Algebra Concepts MCQ Quiz
Test your knowledge of groups, rings, fields, and vector spaces
Answer: c) -1
For \( e \) to be the identity element, we need \( a * e = a \) for all \( a \in \mathbb{R} \).
\( a * e = a + e + 1 = a \) ⇒ \( e + 1 = 0 \) ⇒ \( e = -1 \).
Verification: \( a * (-1) = a + (-1) + 1 = a \), and \( (-1) * a = (-1) + a + 1 = a \).
Answer: a) \( \{1, \omega, \omega^2\} \) where \( \omega \) is a complex cube root of unity
• Option a: Forms a cyclic group of order 3 under multiplication. It satisfies closure (\( \omega \cdot \omega^2 = \omega^3 = 1 \)), associativity, has identity (1), and every element has an inverse (\( \omega^{-1} = \omega^2 \)).
• Option b: Not closed under multiplication (\( 2 \cdot 2 = 4 \) which is not in the set).
• Option c: Most integers don't have multiplicative inverses in \( \mathbb{Z} \).
• Option d: Natural numbers don't have multiplicative inverses or additive identity for multiplication.
Answer: b) \( \frac{a}{a-1} \)
Let \( a^{-1} \) be the inverse of \( a \). Then \( a * a^{-1} = e \), where \( e \) is the identity.
First, find the identity: \( a * e = a + e - ae = a \) ⇒ \( e - ae = 0 \) ⇒ \( e(1-a) = 0 \) ⇒ \( e = 0 \) (since \( a \neq 1 \)).
Now, \( a * a^{-1} = a + a^{-1} - a \cdot a^{-1} = 0 \)
⇒ \( a + a^{-1}(1 - a) = 0 \)
⇒ \( a^{-1}(1 - a) = -a \)
⇒ \( a^{-1} = \frac{-a}{1-a} = \frac{a}{a-1} \)
Answer: b) \( \{e, (1 2 3), (1 3 2)\} \)
\( S_3 = \{e, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)\} \) has order 6.
• Option b is \( A_3 \), the alternating group of degree 3, which is the unique subgroup of order 3 in \( S_3 \).
• Subgroups of order 2 in \( S_3 \) (options a, c, d) are not normal. For example, for \( H = \{e, (1 2)\} \):
Take \( g = (1 3) \), then \( gHg^{-1} = (1 3)\{e, (1 2)\}(1 3) = \{e, (2 3)\} \neq H \).
Thus \( H \) is not normal in \( S_3 \).
Answer: d) Infinite abelian group
• The set is infinite because there are infinitely many real numbers \( a \) and \( b \).
• The group is abelian because:
Let \( A = \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \), \( B = \begin{pmatrix} c & d \\ -d & c \end{pmatrix} \)
\( AB = \begin{pmatrix} ac-bd & ad+bc \\ -(ad+bc) & ac-bd \end{pmatrix} \)
\( BA = \begin{pmatrix} ac-bd & ad+bc \\ -(ad+bc) & ac-bd \end{pmatrix} \)
So \( AB = BA \), hence the group is abelian.
This group is isomorphic to the multiplicative group of nonzero complex numbers.
Answer: c) \( \mathbb{Z} \)
An integral domain is a commutative ring with unity and no zero divisors. A field is a commutative ring with unity where every nonzero element has a multiplicative inverse.
• \( \mathbb{Z}_7 \): Is a field (since 7 is prime)
• \( \mathbb{Q} \): Is a field
• \( \mathbb{Z} \): Is an integral domain but not a field (only ±1 have multiplicative inverses in \( \mathbb{Z} \))
• \( \mathbb{R} \): Is a field
Answer: c) 4
A basis for the vector space of all \( 2 \times 2 \) matrices over \( \mathbb{R} \) is:
\( \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\} \)
This set has 4 elements and any \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) can be written as a linear combination:
\( a\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + c\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + d\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \)
Thus, the dimension is 4.
Answer: d) Both a and b
A polynomial is irreducible over a field if it cannot be factored into polynomials of lower degree with coefficients in that field.
• \( x^2 - 2 \): Irreducible over \( \mathbb{Q} \) (no rational roots)
• \( x^2 + x + 1 \): Irreducible over \( \mathbb{Q} \) (discriminant = 1 - 4 = -3 < 0, no rational roots)
• \( x^2 - 4 \): Reducible over \( \mathbb{Q} \) since \( x^2 - 4 = (x-2)(x+2) \)
Thus, both a and b are irreducible over \( \mathbb{Q} \).
Answer: c) \( \mathbb{Q}(\sqrt{2}, \sqrt{3}) \)
Factor the polynomial: \( x^4 - 5x^2 + 6 = (x^2 - 2)(x^2 - 3) \)
The roots are \( \pm\sqrt{2}, \pm\sqrt{3} \).
The smallest field containing \( \mathbb{Q} \) and all these roots is \( \mathbb{Q}(\sqrt{2}, \sqrt{3}) \).
This is the splitting field because the polynomial factors completely into linear factors over this field:
\( x^4 - 5x^2 + 6 = (x - \sqrt{2})(x + \sqrt{2})(x - \sqrt{3})(x + \sqrt{3}) \)
Answer: b) 2
By the Rank-Nullity Theorem:
\( \text{dim}(V) = \text{rank}(T) + \text{nullity}(T) \)
Here, \( \text{dim}(V) = 3 \) (since \( V = \mathbb{R}^3 \))
\( \text{nullity}(T) = \text{dim}(\text{ker}(T)) = 1 \)
So, \( 3 = \text{rank}(T) + 1 \) ⇒ \( \text{rank}(T) = 2 \)
The rank of \( T \) is the dimension of its range, so the dimension of the range is 2.