention, 0 has arbitrary degree. We will call this expression the graded decomposition of x This notion is of great importance in module theory. statements by keeping track of the grading. The grading helps to prove Definition 1.6. Let R be a Ggraded ring and M be an graded Rmodule. We say that a submodule N in M is graded (or homogeneous) submodule if x N with graded decomposition x = x i, x i M i implies x i N for all i. 1
2 2 Proposition 1.7. Let R be a Ggraded ring and M be an graded Rmodule. Then a submodule N of M is homogeneous if and only if it is generated by homogeneous elements. Proof. Assume first that N is a homogeneous submodule. Let x be an elements from a list of generators of N. One can replace x by its homegenous components in the generating set, and therefore, by doing this with every generator, one will obtain a homogenous generating set. Conversely, let N =< S > where S is a set of homogenous elements for N. Let x N and write x = finite r ix i, x i S. Note that we assume deg(x i ) = i G. In the equality above look in degree α G: x α = s j x j where x α is the homogenous part of x in its graded decomposition, and s j, x j are homogeneous elements of degrees k j and j such that α = k j + j, x j S. Since x j are in N it follows that x α is in N too. LetIN = {0, 1, 2, 3,...}, the set of natural numbers. Let k be a field, and R = k[x 1,..., x n ] be the ring of polynomials in n variables with coefficients in k. Let f R, so f = finite a αx α, where x = x 1 x 2 x n, α = (α 1,..., α n ) IN n, a αi k, and a α 0. We use the notation x α = x α 1 1 x αn n. Definition 1.8. A monomial in R is an element of the form x α 1 1 x αn n = x α. If M = x α, then the multidegree of M is (α 1,..., α n ) N n. One should note that we have the following decomposition as a direct sum of kvector spaces R = α N nm α, where for all α, M α = kx α is an one dimensional kvector space with basis {x α }. Note that k R, and we customarily call an element of k a constant of R Monomial Orderings. We have noticed that for 1 variable polynomials over a field, the notion of degree has allowed one to prove several important results, such as the PID property. An important feature of the degree notion is that it provides a total order on monomials with certain additional properties.
3 For R = k[x 1,..., x n ], k field, we want to define an order on the monomials such that they totally ordered, and the order is preserved if monomials are multiplied by the same element. Also, we want comparability among all elements and to have a smallest element for a given subset. Definition 1.9. A monomial ordering on IN n is an order relation > such that: 3 (1) > is total (everything is comparable), (2) if α > β, then α + γ > β + γ for all γ IN n (this is the compatibility with multiplication), (3) Every subset of IN n has a least element. This monomial ordering on N n induces an ordering on the monomials in R = k[x 1,..., x n ], with k a field. A monomial has the form x α 1 1 x αn n with α = (α 1,..., α n ) IN n and x α x β if and only if α β. We obtain an order on the monomials of R with an order relation > such that it is a total order, it is additive, and every subset has a least element. It is often assumed by many authors that x 1 > x 2 > > x n, but this is not part of the definition. Also, for n 3, x, y, z are often used instead of x 1, x 2, x 3. Consider the following three monomial orderings: Definition Let α, β IN n. (1) Lexicographical Order (lex) In this order, x α > x β if the leftmost nonzero entry of α β is positive. For example, x 2 y > lex xy 5 z because (2, 1, 0) (1, 5, 1) = (1, 4, 1), and the leftmost entry is 1 which is positive. Also, x 2 y 3 z > lex x 2 yz 8 since (2, 3, 1) (2, 1, 8) = (0, 2, 7), and the leftmost nonzero entry, 2, is positive. Here, x > y > z. (2) Graded Lexicographical Order (glex, hlex, grlex) In this order, x α > x β if i α i > i β i or if i α i = i β i and α > lex β. For example, x 2 y 2 z > glex x 3 y, and x 3 yz > glex x 2 x 2 z. Here x > y > z. (3) Graded Reverse Lex Order (grevlex) In this order, x α > x β if i α i > i β i or if i α i = i β i and the rightmost nonzero entry of α β is negative. For example, x 3 y 2 z 5 < x 4 y 6, and x 2 y 2 z 5 < xy 4 z 4. Also, x > y > z.
4 4 Remark The natural question is what about a reverse lex order (instead of graded reverse lex). This would be defined as α > β if the rightmost nonzero entry of α β is negative. This is not a monomial ordering since there is no least element, i.e., consider {xy a } a. As a increases, the corresponding monomial gets smaller and smaller without bound. There is no least element in this set. Fix a monomial ordering on R, say >. Take f R. Then we write f in a standard form such that the monomials appear in decreasing order, i.e., f = a α x α +. Definition We call α the multidegree of f, x α the leading monomial of f, a α the leading coefficient (LC) of f, and a α x α the leading term (LT) of f. Given a polynomial ring, we want to know how to divide two or more polynomials into another polynomial simultaneously. This is done with a division algorithm, after fixing a monomial ordering Division Algorithm. Fix a monomial ordering on IN n (and hence on k[x 1,..., x n ]) and an ordered mtuple (g 1,..., g m ) with g i R = k[x 1,..., x n ] for every i = 1,..., m. The division algorithm allows us to claim thar every f R can be written as f = a 1 g a m g m + r, with a i R, and either r = 0, or r is a klinear combination of monomials, none of which are divisible by LT(g 1 ),..., LT(g m ). Moreover, if a i g i 0, then the multidegree of f is greater or equal to the multidegree of g i for all i = 1,..., m. We present now the division algorithm. Essentially, the algorithm allows us to simultaneously divide f(x) by an ordered set of polynomials {g 1 (x),..., g m (x)}. It proceeds as follows: (1) Compute LT (f). If LT (g 1 ) LT (f), then write h = f hg 1. Repeat with f LT (g 1 ) LT (f). LT (f) LT (g 1 ) and set f = f 1 := (2) Move to g 2 and apply step 1 to f. If LT (g 2 ) LT (f i ), then set f 1 = f hg 2. Now start over with testing g 1 into f 1, and moving through (1) and (2) with the g i s until some LT (g j ) does not divide LT (f 1 ). So for each new f k the algorithm starts again with g 1 and moves through to g i, i = 1,..., m.
5 (3) At some point, no leading terms of any the g i s will divide LT (f k ). Write f = a 1 g a m g m + r where LT (r) is not divisible by any of the LT (g i ). Repeat the previous steps to f k+1 = r LT (r). 5 Example Fix the lex monomial ordering on k[x, y]. We want to divide {g 1 = y 2, g 2 = xy + 1} into f = x 2 y + xy 2 + xy, all taking place in k[x, y]. Since LT (g 1 ) does not divide LT (f) we move to LT (g 2 ), and we write f 1 = x 2 y + xy 2 + xy x(xy + 1) = xy 2 +xy x. Now we can go back to g 1 to get f 2 = xy 2 +xy x xy 2 = xy x. The leading term of g 1 does not divide x 2 y so we move to g 2. We get f 3 = xy x (xy + 1) = x 1. Now neither of the LT (g i ) divide LT (f 3 ), so we work with r = x 1. Note that LT(r) is not divisible by LT(g 1 ), LT(g 2 ), so we stop here. We obtain f = xg 1 + (x + 1)g 2 x Monomial Ideals Definition 2.1. Let R = k[x 1,..., x n ], where x α = x α 1 1 x αn n. monomial ideal if a α x α a x α a for every a α 0. An ideal a R is a In other words, a monomial ideal is a homogeneous ideal of R under the N n grading. If a is a monomial ideal in R, we can set A = {α x α a} IN n. Then A has the property that α A if and only if α + γ A for every γ IN n. This induces a correspondence between monomial ideals a and sets A IN n that satisfy the condition mentioned above. Lemma 2.2. Assume a = (x α α A). Then x β a if and only if α A such that x α x β. Proof. One direction is clear. Now let x β a. Then x β = i I f i x α i, where the sum above is finite, f i R and α i A for all i I.
6 6 Two polynomials are equal if and only if they are equal in each multidegree. So in the above equality look in degree β. This will give x β = M j x α j, j J where M j are monomials, and J I. This implies in particular that for all j J x α j x β. Proposition 2.3. Let a be a monomial ideal. Then there exists A N n such that a = (x α : α A). Proof. Write a = (f i : f i R, i I), where I is a possibly infinite set. By the definition of a monomial ideal, we have that every monomial that appears in f i belongs to a. So we can replace the set of generators {f i } i I by the set of generators consisting of the monomials appearing in each f i, i I. This proves the statement. Theorem 2.4. (Dickson) Any monomial ideal is generated by a finite set of monomials. Proof. We will prove this by induction on the number of variables n in R. When n = 1, R is a PID so the result is clear. Let a be a monomial ideal in R = k[x 1,..., x n ]. By the Proposition above, since a is generated by monomials so we can write a = (x α : α A). For every k A, and i = 1,..., n let us write a {i,k} = (x = x α 1 1 x α 2 2 x α i 1 i 1 xα i+1 i+1 xαn n : x x i + a). These ideals are generated by monomials in n 1 variables so by the induction hypothesis they must be finitely generated. Let us denote a finite generating set of monomials for them by G {i,k}. Now choose α A, so x α a. Let x β a. If β i α i for all i = 1,..., n then x α x β. If not, then there exists i such that α i > β i.
7 7 But then x β = x x β i i z x, so zx β i i x β. with x a {i,βi }. So, there exists an element z in G {i,βi } such that In conclusion a finite set of generators for a is given by x α and all products of the form z x β i i where z G {i,βi } and β i α i, i = 1,..., n. Theorem 2.5. (Hilbert Basis Theoremspecial case)any nonzero ideal I k[x 1,..., x n ] can be generated by a finite set of elements. Proof. Let LT (I) be the ideal generated by the leading terms of all f I. Then by the Dickson lemma, LT (I) = LT (g 1 ),..., LT (g s ). Take f I and apply the division algorithm to f, (g 1,..., g s ) to obtain f = a 1 g a s g s + r, and since r I, then LT (r) LT (I). This implies that k such that LT (g k ) LT (r), a contradiction unless r = 0. So I = (g 1,..., g s ). The set LT (g 1 ),..., LT (g s ) that appeared in the proof of the Hilbert Basis Theorem proved to be an important tool in commutative algebra and we will formally introduce it now. Definition 2.6. Fix a monomial ordering and consider an ideal a R = k[x 1,..., x n ]. Then a finite set S = {g 1,..., g s } a is a Gröbner basis (GB) for a if LT (g) : g a = LT (g 1 ),..., LT (g s ). Theorem 2.7. Any nonzero ideal a of R has a GB and this basis generates the ideal. Proof. Exactly as for the Hilbert basis theorem, it is easy to see that a has a GB. Now if f a, apply the division algorithm to obtain f = a 1 g a s g s + r. If r 0 then r a implies LT (r) LT (g 1 ),..., LT (g s ), a contradiction. It is important to realize that for an arbitrary set of generators for an ideal I their leading terms do not necessarily generate LT (I). Example 2.8. Let g 1 = x 2 y + x 2y 2, g 2 = x 3 2xy, with respect to the grevlex order. Then a = g 1, g 2, and x 2 = xg 1 yg 2 a. So LT (g 1 ), LT (g 2 ) = x 2 y, x 3 which does not contain x 2.
8 8 Proposition 2.9. Let S = {g 1,..., g s } be a GB for a. Then f R,! r R such that f = g + r for some g a and no term of r is divisible by any of the LT (g i ) s. Proof. For existence, use the division algorithm to get the r so that f = a 1 g 1 + +a s g s +r. Let g = f r. For uniqueness, assume f = g 1 + r 1 = g 2 + r 2 for g 1, g 2 a. Then r 2 r 2 = g 2 + g 1 a. This implies LT (r 2 r 1 ) LT (a) = LT (g 1 ),..., LT (g s ). Thus LT (r 2 r 1 ) is some monomial in r 2 or r 1, a contradiction. Corollary With the same notation as above, f a if and only if r = 0 when f is divided by S. With the definition above it is known that Gröbner bases are not unique for a given ideal. Therefore, we will refine the definition in order to obtain this property. Definition A GB = {g 1,..., g s } is minimal if LC (g i ) = 1 and LT (g i ) / (LT (g i j i). Note that an ideal can admit infinitely many minimal GB s, but it is a fact that all minimal GB bases for an ideal a have the same cardinality. Definition A GB is reduced if LC (g i ) = 1 and for every g i GB, no monomial appearing in g i is in LT (g 1 ),..., LT (g i 1 ), LT (g i+1 ),..., LT (g s ). The following important fact is stated here without proof. Proposition Each nonzero ideal a has a unique reduced GB with respect to a given monomial ordering.
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