Question

Let \(\mathbb{Z}_{11}\) be the ring of integers modulo 11. Let \(\mathbb{Z}_{11}[x]\) be the ring of all polynomials with coefficients in \(\mathbb{Z}_{11}\). Which of the following statements is/are TRUE?

• A. The polynomial \(x^2 + x + 4\) is irreducible over \(\mathbb{Z}_{11}\).
• B. \(\mathbb{Z}_{11}[x]/\langle x^2+x+4\rangle\) is a field.
• C. The ideal \(\langle x^2+x+4\rangle\) is maximal.
• D. The ideal \(\langle x^2+x+4\rangle\) is not a prime ideal.

Hint:

Check whether \(x^2+x+4\) has a root in \(\mathbb{Z}_{11}\); no root means irreducible, and in a PID that makes the ideal maximal and the quotient a field.

Answer 1

The Correct Option is A, B, C

Instead of testing every element of $\mathbb{Z}_{11}$ one by one, we can use the quadratic formula idea. For a quadratic $x^2+x+4$ over a field where $2$ is invertible (true for $\mathbb{Z}_{11}$ since $11$ is odd), a root exists exactly when the discriminant $D = 1^2 - 4(1)(4) = 1-16 = -15$ is a perfect square in that field. Reducing mod $11$, $D \equiv -15 + 22 = 7 \pmod{11}$.

Euler's criterion says $7$ is a square mod $11$ exactly when $7^{(11-1)/2} = 7^5 \equiv 1 \pmod{11}$, and a non-square when $7^5 \equiv -1 \pmod{11}$. Computing step by step: $7^2 = 49 \equiv 5$, $7^4 \equiv 5^2 = 25 \equiv 3$, and $7^5 \equiv 3 \times 7 = 21 \equiv 10 \equiv -1 \pmod{11}$. So $7$ is not a square mod $11$, which means the discriminant has no square root in $\mathbb{Z}_{11}$, and the quadratic $x^2+x+4$ has no root there.

1. (A) Irreducible over $\mathbb{Z}_{11}$: True. No root in $\mathbb{Z}_{11}$ means a degree 2 polynomial cannot split into two linear factors over $\mathbb{Z}_{11}$, so it is irreducible.
2. (B) $\mathbb{Z}_{11}[x]/\langle x^2+x+4\rangle$ is a field: True. $\mathbb{Z}_{11}[x]$ is a principal ideal domain because $\mathbb{Z}_{11}$ is a field, and quotienting by the ideal generated by an irreducible element in a PID always gives a field.
3. (C) $\langle x^2+x+4\rangle$ is maximal: True, this is really the same fact as (B) stated differently: an ideal is maximal exactly when the quotient by it is a field.
4. (D) $\langle x^2+x+4\rangle$ is not a prime ideal: False. Every maximal ideal is also prime (a field has no zero divisors), so this ideal is in fact a prime ideal, and (D) is wrong to claim otherwise.

So the true statements are (A), (B), and (C).

Let's summarize:

• Checking the discriminant against quadratic residues is a shortcut for irreducibility that avoids testing all 11 elements directly.
• In a PID, irreducible generator, maximal ideal, and field quotient are three ways of saying the same thing, and maximal always implies prime.

The correct options are (A), (B), and (C).

\[ \boxed{\text{(A), (B) and (C)}} \]