Answer to Question #141829 in Abstract Algebra for anjali

Question

Let R be the ring and R^n= {(x1.......xn)/xi∈ R} be the R-module.
1] If I1,I2....In are ideal of R then N=I1xI2x....xIn={(x1.......xn)/xi∈ I} is a submodule in R^n

Answer 1

Let $R$ be the ring and $R^n= \{(x_1.......x_n)|x_i\in R\}$ be the $R$-module.

Let us show that if $I_1,I_2,...,I_n$ are ideals of $R$ then$N=I_1\times I_2\times\cdots \times I_n=\{(x_1,...,x_n)|x_i\in I\}$ is a submodule in $R^n$.

Taking into account that $I_1,I_2,...,I_n$ are ideals of a ring $R$, we conclude that $I_1,I_2,...,I_n$ are additive subgroups of a ring $R$, and therefore, $N=I_1\times I_2\times\cdots \times I_n$ is a additve subgroup of $R^n=R\times R\times\cdots \times R$. Let $r\in R$ and $(a_1,...,a_n)\in I_1\times I_2\times\cdots \times I_n$ be arbitrary. Since $I_1,I_2,...,I_n$ are ideals of a ring $R$, $ra_1\in I_1,..., ra_n\in I_n$ . Thus, $r(a_1,...a_n)=(ra_1,...,ra_n)\in I_1\times I_2\times\cdots \times I_n$ , and we conclude that $N$ is a submodule in $R^n$.