Rings and Ideals Questions and Answers

Answer: A

The non-Noetherian ring is \(\mathbf{Q}[x_{1}, x_{2}, x_{3}, \ldots]\).

With infinitely many variables, the chain \((x_{1}) \subset (x_{1}, x_{2}) \subset (x_{1}, x_{2}, x_{3}) \subset \cdots\) increases strictly and never stops, so the ascending chain condition fails.

The others are Noetherian. \(\mathbf{Z}\) is Noetherian, so \(\mathbf{Z}[x]\) is, by Hilbert's basis theorem. \(\mathbf{Z}[\sqrt{2}]\) is a finitely generated \(\mathbf{Z}\)-module (indeed a Euclidean domain). And \(\mathbf{R}[x, y]\) is Noetherian, so every quotient of it, including \(\mathbf{R}[x, y]/(x^{2} + y^{2} + 1)\), is Noetherian.

Answer: D

I is the unit ideal, the whole ring.

In \(\mathbf{Z}_{4}[X]\), \(1 + 2X\) is a unit:

\[ (1 + 2X)^{2} = 1 + 4X + 4X^{2} = 1 \]

because \(4 = 0\) in \(\mathbf{Z}_{4}\). An ideal containing a unit contains \(1\) and therefore everything, so \(I = \mathbf{Z}_{4}[X]\).

A prime or maximal ideal must be proper by definition, so A and B fail, and C, 'proper but not prime', fails because I is not proper at all. The trap is that a polynomial of degree 1 cannot be a unit over a field or an integral domain, but \(\mathbf{Z}_{4}\) has the nilpotent element 2.

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].

Answer: A

Only (I) is true.

(I) If \(a \mapsto a^{2}\) is a ring homomorphism, it preserves addition, so \((1 + 1)^{2} = 1^{2} + 1^{2}\). The left side is \(4 \cdot 1\) and the right side is \(2 \cdot 1\), so \(2 \cdot 1 = 0\), that is \(1 + 1 = 0\). (In a commutative ring \((a + b)^{2} = a^{2} + 2ab + b^{2}\), and additivity forces \(2ab = 0\).)

(II) Applying the same idea to \(a \mapsto a^{3}\) gives \((1 + 1)^{3} = 1 + 1\), so \(8 \cdot 1 = 2 \cdot 1\), that is \(6 \cdot 1 = 0\). That does not force \(3 \cdot 1 = 0\). In \(\mathbf{Z}_{2}\), \(a^{3} = a\) for every element, so the cube map is the identity, a ring homomorphism, yet \(1 + 1 + 1 = 1 \neq 0\). So (II) is false, which rules out the options that include it.

Answer: D

The correct statement is any finitely generated ideal of a Boolean ring is principal.

In a Boolean ring every element satisfies \(x^{2} = x\) and \(2x = 0\). For two generators, put \(c = a + b + ab\). Then \(ca = a + ab + ab = a\) and similarly \(cb = b\), so \(a, b \in (c)\), while \(c \in (a, b)\). Hence \((a, b) = (c)\), and by induction every finitely generated ideal is principal.

The others are false. K[x] has infinitely many maximal ideals, such as \((x)\) and \((x - 1)\), so it is not local. The additive group Q is divisible and has no maximal subgroup. And a ring without unity need not have a maximal ideal: the zero-multiplication ring on Q has ideals equal to its additive subgroups, and Q has no maximal subgroup.

Answer: A

The false statement is in a commutative ring with unity, every prime ideal is maximal.

In Z the zero ideal is prime (Z is an integral domain) but not maximal, since \((0) \subset (2) \subset \mathbf{Z}\).

The other three are true. Every maximal ideal M is prime because R/M is a field, hence an integral domain. In a Boolean ring, if P is prime then R/P is a Boolean integral domain; from \(x^{2} = x\), \(x(x - 1) = 0\), so every element is 0 or 1 and R/P is the field with two elements, making P maximal. Maximal implies prime in any commutative ring with unity, Boolean or not.

Answer: B

Only II is true.

II: a ring with unity of order 9 has additive group \(\mathbf{Z}_{9}\) or \(\mathbf{Z}_{3} \times \mathbf{Z}_{3}\). In the first case 1 generates the additive group, so every element is an integer multiple of 1 and the ring is commutative. In the second, the subring generated by 1 is \(\mathbf{Z}_{3}\), and adding one more element a spans the ring as \(\{m + na\}\); any two such elements commute. So every ring with unity of order \(p^{2}\) is commutative.

I is false: the ring of upper triangular 2 × 2 matrices over \(\mathbf{Z}_{2}\) has only 8 elements and is not commutative, and without requiring unity there are non-commutative rings of order 4. So 16 is not a lower bound.