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

Concept:

Quotient Ring and Chinese Remainder Theorem:

• The ring \( \mathbb{R}[x] \) is a Principal Ideal Domain (PID) and a Unique Factorization Domain (UFD).
• The polynomial \( f \) is a product of distinct monic irreducible polynomials \( P_1, P_2, \dots, P_n \), with \( n \geq 2 \). Therefore, \( f \) is reducible.
• The quotient \( \mathbb{R}[x]/(f) \) is not a field because \( f \) is reducible. A quotient ring is a field only when the modulus is irreducible.
• By the Chinese Remainder Theorem (CRT), since \( P_1, P_2, \dots, P_n \) are pairwise coprime:

  \( \mathbb{R}[x]/(f) \cong \mathbb{R}[x]/(P_1) \times \mathbb{R}[x]/(P_2) \times \dots \times \mathbb{R}[x]/(P_n) \)

• Each \( \mathbb{R}[x]/(P_i) \) is a field:
  • If \( P_i \) has degree 1, it is isomorphic to \( \mathbb{R} \).
  • If \( P_i \) has degree 2 and is irreducible over \( \mathbb{R} \), it is isomorphic to \( \mathbb{C} \).
• The quotient ring is a finite-dimensional \( \mathbb{R} \)-vector space and a direct sum of fields. Thus, it is semi simple and contains no non-zero nilpotent elements.

Calculation:

Given,

\( f = P_1 \times P_2 \times \dots \times P_n \) where \( P_i \) are distinct irreducible polynomials.

⇒ \( f \) is reducible because \( n \geq 2 \).

⇒ By CRT, the ring splits as:

\( \mathbb{R}[x]/(f) \cong \mathbb{R}[x]/(P_1) \times \mathbb{R}[x]/(P_2) \times \dots \times \mathbb{R}[x]/(P_n) \)

⇒ Each factor is a field, so \( \mathbb{R}[x]/(f) \) is a product of fields, not a single field.

⇒ The total dimension over \( \mathbb{R} \) is finite since each \( \mathbb{R}[x]/(P_i) \) is finite-dimensional over \( \mathbb{R} \).

⇒ As it is a product of fields, it is reduced: no non-zero nilpotent elements exist.

Option-wise Explanation:

Option 1: \( \mathbb{R}[x]/(f) \) is a field.

False. Since \( f \) is reducible, the quotient is a product of fields, not a single field.

Option 2: \( \mathbb{R}[x]/(f) \) is a finite-dimensional \( \mathbb{R} \)-vector space.

True. Each quotient \( \mathbb{R}[x]/(P_i) \) is finite-dimensional, so the product is finite-dimensional.

Option 3: \( \mathbb{R}[x]/(f) \) is a direct sum of fields, each isomorphic to \( \mathbb{R} \) or \( \mathbb{C} \).

True. By CRT and properties of irreducible polynomials over \( \mathbb{R} \).

Option 4: There are no non-zero elements \( u \in \mathbb{R}[x]/(f) \) such that \( u^m = 0 \) for some \( m \geq 1 \).

True. As a product of fields, the ring has no non-zero nilpotent elements.

∴ The correct statements are 2, 3, and 4.

Concept:

• Divisible Group: A group G is said to be divisible if for every element \( y \in G \) and every positive integer \( n \), there exists an element \( x \in G \) such that \( x^n = y \).
• Group \(\mathbb{Q} \)  under Addition: For any rational \( y \in \mathbb{Q} \) and any \( n \in \mathbb{Z}^+ \), \( x = \frac{y}{n} \in \mathbb{Q} \) satisfies \( nx = y \) ⇒ divisible.
• Group \(\mathbb{C} \setminus \{0\} \)  under Multiplication: Every non-zero complex number has an \( n^{th} \) root in \( \mathbb{C} \setminus \{0\} \) ⇒ divisible.
• Finite Groups: A finite group cannot be divisible, because if \( G \) has finite order, then not all equations like \( x^n = y \) are solvable for arbitrary \( y \in G \).
• Cyclic Group of Order 5: Has only 5 elements. For \( n = 10 \), not every element has an \( x \in G \) such that \( x^{10} = y \) ⇒ not divisible.
• Symmetric Group \(S_5 \) : Non-abelian finite group of permutations of 5 elements. Not divisible for same reason: finite, and not all roots exist for all elements.

Calculation:

Given:

Group must satisfy: for every \( y \in G \) and every \( n \in \mathbb{Z}^+ \), there exists \( x \in G \) such that \( x^n = y \)

Step 1: Check  \(\mathbb{Q} \)  under addition

⇒ Let \( y \in \mathbb{Q}, n \in \mathbb{Z}^+ \Rightarrow x = \frac{y}{n} \in \mathbb{Q} \)

⇒ \( nx = y \Rightarrow \) divisible

Step 2: Check \(\mathbb{C} \setminus \{0\} \)  under multiplication

⇒ Let \( y \in \mathbb{C} \setminus \{0\}, n \in \mathbb{Z}^+ \)

⇒ Take any \( n^{th} \) root of \( y \)

⇒ Roots exist in \( \mathbb{C} \)

⇒ divisible

Step 3: Check cyclic group of order 5

⇒ Order is 5

⇒ For \( n = 2,3,4,\dots \) roots might not exist

⇒ Not divisible

Step 4: Check symmetric group \(S_5 \) 

⇒ Finite group of order 120

⇒ Not every element has an \( n^{th} \) root

⇒ Not divisible

∴ Final Answer: Only options 1 and 2 are correct. 

Concept:

• Idempotent Element: An element r in a ring R is said to be idempotent if r² = r.
• The condition r² = r for all r ∈ R implies that every element is idempotent.
• Such rings are known as Boolean rings.

• Properties of Boolean Rings:
  • Every element satisfies r = −r ⇒ 2r = 0 ⇒ Ring has characteristic 2.
  • Commutativity: All Boolean rings are commutative.
  • Prime ideals: Every nonzero prime ideal is also maximal.
  • Integral domain status: Most Boolean rings are not integral domains because r(r−1) = 0 may have nontrivial zero divisors.
  • Exception: The field 𝔽₂ (with elements 0 and 1 under mod 2 arithmetic) satisfies r² = r and is an integral domain.

Calculation:

Given:

r² = r for all r ∈ R

⇒ r² − r = 0

⇒ r(r − 1) = 0

⇒ This implies possible zero divisors unless r = 0 or r = 1

⇒ But in the ring 𝔽₂, r = 0 or 1, and it has no zero divisors

⇒ So, 𝔽₂ is an example where the ring satisfies r² = r and is an integral domain

⇒ Hence, R is not necessarily a non-integral domain

⇒ r + r = 0

⇒ 2r = 0

⇒ r = −r for all r ∈ R

option 2 is correct 

⇒ Boolean rings are always commutative

⇒ In Boolean rings, every nonzero prime ideal is maximal

∴ Correct statements are:

(2) r = -r for all r ∈ R,

(3) every nonzero prime ideal of R is maximal, and

(4) R must be a commutative ring.

But (1) is incorrect because R can be an integral domain in the case of 𝔽₂.

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

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:

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.

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.

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)

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.