By Kreuzer and Robbiano

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**Extra resources for Computational Commutative Algebra**

**Example text**

We recall that a monoid is a set together with an associative operation on it such that there exists an identity. Since in all the cases considered in this book the operation will be commutative, we shall from now on use the term “monoid” to denote a commutative monoid. 1. Let (Γ, ◦) be a monoid. a) A non-empty subset ∆ ⊆ Γ is called a monoideal (pronounced “monoideal”) in Γ (or a monoid ideal in Γ ) if we have ∆ ◦ Γ ⊆ ∆ . b) A subset B of a monoideal ∆ in Γ is called a system of generators of ∆ (or ∆ is said to be generated by B ) if ∆ is the smallest monoideal in Γ containing B .

M, let fi = ci p∈P pαpi be the factorization of fi . Using the definition and induction on m , we see that gcd(f1 , . . ,αpm } . Thus it follows immediately that gcd(f1 , . . , fm ) divides fi for i = 1, . . , m . Now let g ∈ R be a common divisor of f1 , . . , fm , and let g = c p∈P pβp be the factorization of g . For every i ∈ {1, . . , m} , the condition g | fi implies that βp ≤ αpi for all p ∈ P . Hence we get βp ≤ max{αp1 , . . , αpm } for all p ∈ P , and therefore g | gcd(f1 , .

Show that the following conditions are equivalent. a) The ring R is a field. b) Every finitely generated R -module is free. c) Every cyclic R -module is free. Exercise 8. Let K be a field, P = K[x1 , x2 ] , and I be the ideal in P generated by {x1 , x2 } . Show that I is not a free P -module. Tutorial 1: Polynomial Representation I In what follows we work over the ring K[x, y], where K is one of the fields defined in CoCoA. 8, we see that we can represent every polynomial f ∈ K[x, y] as a list of lists, where a univariate polynomial a0 + a1 x + · · · + ad xd such that a0 , .