Algebra MCQ Quiz - Objective Question with Answer for Algebra - Download Free PDF

Concept:
Principal ideal domain (PID) is an integral domain where every ideal is principal, i.e., generated by a single element.
 

Maximal ideal is an ideal  I such that there are no other ideals strictly between  I  and the whole ring R.

A quotient ring is the ring formed by dividing R by an ideal.

Explanation:

Option 1: This is false. While R itself is a PID, not every quotient ring of a PID is necessarily a PID.

A quotient ring of a PID may fail to satisfy the conditions to be a PID.

Option 2: This is false in a PID. By definition, all ideals in a PID (including quotient rings formed by them)

are principal, so there cannot exist a non-principal ideal in any quotient ring.

Option 3: This is true. A PID has countably many ideals because each ideal is generated by a single element

and the number of elements in R is countable (if R is a ring over the integers or rationals, for example).

Option 4:This is true. In a principal ideal domain, any non-zero quotient ring will have a maximal ideal that is principal,

as the quotient ring inherits many of the properties of the original PID.

The correct answers are Option 1 and Option 2.

Concept:
 

Principal Ideal Domain (PID): A ring in which every ideal is generated by a single element.
 

Unique Factorization Domain (UFD): A domain in which every element can be factored uniquely into

irreducible elements, similar to the factorization of integers.
 

Quotient Ring: The quotient ring S/I is formed by "factoring out" the ideal I from S, making all elements

of I zero in the quotient.

Explanation:

Option 1:  \( S = \mathbb{F}_3[x][y] \) which is a polynomial ring in two variables.

In general, a polynomial ring in two or more indeterminates over a field is not a principal ideal domain.

This is because, in such a ring, not every ideal can be generated by a single element (for example,

the ideal (x, y) cannot be generated by a single element).

False.

Option 2: \(S / (y^2 + x^2)\) is the quotient of S  by the ideal generated by \(y^2 + x^2\).

 For this statement to be true, we need to determine whether the quotient ring retains the property of being

a unique factorization domain.

In general, quotient rings of UFDs may or may not be UFDs, and it depends on the structure of the quotient. However,

in this specific case, without additional structure on the ideal, we cannot guarantee that the quotient is a UFD.

False.

Option 3:  \(S = \mathbb{F}_3[x][y] \) is a polynomial ring in two indeterminates over a field.

It is known that a polynomial ring in any number of variables over a field is a unique factorization domain (UFD).

This is a well-established result in commutative algebra.

True.

Option 4:  The quotient ring \(S/(x) \) is obtained by factoring out the ideal generated by  x  from \( S = \mathbb{F}_3[x][y]\),

leaving us with the ring \(\mathbb{F}_3[y] \), which is a polynomial ring in one variable over \(\mathbb{F}_3\) .

 A polynomial ring in one variable over a field is a principal ideal domain.

True.

Hence, correct options are 1) and 2) , which are false 

Concept:

(i) A principal ideal domain is an integral domain in which every proper ideal can be generated by a single element.

(ii) Ideal generated by irreducible element is primal ideal and maximal ideal

Explanation:

R = ℤ[X]/(x2 + 1) and ψ : ℤ[X] → R be the natural quotient map.

R = ℤ[X]/(x2 + 1) ≅ ℤ(i)

So, R is isomorphic to a subring of ℂ. 

(1) is correct

(2): Let I be the prime ideal generated by (x2 + 1)

Then Ffor any prime number p ∈ ℤ

ψ(p) = p + I, which is not unit so ψ(p) is a proper ideal of R.

(2) is correct.

(3): R = ℤ[X]/(x2 + 1) ≅ ℤ(i) 

ℤ(i) is ED so PID and hence ℤ(i) has infinitely many prime ideals.

Therefore R has infinitely many prime ideals.

(3) is correct

(4): ψ(X) is generated by is not primal ideal

(4) is false

Explanation:

Recall: Eisenstein Criterion

Let f = a0 + a1(x) + a2x2 + … + anxn, ai ∈ ℤ.

If ∃ a prime p such that. p|a0, p|a1, …, p∣an−1, p × an + p2 × a0 then f(x) is irreducible over Q.

Let f(x) = 21x3 − 3x2 + 9

then ∃ p = 3 s.t. 3|0, 3|(−3), 3∣21 but 32 × 21.

Now, By Mod p Test for irreducibility, f(x) is irreducible.

∵ For p = 2, \(\rm\overline{f(x)}\) = x3 + x2 + 1 (f̅(x) = f(x) mod 2)

and f̅(0) = 1 ≠ 0, f̅(1) = 1 ≠ 0 ⇒ f̅(x) has no root in ℤ2.

⇒ f(x) is irreducible over ℤ2.

⇒ f̅(x) is irreducible over Q.

∴option (2) is not correct.

But over R, f(x) is reducible.

∵ f(x) is a polynomial of degree 3.

⇒ f(x) has at least one real root.

Then by reducibility Test for 2 or 3-degree polynomial f(x) is reducible over ℝ.

option (1) is not correct.

∴ f(x) can sometimes be irreducible and can some be reducible.

option (3) is correct.

Recall: One-degree polynomial is always irreducible.

∴ B(x) cart has degree 1.

option (4) is not correct.

Concept:

If xⁿ = 1 then o(x) divides n

Explanation:

ℤ/105ℤ ≅ ℤ₁₀₅

105 = 3 × 5 × 7

So \(U_{ℤ_{105}}\) ≅ U(3) × U(5) × U(7) ≅ ℤ₂ × ℤ₄ × ℤ₆

Given x2 = 1 so o(x) divides 2 Hence o(x) = 1 or 2

Element of ℤ2 of order 1 and 2 is 2

Element of ℤ4 of order 1 and 2 is 2

Element of ℤ6 of order 1 and 2 is 2

Hence total such elements = 2 × 2 × 2 = 8 

Option (4) is correct

Explanation:

Let the operation, Δ i.e., A, B ∈ P(X) ⇒ A Δ B = (A ∪ B) \ (A ∩ B) for this,

(i) Closer: Let A, B ∈ P(x) then A Δ B = (A ∪ B) \ (A ∩ B) ∈ P(X)
So, P(x) is closed under Δ. 

(ii) Associativity: let A, B, C ∈ P(x), then (A Δ B) ΔC = ([(A ∪ B) \ (A ∩ B))] ∪ C) \[([A ∪ B) \ ((A ∩ B))] ∩ C)

A Δ (B Δ C) = (A ∪[(B ∪ C) | (B∩C)]) \ (A∩[(B∪C) | (B∩C)])

form, figures you can see,

(A Δ B) ΔC = A Δ (B Δ C)

(iii) Identity:

AΔϕ = (A ∪ ϕ) \ (A ∩ ϕ) = A \ ϕ = A

So, ϕ ∈ P(x) such that A Δ ϕ = A

(iv) Inverse:

A Δ A = (A ∪ A) \ (A ∩ A) = A \ A = ϕ

So, for A ∈ P(x),  A-1 = A.

∴ P(x) is group under Δ.

Now for * operation, A * B = A ∩ B, A, B ∈ P(x)

let x = {1, 2, 3} then P(x) = {ϕ, x, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}}

Here, if we take, e = x

(∵ x ∩ A = A, A ∈ P(x))

But for e = x, inverse  of any A, A ∈ P(x)

∵ A ∩ B ≠ x (for any A, B ∈ P(x)A, B ≠ x)

So, P(x) is not a group under (*).

option (3) is true.

Concept:

Concept:

If G be a group such that o(G) = 2p where p is odd prime then G ≅ \(\mathbb Z_{2p}\) or G ≅ \(\mathbb D_p\)

(ii) If o(G) = n where G is abelian group then class equation of G is n = 1 + 1 + ... + 1 (n times)

Explanation:

Given o(G) = 10 = 2.5

So here p = 5 which is odd prime.

Hence G ≅ \(\mathbb Z_{10}\) or G ≅ \(\mathbb D_5\)

If G ≅ \(\mathbb Z_{10}\) then it is cyclic so abelian. Therefore class equation of a group of order 10 is 10 = 1 + 1 + … + 1 (10-times)

If G ≅ \(\mathbb D_5\) then there will be 5 rotation and 5 reflection.

So in this case class equation of a group of order 10 is 

10 = 1 + 2 + 2 + 5

Hence option (1) is correct

Solution - Sn denote the group of all permutations on n symbols.  

In \(S_3\) possible Order be lcm (3,1) so maximum possibility be 3 

Therefore, Option 1) is wrong 

In, \(S_4 \)  maximum possibility be 4 

Therefore, Option 2) and Option 3) is also wrong 

In, \(S_5\) has maximum possibility be lcm (3,2) =6 

Therefore, Correct Option is Option 4).

Explanation:

For this type of problems, just try to discard given options by taking suitable counter examples..

For option (1). Let n = 16 then

(n - 1)! + 3 = 15! + 3 = even + odd = odd

⇒ No any even integer divide it ie to (n - 1)! + 3

⇒ option (1) is false. (Because z not true for all)

For option (2). taken n = 17, then

∵ (n - 1)! = 16! , Here n = 17 is prime so it will never divide 16!

⇒ option (2) false.

For option (4). take n = 16, then n² = 16² + n! + 1 = 16 ! + 1 Here 16² is even while 16! + 1 is odd integer So 16² + 16! + 1 

⇒ option (4) is false.

For option (3) is true [consider any n ≥ 16 even]

Note: Proof for this option is too long, so just pay to understand with example.

\(=\frac{p_1^{r_1} p_2^{r_2} \cdots p_n^{r_n}}{p_1^{r_1-1} \cdot\left(p_1-1\right) \cdot p_2^{r_2-1}\left(p_2-1\right) \ldots p_n^{r_{n-1}}\left(p_{n-1}\right)}\)

\(=\frac{p_1 p_2 \cdots p_n}{\left(p_1-1\right)\left(p_2-1\right) \cdots\left(p_n-1\right)}\) = integer (given) = p/n¹/n

∵ (p₁ - 1) × p₁ ⇒ ∃ some other prime p₂ S.t (p₁ - 1)|p₂

But ∵  p₂ is also a prime, so not divisible by any of integer except 1 .

(there of one prime factor. is 2, then n tar as at most two distinct prime faster else one.

thus, option (3) is true

Concept:  

Maximal Ideal: A maximal ideal I in a ring R is an ideal such that the quotient ring R/I is a field. 
 

Prime Ideal: A prime ideal P in a ring R is an ideal such that if the product of two elements is in P ,

then at least one of the elements must be in P .

Explanation:

\(R = \left\{ \sum_{n \in \mathbb{Z}} a_n X^n \mid a_n \in \mathbb{Z}, \text{ and } a_n \neq 0 \text{ for finitely many } n \in \mathbb{Z} \right\}\)

The addition and multiplication in the ring are defined as

\(\left( \sum_{n \in \mathbb{Z}} a_n X^n \right) + \left( \sum_{n \in \mathbb{Z}} b_n X^n \right) = \sum_{n \in \mathbb{Z}} (a_n + b_n) X^n \)

\(\left( \sum_{n \in \mathbb{Z}} a_n X^n \right) \left( \sum_{n \in \mathbb{Z}} b_m X^m \right) = \sum_{k \in \mathbb{Z}} \left( \sum_{n+m=k} a_n b_m \right) X^k\)

Option 1:

The addition in this ring is clearly commutative since the sum of polynomials in any ring is commutative.

Now consider the multiplication. In standard polynomial rings, multiplication is commutative as long as the

coefficients come from a commutative ring (in this case, integers \( \mathbb{Z} \) ).

Since \( \mathbb{Z} \) is commutative under multiplication, and the exponents \(X^n\) follow commutative multiplication rules

(i.e., \(X^n X^m = X^{n+m} \)), the ring R is also commutative under multiplication.

Therefore, the statement R is not commutative is false.

Option 2:

 The ideal \(X-1 \) would be maximal if R/\(X-1 \)  is a field.

However, this is not necessarily true in the ring R as described, since R/\(X-1 \) is unlikely

to be a field (it may reduce to a simpler ring, but not a field).
 

Option 3:

(X - 1, 2) is a standard type of ideal in certain polynomial rings, particularly over integers. For a prime ideal,

the condition that multiplication of elements should remain within the ideal must hold. 

Option 4:

\( (X, 5)\) as a Maximal Ideal: If \( (X, 5)\) is maximal, the quotient R/\( (X, 5)\) should be a field.

While in certain rings this could lead to a field, this needs further validation.

Therefore, option 3) is correct.

Concept:

If R is a commutative ring then,

ab = ba ∀ a,b ∈ R. 

M, which is an ideal of R, will be called the maximal ideal of R,

1) if M ⊂ R, M ≠ R (there is at least one element in R that does not belong to M)

2) There should be no ideal 'N', such that M ⊂ N ⊂ R. (there is no ideal between M and R). 

Analysis:

R/M is a field [∵ every finite integral domain is a field]

∴ R/M is a ring with unity

∴ 1 + M ≠ M

i.e., 1 ∉ M

Now, one belongs to R, but it does not belong to R.

∴ M ≠ R.

Let I be an ideal of R

Such that M ⊆  I ⊆  R

Let, M ≠ I

∃ a ∈ I, such that a ∉ M

∴ a + M ∉ M

Now, R/M is a field.

∴ Every, non-zero of R/M is revertible

∴ a + M is invertible

∴ ∃ b + M ∈ R/M such that

(a + M) (b + M) = 1 + M

ab + M = 1 + M

ab – 1 ∈ M ⊆  I      ---(1)

a ∈ I, b ∈ R

∴ ab ∈ I          ---(2)  (∵ I is an ideal)

From (1) and (2), we can write

ab – (ab – 1) ∈ I

∴ 1 ∈ I

Now, as unity belongs to ideal, so ideal becomes ring

∴ I = R

∴ M is a maximal ideal of R

If R is a commutative ring with unity then every maximal ideal is a prime ideal.

Explanation:

S = \(\rm \begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\) So S² = \(\rm \begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\)\(\rm \begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\) = \(\rm \begin{bmatrix}-1&0\\ 0&-1\end{bmatrix}\) = - I, S3 = -S, S⁴ = I

Also, 𝑇1 = \(\rm \begin{bmatrix}i&0\\ 0&i\end{bmatrix}\)

So, 𝑇1 = i𝑇1, \(𝑇_1^2 =-I\), \(𝑇_1^3 =-iI\), \(𝑇_1^4 =I\)

Let Q₄ = {S, S², S³, I, i, -i, iS} then Q₄ is a quaternion group

Given 

𝐺1, 𝐺2, 𝐺3 be three subgroups of 𝐺𝐿2 (ℂ) given by

𝐺1 = < 𝑆, 𝑇1 >, where 𝑇1 = \(\rm \begin{bmatrix}i&0\\ 0&i\end{bmatrix}\),

𝐺2 = < 𝑆, 𝑇2 >, where 𝑇2 = \(\rm \begin{bmatrix}i&0\\ 0&-i\end{bmatrix}\),

𝐺3 = < 𝑆, 𝑇3 >, where 𝑇3 = \(\rm \begin{bmatrix}0&1\\ 1&0\end{bmatrix}\).

Then 𝐺1 = \(\{S^iT_1^j|i, j\in\mathbb Z\}\) is non-abelian group of order 8

Similarly, 𝐺2 = \(\{S^iT_2^j|i, j\in\mathbb Z\}\) is non-abelian group of order 8

𝐺3 = \(\{S_1, T_3|S^4=I, T_3^2=I, T_3S=T_3S^{-1}\}\) ≈  D₄

So, |𝑍(𝐺3)| = |𝑍(D4)| = 2

Hence 𝑍(𝐺3) = {\(\rm \begin{bmatrix}1&0\\ 0&1\end{bmatrix}\)} false because then |𝑍(𝐺3)| = 1

𝑍(𝐺1) is isomorphic to 𝑍(𝐺2) also 𝐺1 is not somorphic to 𝐺3

Hence (1), (2), (3) are false

(4) is correct

Concept:

A mapping ϕ: \(\mathbb N\) →  \(\mathbb N\) defined by ϕ(n) = {x ∈ \(\mathbb N\) | 1 ≤ x

ϕ (pⁿ) = pⁿ - p^n-1

ϕ(mn) = ϕ(m)ϕ(n) if gcd(m, n) = 1 
Explanation:

ϕ(n) table:

From the table of ϕ(n) we can see that if we take n as a prime number greater than 3, then ϕ(n) > ϕ(n+1) and if we take n + 1 as a prime number greater than 3, then ϕ(n) < ϕ(n+1)  

∴ options (1) and (2) are correct.

ϕ(n) table:

So if we take N = 6, then ∀ n > 6, we have ϕ(N) < ϕ(n)

So option (3) is correct 

So the option that is not true is (4)