By Bernd Sturmfels, Ezra Miller

Contemporary advancements are lined comprises over a hundred figures and 250 workouts comprises whole proofs

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**Additional resources for Combinatorial Commutative Algebra (Graduate Texts in Mathematics, Volume 227)**

**Example text**

Xn−1 ]d−i . Let C[i] denote the lex segment in k[x1 , . . , xn−1 ]d−i of the same cardinality as B[i]. Set C = C[0] ∪ xn · C[1] ∪ x2n · C[2] ∪ · · · ∪ xdn · C[d] . 36 CHAPTER 2. 25 is true in n − 1 variables, so we have inequalities µ≤j (C[i]) ≤ µ≤j (B[i]) for all i, j. 4) We claim that C is a Borel set. Since B is a Borel set, {x1 , . . , xn−1 }B[i] is a subset of B[i−1]. 3) implies n−1 |{x1 , . . , xn−1 } · C[i]| = n−1 µj ({x1 , . . , xn−1 } · C[i]) = j=1 n−1 ≤ µ≤j (C[i]) j=1 µ≤j (B[i]) j=1 n−1 µj ({x1 , .

All six ideals J above are Borel-ﬁxed. Let us conclude this section with one more generality on Gr¨obner bases: they work for submodules of free S-modules. Suppose that F = S β is a free module of rank β, with basis e1 , . . , eβ . There is a general deﬁnition of term order for F, which is a total order on elements of the form mei , for monomials m ∈ S, satisfying appropriate analogues of the multiplicative and artinian properties of term orders for S. Initial modules are deﬁned just as they were for ideals (which constitute the case β = 1).

Xn−1 } · C[i] ⊆ C[i − 1], which means that C is a Borel set in Sd . Since L is a lex segment and since |L| ≤ |B| = |C|, the lexicographically minimal monomials in C and L respectively satisfy min(C) ≤lex lex min(L). lex Since both C and L are Borel-ﬁxed, this implies that min(C[0]) ≤lex lex min(L[0]). lex Thus L[0] ⊆ C[0] since both are lex segments in k[x1 , . . , xn−1 ]d . 5) which completes the proof for i = n − 1. Finally, consider the case i ≤ n − 2. 25 can be applied inductively to the sets B[0] and L[0] to get µ≤i (L) = µ≤i (L[0]) ≤ µ≤i (B[0]) = µ≤i (B) for 1 ≤ i ≤ n − 2.