M x N the following operations are set:
(m1,n1)+(m2,n2)=(m1+m2,n1+n2)r∗(m1,n1)=(r∗m1,r∗n1)r∈R, m1,m2∈M,n1,n2∈N
Since operations on M x N are "distributed" into two parts in M and N separately, it is quite obvious that M x N is an R-module. For example, let's check the commutativity:
(m1,n1)+(m2,n2)=(m1+m2,n1+n2)==(m2+m1,n2+n1)=(m2,n2)+(m1,n1)
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