Question #141829
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
1
Expert's answer
2020-11-02T20:48:19-0500

Let RR be the ring and Rn={(x1.......xn)xiR}R^n= \{(x_1.......x_n)|x_i\in R\} be the RR-module.

Let us show that if I1,I2,...,InI_1,I_2,...,I_n are ideals of RR thenN=I1×I2××In={(x1,...,xn)xiI}N=I_1\times I_2\times\cdots \times I_n=\{(x_1,...,x_n)|x_i\in I\} is a submodule in RnR^n.


Taking into account that I1,I2,...,InI_1,I_2,...,I_n are ideals of a ring RR, we conclude that I1,I2,...,InI_1,I_2,...,I_n are additive subgroups of a ring RR, and therefore, N=I1×I2××InN=I_1\times I_2\times\cdots \times I_n is a additve subgroup of Rn=R×R××RR^n=R\times R\times\cdots \times R. Let rRr\in R and (a1,...,an)I1×I2××In(a_1,...,a_n)\in I_1\times I_2\times\cdots \times I_n be arbitrary. Since I1,I2,...,InI_1,I_2,...,I_n are ideals of a ring RR, ra1I1,...,ranInra_1\in I_1,..., ra_n\in I_n . Thus, r(a1,...an)=(ra1,...,ran)I1×I2××Inr(a_1,...a_n)=(ra_1,...,ra_n)\in I_1\times I_2\times\cdots \times I_n , and we conclude that NN is a submodule in RnR^n.


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