Rings and Ideals Questions and Answers

Q: Let \(R = \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}\) and \(I = \mathbb{Z} \times \mathbb{Z} \times \{0\}\). Then which of the following statement is correct?

\(I\) is a prime ideal but not a maximal ideal of \(R\).The projection \((a, b, c) \mapsto c\) is a surjective ring homomorphism from \(R\) onto \(\mathbb{Z}\) whose kernel is exactly \(I = \mathbb{Z} \times \mathbb{Z} \times \{0\}\). So \(R/I \cong \mathbb{Z}\).An ideal is prime exactly when the quotient is an integral domain, and maximal exactly when the quotient is a field. \(\mathbb{Z}\) is an integral domain but not a field, so \(I\) is prime and not maximal. Concretely, \(I \subsetneq \mathbb{Z} \times \mathbb{Z} \times 2\mathbb{Z} \subsetneq R\), so \(I\) is not maximal. That rules out the three other options.

Q: Let \(R\) be a commutative ring with 1 (unity) which is not a field. Let \(I \subset R\) be a proper ideal such that every element of \(R\) not in \(I\) is invertible in \(R\). Then the number of maximal ideals of \(R\) is

\(R\) has exactly 1 maximal ideal.Let \(\mathfrak{m}\) be any maximal ideal. It is proper, so it contains no unit. Every element outside \(I\) is a unit, so every element of \(\mathfrak{m}\) lies in \(I\): \(\mathfrak{m} \subseteq I\). Since \(I\) is proper and \(\mathfrak{m}\) is maximal, \(\mathfrak{m} = I\).So every maximal ideal equals \(I\), and \(I\) is the only one: \(R\) is a local ring. Such an \(I\) exists, for example \(R = \mathbb{Z}/4\mathbb{Z}\) with \(I = \{0, 2\}\). The condition that \(R\) is not a field only ensures \(I \neq 0\); it does not create more maximal ideals, so 2, 3 and infinite are wrong.

Q: Consider the following conditions on two proper non-zero ideals \(J_{1}\) and \(J_{2}\) of a non-zero commutative ring R.P: For any \(r_{1}, r_{2} \in R\), there exists a unique \(r \in R\) such that \(r - r_{1} \in J_{1}\) and \(r - r_{2} \in J_{2}\).Q: \(J_{1} + J_{2} = R\)Then, which of the following statements is TRUE?

The answer is P implies Q but Q does not imply P.P implies Q: take \(r_{1} = 1\) and \(r_{2} = 0\). P gives an r with \(r - 1 \in J_{1}\) and \(r \in J_{2}\). Then \(1 = r - (r - 1) \in J_{2} + J_{1}\), so \(J_{1} + J_{2} = R\).Q does not imply P: Q gives existence through the Chinese remainder theorem, but not uniqueness. In \(\mathbb{Z}\) with \(J_{1} = 2\mathbb{Z}\) and \(J_{2} = 3\mathbb{Z}\), the ideals are comaximal, yet if r solves the two congruences so does r + 6. Uniqueness in P actually forces \(J_{1} \cap J_{2} = 0\), which Q does not give.So the options saying Q implies P, or that neither implies the other, are wrong.

Q: Which of the following rings need not have identity ?

The ring that need not have an identity is the ring of polynomials \(P_{0}[\mathbf{R}]\) vanishing at the origin.The polynomials with zero constant term are closed under addition and multiplication, so they form a ring. Suppose \(e(x)\) were an identity. Then \(e(x)\cdot x = x\) for the polynomial \(x\), so \(e(x) = 1\), and the constant 1 does not vanish at the origin. So no identity exists in this ring.The other rings all have one. Homomorphisms of a vector space to itself (linear maps) form a ring whose identity is the identity map. Upper triangular \(n \times n\) matrices have the identity matrix \(I\), which is upper triangular. The automorphisms contain the identity map too; as printed this option is loosely worded, because invertible maps are not closed under addition and so are not a ring in the strict sense, but whichever way it is read, the identity map belongs to it.

Q: Let R be a Boolean ring with identity having \(n\) elements, then which of the following values of \(n\) is possible ?

The possible value is \(n = 32\).In a Boolean ring every element satisfies \(x^{2} = x\). Expanding \((x + x)^{2} = x + x\) gives \(4x = 2x\), so \(2x = 0\) and the ring has characteristic 2 and its additive group is a vector space over \(\mathbf{Z}_{2}\). A finite one therefore has \(2^{k}\) elements; in fact a finite Boolean ring with identity is isomorphic to \(\mathbf{Z}_{2}^{k}\).\(32 = 2^{5}\) is a power of 2. 23 is prime, 27 is \(3^{3}\) and 15 is \(3 \times 5\), none of them a power of 2.

Q: Let R be a ring with identity \(1\ (\ne 0)\) and \(m\) be the characteristic of R, then which of the following is true ?

The true statement is wherever R is finite, \(m \ne 0\).In a finite ring the elements \(1, 1 + 1, 1 + 1 + 1, \ldots\) cannot all be distinct, so \(k \cdot 1 = 0\) for some positive \(k\), and the characteristic is positive.The other statements fail. \(\mathbf{Z}_{p}[x]\) is infinite with characteristic \(p\), which refutes A (infinite forces \(m = 0\)) and B (\(m \ne 0\) forces finite). \(\mathbf{Z}_{p}\) itself is finite with \(m = p \ne 0\), which refutes D (\(m \ne 0\) forces infinite).

Q: Which of the following rings is not Noetherian ?

The ring that is not Noetherian is \(\mathbf{R}[x_{1}, x_{2}, \ldots]\).In the polynomial ring in infinitely many variables, the chain of ideals\[ \begin{aligned} (x_{1}) &\subset (x_{1}, x_{2}) \\ &\subset (x_{1}, x_{2}, x_{3}) \subset \cdots \end{aligned} \]is strictly increasing and never stops, so the ascending chain condition fails. The ideal generated by all the variables is not finitely generated.\(\mathbf{Z}\) and \(\mathbf{Z}[i]\) are PIDs, hence Noetherian, and \(\mathbf{Z}_{205}[x]\) is Noetherian by Hilbert's basis theorem, since \(\mathbf{Z}_{205}\) is a finite ring and so Noetherian.

Q: Let \(x\) be the image of X in the quotient ring \(A := \mathbf{R}[X, Y]/\langle X^{2} + Y^{2} - 1\rangle\). Then the element \(x\) is :

\(x\) is irreducible, but not prime in A.Not prime: in A, \(x^{2} = 1 - y^{2} = (1 - y)(1 + y)\), so \(x\) divides the product \((1 - y)(1 + y)\). But \(x\) divides neither factor, since \(A/(x) \cong \mathbf{R}[Y]/(Y^{2} - 1)\), in which \(1 - y\) and \(1 + y\) are non-zero. That quotient is \(\mathbf{R} \times \mathbf{R}\), not a domain, which is another way of saying \((x)\) is not prime.Irreducible: putting \(x = \cos\theta\), \(y = \sin\theta\) identifies A with the ring of real trigonometric polynomials, in which the degree of a product is the sum of the degrees and the units are the non-zero constants (degree 0). \(x\) has degree 1, so any factorisation \(x = gh\) has a factor of degree 0, a unit. This is the textbook example showing that A is not a UFD.So 'not irreducible', 'prime' and 'a unit' are all wrong.

Q: Which of the following is a principal ideal of \(\mathbf{Z}[x]\) ?

The principal ideal is \((2x, 3x, x^{2})\).Since \(3x - 2x = x\), the ideal contains \(x\); and each generator \(2x, 3x, x^{2}\) is a multiple of \(x\). So \((2x, 3x, x^{2}) = (x)\), which is principal.The others are not principal. \((2, x)\) consists of polynomials with even constant term; a single generator would have to divide both 2 and \(x\), so it would be \(\pm 1\), but 1 is not in the ideal. The same argument works for \((3, x)\). And \((x, x^{2} - x, x + 4) = (x, 4)\), which fails for the same reason.

Q: Let \(\mathbf{R}[X]\) be the polynomial ring in one indeterminate over the field of real numbers. Then in the quotient ring \(\mathbf{R}[X]/\lt X^{8} + 1\gt\) number of maximal ideals is :

The quotient has 4 maximal ideals.The roots of \(X^{8} + 1\) are the eight primitive 16th roots of unity, none of them real. They form four conjugate pairs, so over \(\mathbf{R}\) the polynomial factors into four distinct irreducible quadratics \(q_{1}q_{2}q_{3}q_{4}\).Maximal ideals of \(\mathbf{R}[X]/\langle X^{8} + 1 \rangle\) correspond to maximal ideals of \(\mathbf{R}[X]\) containing \(X^{8} + 1\), that is, to its distinct irreducible factors. By the Chinese remainder theorem the ring is \(\mathbf{C}^{4}\), with exactly 4 maximal ideals.

Let \(R = \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}\) and \(I = \mathbb{Z} \times \mathbb{Z} \times \{0\}\). Then which of the following statement is correct?

A.\(I\) is a maximal ideal but not a prime ideal of \(R\).
B.\(I\) is a prime ideal but not a maximal ideal of \(R\).
C.\(I\) is both maximal ideal as well as a prime ideal of \(R\).
D.\(I\) is neither a maximal ideal nor a prime ideal of \(R\).

Exam Year(s)

GATE Mathematics - 2012,

Answer: B

\(I\) is a prime ideal but not a maximal ideal of \(R\).

The projection \((a, b, c) \mapsto c\) is a surjective ring homomorphism from \(R\) onto \(\mathbb{Z}\) whose kernel is exactly \(I = \mathbb{Z} \times \mathbb{Z} \times \{0\}\). So \(R/I \cong \mathbb{Z}\).

An ideal is prime exactly when the quotient is an integral domain, and maximal exactly when the quotient is a field. \(\mathbb{Z}\) is an integral domain but not a field, so \(I\) is prime and not maximal. Concretely, \(I \subsetneq \mathbb{Z} \times \mathbb{Z} \times 2\mathbb{Z} \subsetneq R\), so \(I\) is not maximal. That rules out the three other options.

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Let \(R\) be a commutative ring with 1 (unity) which is not a field. Let \(I \subset R\) be a proper ideal such that every element of \(R\) not in \(I\) is invertible in \(R\). Then the number of maximal ideals of \(R\) is

A.1
B.2
C.3
D.infinite

Exam Year(s)

GATE Mathematics - 2018,

Answer: A

\(R\) has exactly 1 maximal ideal.

Let \(\mathfrak{m}\) be any maximal ideal. It is proper, so it contains no unit. Every element outside \(I\) is a unit, so every element of \(\mathfrak{m}\) lies in \(I\): \(\mathfrak{m} \subseteq I\). Since \(I\) is proper and \(\mathfrak{m}\) is maximal, \(\mathfrak{m} = I\).

So every maximal ideal equals \(I\), and \(I\) is the only one: \(R\) is a local ring. Such an \(I\) exists, for example \(R = \mathbb{Z}/4\mathbb{Z}\) with \(I = \{0, 2\}\). The condition that \(R\) is not a field only ensures \(I \neq 0\); it does not create more maximal ideals, so 2, 3 and infinite are wrong.

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Consider the following conditions on two proper non-zero ideals \(J_{1}\) and \(J_{2}\) of a non-zero commutative ring R.

P: For any \(r_{1}, r_{2} \in R\), there exists a unique \(r \in R\) such that \(r - r_{1} \in J_{1}\) and \(r - r_{2} \in J_{2}\).

Q: \(J_{1} + J_{2} = R\)

Then, which of the following statements is TRUE?

A.P implies Q but Q does not imply P
B.Q implies P but P does not imply Q
C.P implies Q and Q implies P
D.P does not imply Q and Q does not imply P

Exam Year(s)

GATE Mathematics - 2022,

Answer: A

The answer is P implies Q but Q does not imply P.

P implies Q: take \(r_{1} = 1\) and \(r_{2} = 0\). P gives an r with \(r - 1 \in J_{1}\) and \(r \in J_{2}\). Then \(1 = r - (r - 1) \in J_{2} + J_{1}\), so \(J_{1} + J_{2} = R\).

Q does not imply P: Q gives existence through the Chinese remainder theorem, but not uniqueness. In \(\mathbb{Z}\) with \(J_{1} = 2\mathbb{Z}\) and \(J_{2} = 3\mathbb{Z}\), the ideals are comaximal, yet if r solves the two congruences so does r + 6. Uniqueness in P actually forces \(J_{1} \cap J_{2} = 0\), which Q does not give.

So the options saying Q implies P, or that neither implies the other, are wrong.

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Which of the following rings need not have identity ?

A.The ring of homomorphisms of a vector space to itself
B.The ring of automorphisms of a vector space to itself
C.The ring of \(n \times n\) upper triangular matrices over R
D.The ring of polynomials \(P_{0}[\mathbf{R}]\) vanishing at origin

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The ring that need not have an identity is the ring of polynomials \(P_{0}[\mathbf{R}]\) vanishing at the origin.

The polynomials with zero constant term are closed under addition and multiplication, so they form a ring. Suppose \(e(x)\) were an identity. Then \(e(x)\cdot x = x\) for the polynomial \(x\), so \(e(x) = 1\), and the constant 1 does not vanish at the origin. So no identity exists in this ring.

The other rings all have one. Homomorphisms of a vector space to itself (linear maps) form a ring whose identity is the identity map. Upper triangular \(n \times n\) matrices have the identity matrix \(I\), which is upper triangular. The automorphisms contain the identity map too; as printed this option is loosely worded, because invertible maps are not closed under addition and so are not a ring in the strict sense, but whichever way it is read, the identity map belongs to it.

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Let R be a Boolean ring with identity having \(n\) elements, then which of the following values of \(n\) is possible ?

A.\(n = 32\)
B.\(n = 23\)
C.\(n = 27\)
D.\(n = 15\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The possible value is \(n = 32\).

In a Boolean ring every element satisfies \(x^{2} = x\). Expanding \((x + x)^{2} = x + x\) gives \(4x = 2x\), so \(2x = 0\) and the ring has characteristic 2 and its additive group is a vector space over \(\mathbf{Z}_{2}\). A finite one therefore has \(2^{k}\) elements; in fact a finite Boolean ring with identity is isomorphic to \(\mathbf{Z}_{2}^{k}\).

\(32 = 2^{5}\) is a power of 2. 23 is prime, 27 is \(3^{3}\) and 15 is \(3 \times 5\), none of them a power of 2.

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Let R be a ring with identity \(1\ (\ne 0)\) and \(m\) be the characteristic of R, then which of the following is true ?

A.wherever R is infinite, \(m = 0\)
B.wherever \(m \ne 0\), R is finite
C.wherever R is finite, \(m \ne 0\)
D.wherever \(m \ne 0\), R is infinite

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The true statement is wherever R is finite, \(m \ne 0\).

In a finite ring the elements \(1, 1 + 1, 1 + 1 + 1, \ldots\) cannot all be distinct, so \(k \cdot 1 = 0\) for some positive \(k\), and the characteristic is positive.

The other statements fail. \(\mathbf{Z}_{p}[x]\) is infinite with characteristic \(p\), which refutes A (infinite forces \(m = 0\)) and B (\(m \ne 0\) forces finite). \(\mathbf{Z}_{p}\) itself is finite with \(m = p \ne 0\), which refutes D (\(m \ne 0\) forces infinite).

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Which of the following rings is not Noetherian ?

A.\(\mathbf{R}[x_{1}, x_{2}, ........]\)
B.\(\mathbf{Z}[i]\)
C.\(\mathbf{Z}\)
D.\(\mathbf{Z}_{205}[x]\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The ring that is not Noetherian is \(\mathbf{R}[x_{1}, x_{2}, \ldots]\).

In the polynomial ring in infinitely many variables, the chain of ideals

\[ \begin{aligned} (x_{1}) &\subset (x_{1}, x_{2}) \\ &\subset (x_{1}, x_{2}, x_{3}) \subset \cdots \end{aligned} \]

is strictly increasing and never stops, so the ascending chain condition fails. The ideal generated by all the variables is not finitely generated.

\(\mathbf{Z}\) and \(\mathbf{Z}[i]\) are PIDs, hence Noetherian, and \(\mathbf{Z}_{205}[x]\) is Noetherian by Hilbert's basis theorem, since \(\mathbf{Z}_{205}\) is a finite ring and so Noetherian.

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Let \(x\) be the image of X in the quotient ring \(A := \mathbf{R}[X, Y]/\langle X^{2} + Y^{2} - 1\rangle\). Then the element \(x\) is :

A.not irreducible in A
B.irreducible, but not prime in A
C.prime in A
D.a unit in A

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

\(x\) is irreducible, but not prime in A.

Not prime: in A, \(x^{2} = 1 - y^{2} = (1 - y)(1 + y)\), so \(x\) divides the product \((1 - y)(1 + y)\). But \(x\) divides neither factor, since \(A/(x) \cong \mathbf{R}[Y]/(Y^{2} - 1)\), in which \(1 - y\) and \(1 + y\) are non-zero. That quotient is \(\mathbf{R} \times \mathbf{R}\), not a domain, which is another way of saying \((x)\) is not prime.

Irreducible: putting \(x = \cos\theta\), \(y = \sin\theta\) identifies A with the ring of real trigonometric polynomials, in which the degree of a product is the sum of the degrees and the units are the non-zero constants (degree 0). \(x\) has degree 1, so any factorisation \(x = gh\) has a factor of degree 0, a unit. This is the textbook example showing that A is not a UFD.

So 'not irreducible', 'prime' and 'a unit' are all wrong.

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Which of the following is a principal ideal of \(\mathbf{Z}[x]\) ?

A.\((2, x)\)
B.\((3, x)\)
C.\((x, x^{2} - x, x + 4)\)
D.\((2x, 3x, x^{2})\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The principal ideal is \((2x, 3x, x^{2})\).

Since \(3x - 2x = x\), the ideal contains \(x\); and each generator \(2x, 3x, x^{2}\) is a multiple of \(x\). So \((2x, 3x, x^{2}) = (x)\), which is principal.

The others are not principal. \((2, x)\) consists of polynomials with even constant term; a single generator would have to divide both 2 and \(x\), so it would be \(\pm 1\), but 1 is not in the ideal. The same argument works for \((3, x)\). And \((x, x^{2} - x, x + 4) = (x, 4)\), which fails for the same reason.

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10.

Let \(\mathbf{R}[X]\) be the polynomial ring in one indeterminate over the field of real numbers. Then in the quotient ring \(\mathbf{R}[X]/\lt X^{8} + 1\gt\) number of maximal ideals is :

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The quotient has 4 maximal ideals.

The roots of \(X^{8} + 1\) are the eight primitive 16th roots of unity, none of them real. They form four conjugate pairs, so over \(\mathbf{R}\) the polynomial factors into four distinct irreducible quadratics \(q_{1}q_{2}q_{3}q_{4}\).

Maximal ideals of \(\mathbf{R}[X]/\langle X^{8} + 1 \rangle\) correspond to maximal ideals of \(\mathbf{R}[X]\) containing \(X^{8} + 1\), that is, to its distinct irreducible factors. By the Chinese remainder theorem the ring is \(\mathbf{C}^{4}\), with exactly 4 maximal ideals.

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