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UPSC 2015 Maths Optional Paper 2 Q1b — Step-by-Step Solution

10 marks · Section A

Subrings and ideals · Algebra · asked 4× in 13 yrs · Read the full method →

Question

Give an example of a ring having identity but a subring of this having a different identity.

Technique

Pick a ring with zero divisors ($\mathbb Z_n$ for composite $n$); look for an idempotent $e\ne 1$ (i.e., $e^2=e$, $e\ne 0,1$); the principal ideal $(e)$ becomes a subring with identity $e$.

Solution

Strategy. A “subring” here means a subset closed under addition and multiplication, with its own multiplicative identity that differs from the parent ring’s identity.

Example: $R=\mathbb Z_6$, subring $S=\{0,2,4\}$

Consider $R=\mathbb Z_6=\{0,1,2,3,4,5\}$ with addition and multiplication mod 6.

• $R$ has multiplicative identity $1$, since $1\cdot x=x$ for all $x\in R$.

Let $S=\{0,2,4\}\subset R$. Check $S$ is a subring:

Closure under addition (mod 6):

• $2+2=4\in S$, $2+4=0\in S$, $4+4=8\equiv 2\in S$. ✓

Closure under multiplication (mod 6):

• $2\cdot 2=4\in S$, $2\cdot 4=8\equiv 2\in S$, $4\cdot 4=16\equiv 4\in S$. ✓

So $S$ is a subring of $R$.

Identity of $S$: Test each element to find $e_S$ with $e_S\cdot x=x$ for all $x\in S$:

• $4\cdot 2=8\equiv 2$ ✓
• $4\cdot 4=16\equiv 4$ ✓
• $4\cdot 0=0$ ✓

So $4$ is the multiplicative identity of $S$.

But $4\ne 1$ in $R$. So $R$ has identity $1$ while subring $S$ has identity $4$.

Answer