By Rafael Villarreal

**Read Online or Download Monomial Algebras Villarreal PDF**

**Best algebra books**

**Structure and Representation of Jordan Algebras**

###############################################################################################################################################################################################################################################################

- Algebra Arrayán
- The Algebra of Logic
- A supplement for Category theory for computing science
- Lineare Algebra für Wirtschaftswissenschaftler: Mit Aufgaben und Lösungen
- Math Triumphs Foundations for Algebra 1 Teacher Edition Glencoe McGraw-Hill 2010

**Additional resources for Monomial Algebras Villarreal**

**Example text**

H i ) = d -i, for i = 1, . . , d, where dim(R/I) = d. If k is an infinite field and deg(a;j) = 1 for all i, then hi, . . ,hd can be chosen in RI . 38_______________________________________ Chapter 2 Proof. Assume d > 0. Let pi, . . , pr be the set of minimal primes of I of height g = ht (/). We claim that there is a homogeneous polynomial hi not in U[ =1 pj. To show it we use induction on r. Since pi • • - p r _ i <£. pr and because pj is graded we can pick / G pi • • -p r -i l~) R^ and / 0 p r , di > 0.

Ht(7). Proof. One may assume that there are Zi, . . , zn in 7? 6 are satisfied. We will show that g is equal to the height of 7. 5 the Zi's are algebraically independent. ,zn]) =g. /7). ,xn]/I. 8 (Noether Normalization Lemma) If R = k\x\ is a polynomial ring over a field k and I ^ R is an ideal, then there is an integral extension where hi,... ,hd are in R and d = dim(_R/7). Proof. ,zg) A R/I. To conclude the argument set hi = zg+i for i = 1 , . . , d. 9 If R = k[xi,... ,xn] is a polynomial ring over a field k, then R is a catenary ring.

0r is an M-regular sequence in m, then r < dim(M). Proof. By induction on dim(M). If dim(M) = 0, then m is an associated prime of M and every element of m is a zero divisor of M. 10 one has dim(M/0iM) < dim(M). Since 9%, . . ,9r is a regular sequence on M/OiM by induction one derives r < dim(M). 7 Let M be an R-module and let I be an ideal of R. (a) HomR(R/I, M) = (0) iff there is x 6 7 which is regular on M. (b) ExtrR(R/I,M) ~ HomR(R/I,M/0M), M-regular sequence in I. where 6 = 0 1 , . . , 0 r is any Proof, (a) =>) Assume 7 C Z(M).