CUET PG Mathematics Chapter Practice

To determine if a non-empty subset $S$ of a ring $R$ is a subring, we check two conditions: 1. For all $a, b \in S$, $a-b \in S$. 2. For all $a, b \in S$, $a \cdot b \in S$. Let's examine each option: **1. The set $S_1$ of $M_2(\mathbb{Z})$:** * $S_1$ is non-empty, as * Let $A$ and $B$ be in $S_1$, where Since $a_1-a_2 \in \mathbb{Z}$ and $b_1-b_2 \in \mathbb{Z}$, $A-B \in S_1$. * Since $a_1a_2 \in \mathbb{Z}$ and $b_1b_2 \in \mathbb{Z}$, $A \cdot B \in S_1$. Therefore, $S_1$ is a subring. **2. The set $S_2$ of $\mathbb{Z} \times \mathbb{Z}$:** * $S_2$ is non-empty, as $(0,0) \in S_2$. * Let $(a_1, a_1)$ and $(a_2, a_2)$ be in $S_2$. Since $a_1-a_2 \in \mathbb{Z}$, $(a_1-a_2, a_1-a_2) \in S_2$. * Since $a_1a_2 \in \mathbb{Z}$, $(a_1a_2, a_1a_2) \in S_2$. Therefore, $S_2$ is a subring. **3. The set $S_3$ of $M_2(\mathbb{Z})$:** * $S_3$ is non-empty, as the zero matrix has a sum of entries $0+0+0+0=0$. * Let $A$ and $B$ be in $S_3$, where This means $a+b+c+d=0$ and $e+f+g+h=0$. The sum of entries of $A-B$ is So $A-B \in S_3$. * Now check closure under multiplication. Let's take two matrices in $S_3$: Let $A$ be The sum of entries is $1-1+0+0=0$, so $A \in S_3$. Let $B$ be The sum of entries is $1+0-1+0=0$, so $B \in S_3$. Now compute their product: The sum of entries of $A \cdot B$ is $2+0+0+0=2$. Since $2 \neq 0$, $A \cdot B \notin S_3$. Therefore, $S_3$ is NOT a subring. **4. The set $S_4$ of $\mathbb{R}$:** * $S_4$ is non-empty, as $0 = 0+0\sqrt{3} \in S_4$. * Let $(a_1+b_1\sqrt{3})$ and $(a_2+b_2\sqrt{3})$ be in $S_4$. Since $a_1-a_2 \in \mathbb{Z}$ and $b_1-b_2 \in \mathbb{Z}$, this difference is in $S_4$. * Since $(a_1a_2+3b_1b_2) \in \mathbb{Z}$ and $(a_1b_2+b_1a_2) \in \mathbb{Z}$, this product is in $S_4$. Therefore, $S_4$ is a subring. The only set that is not a subring is $S_3$ because it is not closed under multiplication. Answer: $\boxed{S_3}$

A set $R$ with two binary operations $(+, \cdot)$ is a ring if: 1. $(R, +)$ is an abelian group. 2. Multiplication $(\cdot)$ is associative. 3. The distributive laws hold: $a \cdot (b+c) = (a \cdot b) + (a \cdot c)$ and $(a+b) \cdot c = (a \cdot c) + (b \cdot c)$ for all $a, b, c \in R$. Let's examine each option: * **Option 1: The set of integers $\mathbb{Z}$ with standard addition $(+)$ and multiplication $(\cdot)$.** $\mathbb{Z}$ is a canonical example of a ring, satisfying all the ring axioms. $( \mathbb{Z}, +)$ is an abelian group, multiplication is associative, and multiplication distributes over addition. * **Option 2: The set of $2 \times 2$ matrices with integer entries, $M_2(\mathbb{Z})$, with standard matrix addition $(+)$ and multiplication $(\cdot)$.** $M_2(\mathbb{Z})$ forms a ring under standard matrix addition and multiplication. The set of $2 \times 2$ matrices with entries from a ring (like $\mathbb{Z}$) always forms a ring. All ring axioms are satisfied. * **Option 3: The set of even integers $2\mathbb{Z} = \{2k \mid k \in \mathbb{Z}\}$ with standard addition $(+)$ and multiplication $(\cdot)$.** 1. $(2\mathbb{Z}, +)$ is an abelian group: It is closed under addition ($2k_1 + 2k_2 = 2(k_1+k_2) \in 2\mathbb{Z}$), addition is associative, $0 = 2 \cdot 0 \in 2\mathbb{Z}$ is the additive identity, for any $2k \in 2\mathbb{Z}$, its additive inverse $-2k \in 2\mathbb{Z}$, and addition is commutative. 2. Multiplication is associative: Standard integer multiplication is associative. 3. Distributive laws hold: Standard integer multiplication distributes over addition. Thus, $2\mathbb{Z}$ is a ring (it is a subring of $\mathbb{Z}$). * **Option 4: The set of integers $\mathbb{Z}$ with standard addition $(+)$ and a new multiplication operation $(\ast)$ defined as $a \ast b = a+b$ for all $a, b \in \mathbb{Z}$.** 1. $(\mathbb{Z}, +)$ is an abelian group. This part is satisfied. 2. Check associativity of $(\ast)$: For any $a, b, c \in \mathbb{Z}$, Since addition in $\mathbb{Z}$ is associative, $(a+b)+c = a+(b+c)$, so the operation $(\ast)$ is associative. 3. Check distributive laws: We need to verify if $a \ast (b+c) = (a \ast b) + (a \ast c)$ (left distributive law) and $(a+b) \ast c = (a \ast c) + (b \ast c)$ (right distributive law). Let's check the left distributive law: For the left distributive law to hold, we must have $a+(b+c) = a+b+a+c$. This simplifies to: This condition must be true for *all* $a \in \mathbb{Z}$, but it is not (for example, if we choose $a=1$, then $1 \neq 0$). Since the left distributive law does not hold for all elements (e.g., for $a=1, b=2, c=3$): Since $6 \neq 7$, the law fails. Therefore, $(\mathbb{Z}, +, \ast)$ is not a ring. Therefore, the set described in Option 4 does NOT form a ring. Answer: $\boxed{\text{Option 4}}$

**Step 1: Understand the definition of a general ring.** A ring $(R, +, \cdot)$ is a set with two binary operations, addition and multiplication, satisfying specific axioms: 1. $(R, +)$ is an abelian group (closure, associativity, identity $0$, inverse $-a$, commutativity). 2. Multiplication is associative. 3. Distributive laws hold: It is crucial to remember that a general ring does **not** necessarily have a multiplicative identity (unity), nor is multiplication necessarily commutative, nor do non-zero elements necessarily have multiplicative inverses. **Step 2: Evaluate each option based on the definition of a general ring.** * **Option 1: "For all $a, b \in R$, $a \cdot b = b \cdot a$."** This statement describes a commutative ring. However, not all rings are commutative. For example, the ring of $2 \times 2$ matrices with real entries, denoted by $M_2(\mathbb{R})$, is a non-commutative ring. If we take and then while so $AB \neq BA$. Therefore, this statement is not always true for any arbitrary ring. * **Option 2: "There exists an element $1 \in R$ such that $a \cdot 1 = 1 \cdot a = a$ for all $a \in R$."** This statement describes a ring with unity (or a ring with identity). Not all rings have a multiplicative identity. For instance, the ring of even integers $(2\mathbb{Z}, +, \cdot)$ does not contain an element $1$ such that $a \cdot 1 = a$ for all $a \in 2\mathbb{Z}$. The only integer that acts as a multiplicative identity is $1$, but $1 \notin 2\mathbb{Z}$. Therefore, this statement is not always true for any arbitrary ring. * **Option 3: "For all $a, b, c \in R$, $a \cdot (b - c) = (a \cdot b) - (a \cdot c)$."** In any ring $R$, subtraction is defined using the additive inverse: Let's expand the left-hand side of the equation: By the left distributive law (an axiom of any ring): From the basic properties of a ring (as stated in the topic notes), we know that for any $a, x \in R$, Applying this property to $a \cdot (-c)$: By the definition of subtraction, $x + (-y) = x - y$: Thus, is always true for any ring. * **Option 4: "For every non-zero $a \in R$, there exists $a^{-1} \in R$ such that $a \cdot a^{-1} = a^{-1} \cdot a = 1$."** This statement describes a division ring (or a field, if also commutative). A general ring does not require every non-zero element to have a multiplicative inverse. For example, in the ring of integers $(\mathbb{Z}, +, \cdot)$, elements like $2$ do not have a multiplicative inverse within $\mathbb{Z}$. Therefore, this statement is not always true for any arbitrary ring. **Step 3: Conclusion.** Only Option 3 is always true for any arbitrary ring, as it directly follows from the ring axioms (distributivity) and the definition of subtraction in terms of additive inverses. Answer: \boxed{\text{Option 3}}

A non-empty subset $S$ of a ring $R$ is a subring if it is closed under subtraction and multiplication. That is, for all $x, y \in S$: 1. $x - y \in S$ 2. $x \cdot y \in S$ Let's analyze each option: **Option 1: The center of $R$, $Z(R) = \{x \in R \mid xr = rx \text{ for all } r \in R\}$.** * **Non-empty**: $0 \in Z(R)$ since $0 \cdot r = 0 = r \cdot 0$ for all $r \in R$. * **Closure under subtraction**: Let $x, y \in Z(R)$. Then $xr = rx$ and $yr = ry$ for all $r \in R$. We have Thus, $x-y \in Z(R)$. * **Closure under multiplication**: Let $x, y \in Z(R)$. Then $xr = rx$ and $yr = ry$ for all $r \in R$. We have Thus, $xy \in Z(R)$. Therefore, $Z(R)$ is a subring of $R$. **Option 2: The set of all elements $x \in R$ such that $xa = ax$ for a fixed element $a \in R$.** This set is the centralizer of $a$, denoted $C(a)$. * **Non-empty**: $0 \in C(a)$ since $0 \cdot a = 0 = a \cdot 0$. * **Closure under subtraction**: Let $x, y \in C(a)$. Then $xa = ax$ and $ya = ay$. We have Thus, $x-y \in C(a)$. * **Closure under multiplication**: Let $x, y \in C(a)$. Then $xa = ax$ and $ya = ay$. We have Thus, $xy \in C(a)$. Therefore, $C(a)$ is a subring of $R$. **Option 3: The set of all idempotent elements in $R$, $E = \{x \in R \mid x^2 = x\}$.** * **Non-empty**: $0 \in E$ since $0^2 = 0$. If $R$ has a multiplicative identity $1$, then $1 \in E$ since $1^2 = 1$. * **Closure under subtraction**: Let $x, y \in E$. Then $x^2 = x$ and $y^2 = y$. Consider For $x-y$ to be in $E$, we must have $(x-y)^2 = x-y$. This would imply This condition is not generally true for all idempotent elements in a ring. **Counterexample**: Consider the ring of $2 \times 2$ matrices with integer entries, $M_2(\mathbb{Z})$. Let Both $X$ and $Y$ are idempotent elements in $M_2(\mathbb{Z})$ because $X^2=X$ and $Y^2=Y$. Now consider their difference: Let's check if $X-Y$ is idempotent: Since $X-Y$ is not an idempotent element. Thus, $E$ is not closed under subtraction and therefore is not necessarily a subring of $R$. **Option 4: The intersection of any two subrings of $R$.** Let $S_1$ and $S_2$ be two subrings of $R$. * **Non-empty**: Since $S_1$ and $S_2$ are subrings, they must contain the additive identity $0$ of $R$. So $0 \in S_1 \cap S_2$. * **Closure under subtraction**: Let $x, y \in S_1 \cap S_2$. This means $x, y \in S_1$ and $x, y \in S_2$. Since $S_1$ is a subring, $x-y \in S_1$. Since $S_2$ is a subring, $x-y \in S_2$. Therefore, $x-y \in S_1 \cap S_2$. * **Closure under multiplication**: Let $x, y \in S_1 \cap S_2$. This means $x, y \in S_1$ and $x, y \in S_2$. Since $S_1$ is a subring, $x \cdot y \in S_1$. Since $S_2$ is a subring, $x \cdot y \in S_2$. Therefore, $x \cdot y \in S_1 \cap S_2$. Thus, the intersection of any two subrings of $R$ is a subring. Based on the analysis, the set of all idempotent elements in $R$ is not necessarily a subring of $R$. The final answer is $\boxed{\text{The set of all idempotent elements in } R, E = \{x \in R \mid x^2 = x\}}$.

To check if a non-empty subset $S$ of a ring $R$ is a subring, we must verify two conditions: 1. For any $A, B \in S$, $A - B \in S$. 2. For any $A, B \in S$, $A \cdot B \in S$. Also, the additive identity ($0$) of $R$ must be in $S$. Let's examine each option: **Option 1: The set of all matrices $A \in M_2(\mathbb{Z})$ such that $\det(A) = 0$.** Let $S_1 = \{ A \in M_2(\mathbb{Z}) \mid \det(A) = 0 \}$. Consider Both $A, B \in S_1$ since $\det(A) = 0$ and $\det(B) = 0$. However, The determinant of $A-B$ is $\det(A-B) = (1)(-1) - (0)(0) = -1 \neq 0$. Thus, $A-B \notin S_1$. Therefore, $S_1$ is not a subring. **Option 2: The set of all matrices $A \in M_2(\mathbb{Z})$ such that $A$ is invertible.** Let $S_2 = \{ A \in M_2(\mathbb{Z}) \mid A \text{ is invertible} \}$. For matrices with integer entries, invertible typically means $\det(A) = \pm 1$. First, the zero matrix is not in $S_2$ because it is not invertible. A subring must contain the additive identity of the parent ring. Thus, $S_2$ cannot be a subring. Also, if we take both are in $S_2$. But which is not in $S_2$. **Option 3: The set of all matrices of the form** **where $a, b, c \in \mathbb{Z}$.** Let $S_3 = \left\{ \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \mid a, b, c \in \mathbb{Z} \right\}$. 1. **Non-empty:** The zero matrix is in $S_3$ (take $a=0, b=0, c=0$). 2. **Closed under subtraction:** Let be two matrices in $S_3$. Then: Since $a_1-a_2, b_1-b_2, c_1-c_2$ are all integers, $A-B$ is of the required form and thus $A-B \in S_3$. 3. **Closed under multiplication:** Let be two matrices in $S_3$. Then: Since $a_1a_2$, $a_1b_2+b_1c_2$, and $c_1c_2$ are all integers, $A \cdot B$ is of the required form and thus $A \cdot B \in S_3$. Since all conditions are satisfied, $S_3$ is a subring of $M_2(\mathbb{Z})$. **Option 4: The set of all matrices $A \in M_2(\mathbb{Z})$ such that $A^2 = A$.** Let $S_4 = \{ A \in M_2(\mathbb{Z}) \mid A^2 = A \}$. These are idempotent matrices. Consider Both $A, B \in S_4$ because $A^2=A$ and $B^2=B$. However, Let's check if $A-B$ is idempotent: Since $(A-B)^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \neq \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = A-B$, $A-B \notin S_4$. Therefore, $S_4$ is not a subring. Answer: \boxed{\text{The set of all matrices of the form } \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \text{ where } a, b, c \in \mathbb{Z}.}