Discussion
Suppose the ideal \(A = (p(x))\) in F[x] F-field is a maximal ideal. Then the polynomial \(p(x)\) :
Answer: C
\(p(x)\) is irreducible.
F[x] is a principal ideal domain, and in a PID a nonzero ideal \((p)\) is maximal exactly when \(p\) is irreducible. If \(p = gh\) with both factors of positive degree, then \((p) \subsetneq (g) \subsetneq F[x]\), so \((p)\) would not be maximal. Equivalently, \(F[x]/(p(x))\) is a field exactly when \(p\) is irreducible.
The other options do not follow. 'Has no root in F' fails for degree 1: \(x - a\) generates a maximal ideal but has the root \(a\). 'Has all roots in F' fails for \(x^{2} + 1\) over R. Being monic is a normalisation, not a consequence: \(2x^{2} + 2\) generates the same maximal ideal as \(x^{2} + 1\) in R[x].
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