Question #4998

Let R and S be rings. The direct product of R and S,
R ⊕ S = {(r, s) : r ∈ R, s ∈ S}
is a ring, where
(r1, s1) + (r2, s2) = (r1 + r2, s1 + s2)
(r1, s1) x (r2, s2) = (r1 x r2, s1 x s2)
(a) List the elements of Z2 ⊕ Z3 and Z3 ⊕ Z2.
(b) Are they structurally the same (that is isomorphic)? If so, how should an
element in Z2 ⊕ Z3 be identified in Z3 ⊕ Z2?

Expert's answer

the elements of Z_{2}is 0 and 1 let denote it by 0_{Z2} and 1_{Z2}.

the elements of Z_{3}is 0, 1 and 2 let denote it by 0_{Z3}, 1_{Z3}.,2_{Z3}

a) elements of Z_{2}⊕ Z_{3} are the pairs

(0_{Z2 },0_{Z3}), (0_{Z2 }, 1_{Z3}), (0_{Z2 }, 2_{Z3}),(1_{Z2 }, 0_{Z3}), (1_{Z2 }, 1_{Z3}), (1_{Z2}, 2_{Z3})

elements of Z_{3}⊕ Z_{2} are the pairs

(0_{Z3 },0_{Z2}), (0_{Z3 }, 1_{Z2}), (1_{Z3 }, 0_{Z2}),(1_{Z3 }, 1_{Z2}), (2_{Z3 }, 0_{Z2}), (2_{Z3}, 1_{Z2})

b) this rings are isomorphic because we can show function

f: Z_{2}⊕ Z_{3}_{} -->

Z_{3}⊕ Z_{2 }that permute coordinates

for x from Z_{2}and y from Z_{3} f((x,y))=(y,x).

this function is a injection and surjection and it preserve operations:

for x,z from Z_{2}and y,t from Z_{3}

f((x,y)+(z,t))=f((x+z,y+t))=(y+t,x+z)=(y,x)+(t,z)=f((x,y))+f((z,t)).

f((x,y)*(z,t))=f((x*z,y*t))=(y*t,x*z)=(y,x)*(t,z)=f((x,y))*f((z,t))

So we have that rings isomorphic

the elements of Z

a) elements of Z

(0

elements of Z

(0

b) this rings are isomorphic because we can show function

f: Z

Z

for x from Z

this function is a injection and surjection and it preserve operations:

for x,z from Z

f((x,y)+(z,t))=f((x+z,y+t))=(y+t,x+z)=(y,x)+(t,z)=f((x,y))+f((z,t)).

f((x,y)*(z,t))=f((x*z,y*t))=(y*t,x*z)=(y,x)*(t,z)=f((x,y))*f((z,t))

So we have that rings isomorphic

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