← 2014 Paper 2

UPSC 2014 Maths Optional Paper 2 Q3a — Step-by-Step Solution

15 marks · Section A

Fields and finite fields · Algebra · asked 6× in 13 yrs · Read the full method →

Question

Show that the set $\{a+b\omega:\omega^{3}=1\}$, where $a$ and $b$ are real numbers, is a field with respect to usual addition and multiplication.

Technique

Show the set equals $\mathbb C$ (real-linear bijection); invoke that $\mathbb C$ is a field.

Solution

Strategy. Interpret $\omega$ as a non-real cube root of unity (the trivial case $\omega=1$ gives $\mathbb R$); show the set equals $\mathbb C$, which is a field.

Step 1 — Identify the set

Take $\omega=e^{2\pi i/3}=-\dfrac{1}{2}+i\dfrac{\sqrt 3}{2}$ (the standard non-real cube root of unity; the other is $\omega^{2}=\bar\omega$).

For $a,b\in\mathbb R$:

Claim: As $(a,b)$ ranges over $\mathbb R^{2}$, this expression ranges over all of $\mathbb C$.

Proof. The map $(a,b)\mapsto(a-b/2,\;b\sqrt 3/2)$ from $\mathbb R^{2}$ to $\mathbb R^{2}$ has Jacobian determinant

so it’s a linear bijection from $\mathbb R^{2}$ onto $\mathbb R^{2}=\mathbb C$ (identifying $\mathbb C$ as a 2-dim real vector space).

So the set $S=\{a+b\omega:a,b\in\mathbb R\}=\mathbb C$.

Step 2 — $\mathbb C$ is a field

$\mathbb C$ is the standard complex-number field, with:

• Commutative addition with identity $0$ and inverses $-z$.
• Commutative multiplication with identity $1$ and inverse $z^{-1}=\bar z/|z|^{2}$ for $z\ne 0$.
• Distributivity.

These are standard and well-known.

Answer