UPSC 2023 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

Prove that $x^2+1$ is an irreducible polynomial in $\mathbb{Z}_3[x]$ . Further show that the quotient ring $\dfrac{\mathbb{Z}_3[x]}{\langle x^2+1\rangle}$ is a field of 9 elements.

Technique

Degree-2 irreducibility test by root search; PID $\Rightarrow$ irreducible-generated ideal is maximal; division algorithm to enumerate cosets.

Solution

Part 1 — Irreducibility

Strategy. A degree-2 polynomial over a field is irreducible iff it has no root in that field. Test all elements of $\mathbb{Z}_3=\{0,1,2\}$ .

Evaluate $p(x)=x^{2}+1$ at every element of $\mathbb{Z}_3$ :

p(0)=0+1=1\ne 0,\quad p(1)=1+1=2\ne 0,\quad p(2)=4+1=5\equiv 2\pmod 3\ne 0.

No root in $\mathbb{Z}_3$ . Since $\deg p=2$ , any factorisation over $\mathbb{Z}_3[x]$ must split it into two linear factors, each contributing a root in $\mathbb{Z}_3$ — impossible. Hence $x^{2}+1$ is irreducible in $\mathbb{Z}_3[x]$ .

Part 2 — Quotient is a field of 9 elements

Strategy. $\mathbb{Z}_3[x]$ is a PID (Euclidean domain via degree); irreducibles generate maximal ideals; quotient by a maximal ideal is a field. Count cosets explicitly.

Step 1 — $\langle x^{2}+1\rangle$ is maximal.

$\mathbb{Z}_3$ is a field, so $\mathbb{Z}_3[x]$ is a Euclidean domain (with $\deg$ as the Euclidean function), hence a PID. In a PID, every non-zero prime ideal is maximal; equivalently, $(p(x))$ is maximal iff $p(x)$ is irreducible. By Part 1, $\langle x^{2}+1\rangle$ is maximal. Therefore the quotient

F=\mathbb{Z}_3[x]/\langle x^{2}+1\rangle

is a field.

Step 2 — Count the elements.

For each $f(x)\in\mathbb{Z}_3[x]$ , divide by $x^{2}+1$ (which has leading coefficient $1$ , a unit) using the Euclidean algorithm:

f(x)=q(x)(x^{2}+1)+r(x),\quad\deg r<2.

So every coset has a unique representative of the form $a+bx$ with $a,b\in\mathbb{Z}_3$ . Uniqueness: if $a+bx\equiv a'+b'x\pmod{x^{2}+1}$ then $(a-a')+(b-b')x\in\langle x^{2}+1\rangle$ ; but any non-zero element of this ideal has degree $\ge 2$ , forcing $a=a',b=b'$ .

The representatives are exactly the pairs $(a,b)\in\mathbb{Z}_3\times\mathbb{Z}_3$ , giving

|F|=3\times 3=9.

Answer

\boxed{\;F=\mathbb{Z}_3[x]/\langle x^{2}+1\rangle\text{ is a field with 9 elements.}\;}