Let be a monomial order. Let denote the leading term of with respect to and let denote the multidegree of .
- Prove that and for any nonzero .
- Prove that with equality if .
- Prove that for any monomial .
- Prove that if divides , then . Deduce that the leading term of a polynomial does not divide any of its lower terms.
We will begin with a more concrete definition of monomial order.
Let be a field and let . We define the monomials in to be the set . Note that is closed under multiplication in , so that is a subsemigroup (in facr a submonoid) of . Now is also a semigroup under pointwise addition. Define by ; certainly is surjective, and moreover is a semigroup homomorphism. Also, we have if and only if and are associates in (and thus associates in ). (In the language of semigroups, the kernel of is the associate congruence.)
Note that every relation on induces a relation on as follows: if and only if .
Definition: We will say that a relation on is an m-order if the following hold.
- is a linear order; that is, reflexive, antisymmetic, transitive, and total.
- Every nonempty subset of contains a -minimal element. That is, for every nonempty , there exists such that for all , .
- If , then .
A relation on is called a monomial order if it is induced by some m-order on .
Lemma: Let be a monomial order on . Then the following hold for all .
- is reflexive, transitive, and total.
- and if and only if and are associates in .
- Every nonempty subset of contains a -minimal element. That is, for every nonempty subset , there exists such that for all , .
- If , then .
These follow directly from the definitions of m-order and .
Lemma: Let be an m-order on . Then 0 is minimal with respect to . Proof: Suppose we have . Note that for all , where we interpret as the -fold sum of . Now the multiples of form a nonempty ssubset of , which must contain a minimal element . That is, . Because is cancellative, we have . Thus is minimal with respect to .
Corollary: If is a monomial order on , then is minimal with respect to for all . In particular, for all .
Definition: Let be a field, a positive natural number, , and let be a monomial order on . Let . Note that contains a unique maximal element with respect to ; this is called the leading term of and denoted . The multidegree of , denoted , is where the right hand delta is the semigroup homomorphism defined above.
Let and be in , with a field, and let be a monomial order on . Note that . Say and . Then for all and for all . So for all , we have ; even after is simplified, is maximal among the terms. Thus . By definition, then, .
Now suppose ; say . Since , we have ; thus if then . If is greatest among a set of terms, then, it cannot divide any of them.