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UPSC 2015 Maths Optional Paper 2 Q4a — Step-by-Step Solution

15 marks · Section A

Integral domains; characteristic · Algebra · asked 2× in 13 yrs · Read the full method →

Question

Do the following sets form integral domains with respect to ordinary addition and multiplication? If so, state if they are fields: (i) The set of numbers of the form $b\sqrt 2$ with $b$ rational. (ii) The set of even integers. (iii) The set of positive integers.

Technique

Verify ring axioms (closure, identities, inverses, zero divisors) one at a time for each set.

Solution

Recall. An integral domain is a commutative ring with identity having no zero divisors. A field is an integral domain in which every nonzero element has a multiplicative inverse.

Part (i) — $S=\{b\sqrt 2:b\in\mathbb Q\}$

Closure under addition: $b_1\sqrt 2+b_2\sqrt 2=(b_1+b_2)\sqrt 2\in S$ ✓.

Closure under multiplication: $(b_1\sqrt 2)(b_2\sqrt 2)=2b_1b_2$, which is rational. Is $2b_1b_2\in S$? Only if it can be written as $b\sqrt 2$, i.e. $2b_1b_2=b\sqrt 2$ for some rational $b$. Solving: $b=2b_1b_2/\sqrt 2=\sqrt 2\cdot b_1 b_2$, which is irrational (for $b_1,b_2\ne 0$). So in general, $2b_1b_2\notin S$.

So $S$ is not closed under multiplication ⇒ not even a ring.

Conclusion (i): Not an integral domain (not closed under multiplication).

Part (ii) — $S=2\mathbb Z=\{\dots,-4,-2,0,2,4,\dots\}$

Closure under +,×: Sum and product of even integers are even ✓.

Commutative ring: inherits commutativity and distributivity from $\mathbb Z$ ✓.

Multiplicative identity: Need $e\in S$ with $e\cdot x=x$ for all $x\in S$. But $1\notin S$ (1 is odd), and no even integer satisfies $e\cdot 2=2$ (would need $e=1$). No identity.

So $S$ is a rng (ring without identity), not a ring with identity. Hence not an integral domain in the standard definition that requires identity.

(Even-integer set has no zero divisors and is a commutative ring, just lacking identity.)

Conclusion (ii): Not an integral domain (no multiplicative identity).

Part (iii) — $S=\mathbb Z^+=\{1,2,3,\dots\}$

Closure under addition: Sum of positives is positive ✓.

Additive identity ($0$)? $0\notin S$. So no additive identity.

Additive inverses? $-1$ is the additive inverse of $1$, but $-1\notin S$. So no inverses.

So $S$ is not even a group under addition, hence not a ring.

Conclusion (iii): Not an integral domain (not closed under additive inverses; no zero).

Summary

Answer