The Annihilator Graph of a Ring一个环零化子图.pptVIP

New Material The proximity of the points x2y4 and x4y2 causes Ann(x2y4)=P(MI)-{1}=Ann(x4y2). The main consequence of this is that we lose the one to one correspondence between the antichains and the nodes of the graph when we add mixed monomials to the monomial ideal. New Material Definition: In R = D[x1,…, xn]/I, where I is a monomial ideal such that each xik exists as one of the generators of I. We say a corner is any monomial, X, such that xiX=0 for all i. As such, Ann(X) = P(MI) – {1}. New Material Definition: In R = D[x1,…, xn]/I, where I is a monomial ideal such that each xik exists as one of the generators of I. We say a corner is any monomial, X, such that xiX=0 for all i. As such, Ann(X) = P(MI) – {1}. This leads us to the following lemma. New Material Lemma: Suppose R = D[x1,…, xn]/I, with I as before, then in Гa(R), the antichains made from the set of the corners are represented by the vertex with maximal degree. New Material Lemma: Suppose R = D[x1,…, xn]/I, with I as before, then in Гa(R), the antichains made from the set of the corners are represented by the vertex with maximal degree. Furthermore, for X a corner, and xk the variable of highest degree in X, if X=xkXk, then Xk is the monomial whose vertex has the second highest degree in Гa(R). New Material Proof outline: The annihilator of a corner is P(MI) - {1}, and as such, the degree of the vertex represented by a corner is |Гa(R)|. For the second half of the lemma, we need to look at the monomial Xk. We prove this is in fact the second highest degree by looking at how much less than the maximal degree it is. Proof Outline (cont.) We can’t calculate directly how much less it is, but we can describe it in terms of antichains. If deg(X)=D, then deg(Xk)=D-Dk for some Dk. The way Xk is defined, we can see that Dk is in fact exactly the number of antichains in MI which do not contain the variable xk. Since xk is maximal, this minimizes Dk. Theorem Suppose R1 = D[x1,…, xn]/I, and R2 =