← 2015 Paper 2

UPSC 2015 Maths Optional Paper 2 Q2a — Step-by-Step Solution

15 marks · Section A

Ring homomorphisms; quotient rings · Algebra · asked 3× in 13 yrs · Read the full method →

Question

If RR is a ring with unit element 11 and ϕ\phi is a homomorphism of RR onto R′R', prove that ϕ(1)\phi(1) is the unit element of R′R'.

Technique

Pick arbitrary r′∈R′r'\in R'; use surjectivity to find pre-image; apply homomorphism property to 1⋅r1\cdot r and r⋅1r\cdot 1.

Solution

Setup. RR is a ring with 11. ϕ:R→R′\phi:R\to R' is a surjective ring homomorphism. To show: ϕ(1)\phi(1) is the multiplicative identity of R′R'.

A ring homomorphism preserves both operations:

Step 1 — Let r′∈R′r'\in R' be arbitrary

By surjectivity, there exists r∈Rr\in R with ϕ(r)=r′\phi(r)=r'.

Step 2 — Compute ϕ(1)⋅r′\phi(1)\cdot r'

ϕ(1)⋅r′=ϕ(1)⋅ϕ(r)=ϕ(1⋅r)=ϕ(r)=r′.\phi(1)\cdot r'=\phi(1)\cdot\phi(r)=\phi(1\cdot r)=\phi(r)=r'.

Step 3 — Compute r′⋅ϕ(1)r'\cdot\phi(1)

r′⋅ϕ(1)=ϕ(r)⋅ϕ(1)=ϕ(r⋅1)=ϕ(r)=r′.r'\cdot\phi(1)=\phi(r)\cdot\phi(1)=\phi(r\cdot 1)=\phi(r)=r'.

Step 4 — Conclusion

Since ϕ(1)⋅r′=r′=r′⋅ϕ(1)\phi(1)\cdot r'=r'=r'\cdot\phi(1) for every r′∈R′r'\in R', ϕ(1)\phi(1) acts as the (two-sided) multiplicative identity in R′R'. Identities are unique when they exist, so ϕ(1)=1R′\phi(1)=1_{R'}.

Answer

  ϕ(1) is the unit element of R′.  \boxed{\;\phi(1)\text{ is the unit element of }R'.\;}
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