CUET PG Mathematics Chapter-wise PYQs

**Given:** The cardinalities for a potential integral domain are: A. $5$ B. $6$ C. $7$ D. $10$ **Required:** To determine for which of the given cardinalities an integral domain is not possible. **Solution:** *Step 1:* State the fundamental theorem regarding finite integral domains. A key theorem in ring theory states that every finite integral domain is a field. *Step 2:* State the property for the order of a finite field. The order, or cardinality, of any finite field must be of the form $p^n$, where $p$ is a prime number and $n$ is a positive integer. *Step 3:* Combine the results from Step 1 and Step 2. From the two theorems, it follows that the cardinality of any finite integral domain must be a power of a prime number, $p^n$. We must check which of the given cardinalities do not satisfy this condition. *Step 4:* Analyze each given cardinality. - **A. 5:** The cardinality is $5$, which can be written as $5^1$. This is a prime power ($p=5, n=1$). Thus, an integral domain of order $5$ exists (e.g., the field $\mathbb{Z}_5$). - **B. 6:** The cardinality is $6$. The prime factorization of $6$ is $2 \times 3$. This is not a power of a single prime. Thus, an integral domain of order $6$ is not possible. - **C. 7:** The cardinality is $7$, which can be written as $7^1$. This is a prime power ($p=7, n=1$). Thus, an integral domain of order $7$ exists (e.g., the field $\mathbb{Z}_7$). - **D. 10:** The cardinality is $10$. The prime factorization of $10$ is $2 \times 5$. This is not a power of a single prime. Thus, an integral domain of order $10$ is not possible. *Step 5:* Conclude based on the analysis. The cardinalities for which an integral domain is not possible are $6$ (B) and $10$ (D). **Answer:** $\boxed{\text{B and D only}}$

**Given:** A ring $(R, +, \cdot)$ where every element is idempotent. This means for any $x \in R$, the property $x^2 = x$ holds. **Required:** To determine which of the given properties is always true for such a ring. **Solution:** *Step 1:* Analyze the consequence of the idempotent property on the sum of two elements. Let $a, b$ be any two elements in $R$. Since $R$ is closed under addition, $a+b \in R$. Therefore, $a+b$ must also be idempotent. Using the distributive laws to expand the left side: By the idempotent property, $a^2 = a$ and $b^2 = b$. Substituting these into the expansion gives: Equating the two expressions for $(a+b)^2$: Subtracting $a$ and $b$ from both sides (which is possible as $(R,+)$ is a group) yields: This implies that for any $a, b \in R$: *Step 2:* Determine the characteristic of the ring. For any element $x \in R$, we know $x^2 = x$. From the result in Step 1, we can set $a=x$ and $b=x$: Since $x^2=x$, this becomes: This means every element is its own additive inverse. This is equivalent to $x+x=0$, or $2x=0$, for all $x \in R$. This shows that a non-trivial Boolean ring has a characteristic of 2. *Step 3:* Prove commutativity. From Step 1, we have $ab = -ba$. From Step 2, we know that for any element $y \in R$, $y = -y$. Let $y = ba$. Then $ba = -ba$. Substituting this into the first equation: Since this holds for all $a, b \in R$, the ring $R$ is always a commutative ring. *Step 4:* Evaluate the other options using counterexamples. - **Option 2 (not an integral domain) and Option 4 (an integral domain with unity):** Consider the ring $\mathbb{Z}_2 = \{0, 1\}$. Here, $0^2=0$ and $1^2=1$, so it is a Boolean ring. $\mathbb{Z}_2$ is also an integral domain. This serves as a counterexample to the claim that a Boolean ring is "not an integral domain". - **Option 3 (a field):** Consider the direct product ring $R = \mathbb{Z}_2 \times \mathbb{Z}_2$. Every element in this ring is idempotent. However, it is not a field because it has zero divisors. For example, $(1,0) \neq (0,0)$ and $(0,1) \neq (0,0)$, but their product is $(1,0) \cdot (0,1) = (0,0)$. Based on the derivation, the only property that is always true is that the ring is commutative. **Answer:** $\boxed{\text{a commutative ring}}$

**Given:** Two lists, List-I containing sets with algebraic operations and List-II containing types of algebraic structures. **Required:** To match each item in List-I with the correct item in List-II. **Solution:** *Step 1:* Analyze item (A) from List-I. The set of all even integers, $(2\mathbb{Z}, +, \cdot)$, forms a ring. Multiplication is commutative since for any $2m, 2n \in 2\mathbb{Z}$, $(2m)(2n) = 4mn = (2n)(2m)$. However, this ring does not have a multiplicative identity (unity), because $1 \notin 2\mathbb{Z}$. Since it lacks unity, it cannot be an integral domain or a field. It is a commutative ring. Thus, (A) matches with (IV). *Step 2:* Analyze item (B) from List-I. The set of Gaussian integers, $\mathbb{Z}[i] = \{a + ib : a, b \in \mathbb{Z}\}$, is a subring of the complex numbers $\mathbb{C}$. 1. It is a commutative ring because complex multiplication is commutative. 2. It has unity, $1 = 1 + 0i \in \mathbb{Z}[i]$. 3. It has no zero divisors, because if $(a+ib)(c+id)=0$, then since $\mathbb{C}$ is a field, either $a+ib=0$ or $c+id=0$. Therefore, $\mathbb{Z}[i]$ is an integral domain. It is not a field because not every non-zero element has an inverse in $\mathbb{Z}[i]$ (e.g., the inverse of $2$ is $1/2$, which is not in $\mathbb{Z}[i]$). Thus, (B) matches with (II). *Step 3:* Analyze item (C) from List-I. The set of rational numbers, $(\mathbb{Q}, +, \cdot)$, is a commutative ring with unity $1$. For every non-zero element $a/b \in \mathbb{Q}$ (where $a,b \neq 0$), its multiplicative inverse is $b/a$, which is also in $\mathbb{Q}$. By definition, this makes $\mathbb{Q}$ a field. Thus, (C) matches with (I). *Step 4:* Analyze item (D) from List-I. The set is a ring under matrix addition and multiplication. To check for commutativity, consider two elements from $S$: Let $A$ and $B$ be two elements from $S$: Now, we compute their products: In general, $x_1 y_2 \neq x_2 y_1$. For example, if $x_1=1, y_1=2, x_2=3, y_2=4$, then: Since $A \cdot B \neq B \cdot A$, the ring $S$ is non-commutative. Thus, (D) matches with (III). *Step 5:* Consolidate the matches. (A) $\to$ (IV) (B) $\to$ (II) (C) $\to$ (I) (D) $\to$ (III) **Answer:** $\boxed{\text{(A) - (IV), (B) - (II), (C) - (I), (D) - (III)}}$

**Given:** A field $F$ with order $|F| = 16384$. **Required:** The number of proper subfields of $F$. **Solution:** *Step 1:* Express the order of the field in the form $p^n$, where $p$ is a prime number and $n$ is a positive integer. The order of the field is $16384$. We recognize this as a power of $2$. Thus, the field $F$ is isomorphic to the finite field $\mathbb{F}_{2^{14}}$. Here, $p=2$ and $n=14$. *Step 2:* Determine the total number of subfields of $F$. The number of subfields of a finite field $\mathbb{F}_{p^n}$ is equal to the number of positive divisors of the exponent $n$. The positive divisors of $n=14$ are $1, 2, 7,$ and $14$. The number of divisors is $4$. Therefore, $F$ has exactly $4$ subfields. These subfields have orders $2^1, 2^2, 2^7,$ and $2^{14}$. *Step 3:* Calculate the number of proper subfields. A proper subfield is any subfield of $F$ that is not equal to $F$ itself. The field $F$ corresponds to the subfield of order $2^{14}$. **Answer:** $\boxed{3}$

**Given:** - A polynomial $P(x)$ of degree 10. - The leading coefficient of $P(x)$ is 11. - The factors of $P(x)$ are $(x-1), (x-2), (x-3), \dots, (x-10)$. **Required:** - The coefficient of the $x^9$ term in $P(x)$. **Solution:** *Step 1:* Construct the polynomial $P(x)$ using its factors and leading coefficient. Since the polynomial has degree 10 and its factors are $(x-1), (x-2), \dots, (x-10)$, it can be expressed in the form: The leading coefficient $A$ is given as 11. *Step 2:* Relate the coefficient of $x^9$ to the sum of the roots of the polynomial. The roots of $P(x)$ are $r_1=1, r_2=2, \dots, r_{10}=10$. For a polynomial of degree $n$, the coefficient of the $x^{n-1}$ term is $-A \times (\text{sum of the roots})$. Here, $n=10$ and $A=11$. The coefficient of $x^9$ is $-11 \times (1+2+\dots+10)$. *Step 3:* Calculate the sum of the roots. The sum of the roots is the sum of the first 10 positive integers. Using the formula for the sum of the first $k$ integers, $S_k = \frac{k(k+1)}{2}$: *Step 4:* Determine the coefficient of $x^9$. The coefficient of $x^9$ in $P(x)$ is $-11$ multiplied by the sum of the roots. **Answer:** $\boxed{-605}$

**Given:** The rings $\mathbb{Z}_{100}$, $\mathbb{Z}_{102}$, $\mathbb{Z}_{113}$, and $\mathbb{Z}_{153}$. **Required:** To determine which of the given rings is an integral domain. **Solution:** *Step 1:* Recall the definition of an integral domain and the condition for $\mathbb{Z}_n$. An integral domain is a commutative ring with unity that has no zero divisors. A ring of integers modulo $n$, denoted $\mathbb{Z}_n$, is an integral domain if and only if $n$ is a prime number. This is because if $n$ is composite, say $n = ab$ for $1 < a, b < n$, then $a$ and $b$ are non-zero elements in $\mathbb{Z}_n$ whose product $ab \equiv 0 \pmod n$, making them zero divisors. *Step 2:* Check the primality of $n$ for each option. 1. For $\mathbb{Z}_{100}$, $n = 100$. Since $100 = 10 \times 10$, $n$ is composite. Thus, $\mathbb{Z}_{100}$ is not an integral domain. 2. For $\mathbb{Z}_{102}$, $n = 102$. Since $102 = 2 \times 51$, $n$ is composite. Thus, $\mathbb{Z}_{102}$ is not an integral domain. 3. For $\mathbb{Z}_{113}$, $n = 113$. To test for primality, we check for divisibility by primes up to $\sqrt{113} \approx 10.6$. The primes to check are 2, 3, 5, 7. * $113$ is not divisible by 2 (it is odd). * The sum of digits is $1+1+3=5$, which is not divisible by 3. * The number does not end in 0 or 5, so it is not divisible by 5. * $113 \div 7 = 16$ with a remainder of $1$. Since $113$ is not divisible by any prime less than or equal to its square root, $113$ is a prime number. 4. For $\mathbb{Z}_{153}$, $n = 153$. The sum of digits is $1+5+3=9$, which is divisible by 3 and 9. Thus, $n$ is composite ($153 = 9 \times 17$). Thus, $\mathbb{Z}_{153}$ is not an integral domain. *Step 3:* Conclude which ring is an integral domain. Since $n = 113$ is the only prime number among the choices, $\mathbb{Z}_{113}$ is the only ring that is an integral domain. **Answer:** $\boxed{\mathbb{Z}_{113}}$

**Given:** The rings $\mathbb{Z}_{100}$, $\mathbb{Z}_{102}$, $\mathbb{Z}_{113}$, and $\mathbb{Z}_{153}$. **Required:** To determine which of the given rings is an integral domain. **Solution:** *Step 1:* Recall the definition of an integral domain and the property of the ring $\mathbb{Z}_n$. An integral domain is a commutative ring with unity that has no zero divisors. A fundamental theorem states that the ring of integers modulo $n$, $(\mathbb{Z}_n, +_n, \times_n)$, is an integral domain if and only if $n$ is a prime number. *Step 2:* Analyze the modulus $n$ for each option to check for primality. * For $\mathbb{Z}_{100}$: The modulus is $n=100$. Since $100 = 10 \times 10$, $100$ is a composite number. Therefore, $\mathbb{Z}_{100}$ is not an integral domain. For example, $10 \neq 0$ and $10 \neq 0$, but $10 \times_{100} 10 = 100 \equiv 0 \pmod{100}$. * For $\mathbb{Z}_{102}$: The modulus is $n=102$. Since $102$ is an even number, it is divisible by $2$. $102 = 2 \times 51$, so $102$ is composite. Therefore, $\mathbb{Z}_{102}$ is not an integral domain. * For $\mathbb{Z}_{113}$: The modulus is $n=113$. To check if $113$ is prime, we test for divisibility by prime numbers up to $\sqrt{113} \approx 10.6$. The primes to check are $2, 3, 5, 7$. * $113$ is not divisible by $2$ (it's odd). * The sum of digits is $1+1+3=5$, which is not divisible by $3$. * The last digit is not $0$ or $5$, so it's not divisible by $5$. * $113 = 7 \times 16 + 1$, so it's not divisible by $7$. Since $113$ is not divisible by any prime less than or equal to its square root, $113$ is a prime number. Therefore, $\mathbb{Z}_{113}$ is an integral domain. * For $\mathbb{Z}_{153}$: The modulus is $n=153$. The sum of digits is $1+5+3=9$, which is divisible by $3$ and $9$. $153 = 3 \times 51$, so $153$ is composite. Therefore, $\mathbb{Z}_{153}$ is not an integral domain. *Step 3:* Conclude based on the analysis. Only the ring with a prime modulus, $\mathbb{Z}_{113}$, is an integral domain. **Answer:** $\boxed{\mathbb{Z}_{113}}$

**Given:** * Statement I: For each positive integer $n$, the set $n\mathbb{Z} = \{0, \pm n, \pm 2n, \pm 3n, ...\}$ is a subring of the ring of integers $\mathbb{Z}$. * Statement II: $(\mathbb{Z}_6, +_6, \times_6)$ is an integral domain. **Required:** To determine the correctness of each statement and choose the correct option. **Solution:** *Step 1:* Analyze Statement I using the subring test. A non-empty subset $S$ of a ring $R$ is a subring if for any elements $a, b \in S$, both $a - b \in S$ and $a \cdot b \in S$. Let $R = \mathbb{Z}$ and $S = n\mathbb{Z}$. 1. The set $n\mathbb{Z}$ is non-empty because $0 = n \cdot 0 \in n\mathbb{Z}$. 2. Let $a, b \in n\mathbb{Z}$. By definition, $a = nk_1$ and $b = nk_2$ for some integers $k_1, k_2 \in \mathbb{Z}$. 3. Check for closure under subtraction: Since $k_1 - k_2$ is an integer, $a - b \in n\mathbb{Z}$. 4. Check for closure under multiplication: Since $nk_1k_2$ is an integer, $a \cdot b \in n\mathbb{Z}$. Both conditions of the subring test are satisfied. Therefore, $n\mathbb{Z}$ is a subring of $\mathbb{Z}$. Statement I is correct. *Step 2:* Analyze Statement II by checking the definition of an integral domain. An integral domain is a commutative ring with unity that has no zero divisors. The ring $(\mathbb{Z}_6, +_6, \times_6)$ is a commutative ring with unity $1$. A zero divisor is a non-zero element $a$ for which there exists a non-zero element $b$ such that $a \cdot b = 0$. Consider the elements $2$ and $3$ in $\mathbb{Z}_6$. Both are non-zero. Their product is: Since the product of two non-zero elements is zero, $\mathbb{Z}_6$ contains zero divisors. A ring with zero divisors cannot be an integral domain. Therefore, Statement II is incorrect. This aligns with the theorem that $\mathbb{Z}_n$ is an integral domain if and only if $n$ is a prime number. Since $6$ is not prime, $\mathbb{Z}_6$ is not an integral domain. *Step 3:* Conclude based on the analysis. Statement I is correct and Statement II is incorrect. This corresponds to the third option. **Answer:** $\boxed{\text{Statement I is correct but statement II is incorrect}}$