← 2014 Paper 2

UPSC 2014 Maths Optional Paper 2 Q2a — 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 $\mathbb Z_7$ is a field. Then find $([5]+[6])^{-1}$ and $(-[4])^{-1}$ in $\mathbb Z_7$.

Technique

Standard ”$\mathbb Z_p$ is a field iff $p$ prime” theorem; computing inverses via brute-force or extended Euclidean.

Solution

Part 1 — $\mathbb Z_7$ is a field

General theorem: $\mathbb Z_n$ is a field if and only if $n$ is prime.

Proof outline:

1. $\mathbb Z_n$ is a commutative ring with unity (additive and multiplicative inheritance from $\mathbb Z$, with operations modulo $n$). Standard.

2. Non-trivial unity: $[1]\ne[0]$ in $\mathbb Z_n$ for $n\ge 2$. ✓

3. Every non-zero element has a multiplicative inverse: if $[a]\ne[0]$ in $\mathbb Z_n$ with $n$ prime, then $\gcd(a,n)=1$, so by Bézout’s identity there exist integers $b,c$ with $ab+nc=1$. Reducing mod $n$: $[a][b]=[1]$, so $[b]=[a]^{-1}$.

For $n=7$ (prime): non-zero elements $\{[1],[2],[3],[4],[5],[6]\}$ all have inverses in $\mathbb Z_7$. Hence $\mathbb Z_7$ is a field.

Explicit inverse table in $\mathbb Z_7$:

(Verify: $[2][4]=[8]=[1]$, $[3][5]=[15]=[1]$, $[6][6]=[36]=[1]$.)

Part 2 — $([5]+[6])^{-1}$

$[5]+[6]=[11]=[4]$ in $\mathbb Z_7$.

From the table, $[4]^{-1}=[2]$.

Part 3 — $(-[4])^{-1}$

$-[4]=[-4]=[3]$ in $\mathbb Z_7$ (since $-4+7=3$).

From the table, $[3]^{-1}=[5]$.

Answer