Rings and Ideals Questions and Answers

Q: Which of the following rings is not noetherian ?

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.

Q: Let I be the ideal in the polynomial ring \(\mathbf{Z}_{4}[X]\) generated by the polynomial 1 + 2X. Then :

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.

Q: Suppose the ideal \(A = (p(x))\) in F[x] F-field is a maximal ideal. Then the polynomial \(p(x)\) :

\(p(x)\) 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].

Q: Consider the following statements :(I) If R is a commutative ring with unity and \(f : R \to R\) be a ring homomorphism defined by \(f(a) = a^{2}\), then 1 + 1 = 0.(II) If R is a commutative ring with unity and \(f : R \to R\) be a ring homomorphism defined by \(f(a) = a^{3}\), then 1 + 1 + 1 = 0.Which of the following is true ?

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.

Q: Which is correct ?

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.

Q: Which of the following is false ?

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.

Q: Consider the following statements :I : Any finite non-commutative ring must have at least 16 elements.II : Let R be a ring with unity such that |R| = 9. Then R is commutative.Which of the following is true ?

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.

11.

Which of the following rings is not noetherian ?

A.\(\mathbf{Q}[x_{1}, x_{2}, x_{3}, \ldots]\)
B.\(\mathbf{Z}[x]\)
C.\(\mathbf{Z}[\sqrt{2}]\)
D.\(\mathbf{R}[x, y]/(x^{2} + y^{2} + 1)\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

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12.

Let I be the ideal in the polynomial ring \(\mathbf{Z}_{4}[X]\) generated by the polynomial 1 + 2X. Then :

A.I is a prime ideal
B.I is a maximal ideal
C.I is a proper ideal which is not a prime ideal
D.I is a unit ideal

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

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.

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13.

Suppose the ideal \(A = (p(x))\) in F[x] F-field is a maximal ideal. Then the polynomial \(p(x)\) :

A.has no root in F
B.is monic
C.is irreducible
D.has all roots in F

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

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].

Discuss in Forum

14.

Consider the following statements :

(I) If R is a commutative ring with unity and \(f : R \to R\) be a ring homomorphism defined by \(f(a) = a^{2}\), then 1 + 1 = 0.
(II) If R is a commutative ring with unity and \(f : R \to R\) be a ring homomorphism defined by \(f(a) = a^{3}\), then 1 + 1 + 1 = 0.

Which of the following is true ?

A.Only (I) is true
B.Only (II) is true
C.Both (I) and (II) are true
D.Neither (I) nor (II) is true

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

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.

Discuss in Forum

15.

Which is correct ?

A.The polynomial ring K[x] over a field K is local
B.The additive group of rationals has a maximal subgroup
C.Any ring (may not have unity) has a maximal ideal
D.Any finitely generated ideal of a Boolean ring is principal

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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.

Discuss in Forum

16.

Which of the following is false ?

A.In a commutative ring with unity, every prime ideal is maximal
B.In a commutative ring with unity, every maximal ideal is prime
C.In a Boolean ring with unity, every prime ideal is maximal
D.In a Boolean ring with unity, every maximal ideal is prime

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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.

Discuss in Forum

17.

Consider the following statements :

I : Any finite non-commutative ring must have at least 16 elements.
II : Let R be a ring with unity such that |R| = 9. Then R is commutative.

Which of the following is true ?

A.Only I is true
B.Only II is true
C.Both I and II are true
D.Neither I nor II is true

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

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.

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