M x N the following operations are set:
"(m_1,n_1)+(m_2,n_2)=(m_1+m_2,n_1+n_2)\\\\\nr*(m_1,n_1)=(r*m_1,r*n_1)\\\\\nr\\in R,\\ m_1,m_2\\in M,n_1,n_2\\in 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:
"(m_1,n_1)+(m_2,n_2)=(m_1+m_2,n_1+n_2)=\\\\\n=(m_2+m_1,n_2+n_1)=(m_2,n_2)+(m_1,n_1)"
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