Answer to Question #109443 in Abstract Algebra for Suraj Singh

Question

Prove that (R^5)/R ~_ R^4 as rings

Answer 1

Let's consider the following structure of R⁵, R⁴ and R:

R⁵ = {(r₁, r₂, r₃, r₄, r₅) | r_i ∈ R for i = 1,2,3,4,5},

R⁴ = {(r₁, r₂, r₃, r₄) | r_i ∈ R for i= 1,2,3,4},

R = {(0, 0, 0, 0, r) | r ∈ R }, with ring operations applied to each component(i.e. (r₁, r₂, r₃, r₄, r₅) +(or *) (p₁, p₂, p₃, p₄, p₅) = (r₁+(*)p₁, ..., r₅+(*)p₅) and same for R⁴ and R).

It could be inferred(from the expressions, represented above) that R⁵/R = {(r₁, r₂, r₃, r₄, 0) | r_i ∈ R for i = 1, 2, 3, 4}. Here is the isomorphism between the R⁵/R and R⁴:

f(r₁, r₂, r₃, r₄, 0) = (r₁, r₂, r₃, r₄).

f^-1(r₁, r₂, r₃, r₄) = (r₁, r₂, r₃, r₄, 0).

Let's check f and f^-1 are homomorphisms of rings:

Since the last component of the element from R⁵/R always equals to zero(and due to our definition of ring operations) f is a homomorphism:

f(a + b) = f(a) + f(b)

f(a*b) = f(a)*f(b) where a,b belong to the R⁵/R

f^-1 = g is just adding the 0 as the last componet. So, because of the definition of ring operations =>

g(a + b) = g(a) + g(b)

g(a*b) = g(a)*g(b) where a,b belong to the R⁴

This means, that f is invertible homomorphism => R⁵/R ~^f R⁴.