Question

Which one of the following options is incorrect?

• A. In an integral domain, every prime element is irreducible.
• B. In a unique factorization domain, every irreducible element is prime.
• C. The polynomial \(p(x) = x^3 - 6x + 9\) is irreducible over \(\mathbb{Q}\).
• D. The polynomial \(p(x) = x^3 + 4x + 7\) is irreducible over \(\mathbb{Z}_5\).

Hint:

Test rational roots for (C) and check for roots in Z5 for (D).

Answer 1

The Correct Option is C

Step 1: Check option (A). In any integral domain, if an element p is prime, then whenever p divides ab, p divides a or p divides b, and p is nonzero and not a unit. Suppose p = ab. Since p divides ab, p divides a or p divides b, say p divides a, so a = pc for some c. Then p = ab = pcb, so p(1 - cb) = 0, and since the domain has no zero divisors, cb = 1, making b a unit. Hence p is irreducible, so (A) is true.
Step 2: Check option (B). This is the defining feature of a unique factorization domain: in a UFD, every irreducible element is prime. This is a standard theorem, so (B) is true.
Step 3: Check option (C). Test rational roots of \(p(x) = x^3 - 6x + 9\) using the rational root theorem, with candidates \(\pm1, \pm3, \pm9\): \[p(-3) = (-3)^3 - 6(-3) + 9 = -27 + 18 + 9 = 0\] So \(x = -3\) is a root, and \(p(x) = (x+3)(x^2-3x+3)\). This means \(p(x)\) is reducible over \(\mathbb{Q}\), so statement (C) is false.
Step 4: Check option (D). Reduce \(p(x) = x^3+4x+7\) modulo 5 (\(7 \equiv 2\)) and test every element of \(\mathbb{Z}_5\): \[p(0)\equiv2,\ p(1)\equiv2,\ p(2)\equiv3,\ p(3)\equiv1,\ p(4)\equiv2 \pmod5\] None of these is 0, so \(p(x)\) has no root in \(\mathbb{Z}_5\). A cubic with no root cannot have a linear factor, so it is irreducible over \(\mathbb{Z}_5\). Statement (D) is true.
Step 5: Only statement (C) is false, so it is the incorrect option. \[\boxed{\text{Option (C)}}\]