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

Step 1: Check whether \(x^2+x+4\) has a root in \(\mathbb{Z}_{11}\).
A degree 2 polynomial over a field is irreducible exactly when it has no root in that field (if it had a root \(r\), it would factor as \((x-r)\) times a linear factor). So test every element of \(\mathbb{Z}_{11} = \{0,1,2,\dots,10\}\) in \(x^2+x+4\):
\(0\to4,\ 1\to6,\ 2\to10,\ 3\to5,\ 4\to2,\ 5\to1,\ 6\to2,\ 7\to5,\ 8\to10,\ 9\to6,\ 10\to4\) (all values taken mod 11).
None of these equal \(0\), so \(x^2+x+4\) has no root in \(\mathbb{Z}_{11}\), which means it is irreducible over \(\mathbb{Z}_{11}\). Option (A) is TRUE.

Step 2: Use the fact that \(\mathbb{Z}_{11}[x]\) is a principal ideal domain (PID).
Since \(\mathbb{Z}_{11}\) is a field, \(\mathbb{Z}_{11}[x]\) behaves like the integers do for prime numbers: every ideal is generated by a single polynomial, and an ideal generated by an irreducible polynomial is a maximal ideal. Since Step 1 showed \(x^2+x+4\) is irreducible, the ideal \(\langle x^2+x+4\rangle\) is maximal. Option (C) is TRUE.

Step 3: Use the maximal ideal to conclude the quotient is a field.
For a commutative ring \(R\) with unity, a basic theorem says an ideal \(M\) is maximal if and only if the quotient ring \(R/M\) is a field. Since Step 2 showed \(\langle x^2+x+4\rangle\) is maximal in \(\mathbb{Z}_{11}[x]\), the quotient \(\mathbb{Z}_{11}[x]/\langle x^2+x+4\rangle\) must be a field. Option (B) is TRUE. In fact this field has \(11^2=121\) elements, since every element can be written uniquely as \(a+bx\) with \(a,b\in\mathbb{Z}_{11}\).

Step 4: Check option (D).
Another basic fact is that every maximal ideal in a commutative ring with unity is also a prime ideal (a field, which \(R/M\) is, has no zero divisors, and that is exactly what makes \(M\) prime). Since we showed \(\langle x^2+x+4\rangle\) is maximal, it is automatically prime. So the claim in (D), that this ideal is not prime, is FALSE.

Final Answer:
\(x^2+x+4\) is irreducible over \(\mathbb{Z}_{11}\), so the ideal it generates is maximal and the quotient ring is a field; that same ideal is therefore also prime, making (D) false.
\[ \boxed{\text{(A), (B) and (C)}} \]