| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | B. 2 Explanation: For \(Z_3[x]/(x^3 + cx^2 + 1)\) to be a field, the polynomial \(x^3 + cx^2 + 1\) must be irreducible over \(Z_3\). This means that the polynomial should not have any roots in \(Z_3\) and should not factor into polynomials of lower degree with coefficients in \(Z_3\). Let's check each option: A. If \(c = 0\), the polynomial becomes \(x^3 + 1\). This polynomial has a root in \(Z_3\), namely \(x = 2\), since \(2^3 + 1 = 9 = 0\) in \(Z_3\). Therefore, the polynomial is not irreducible, and \(Z_3[x]/(x^3 + 1)\) is not a field. B. If \(c = 2\), the polynomial becomes \(x^3 + 2x^2 + 1\). This polynomial does not have any roots in \(Z_3\), and it cannot be factored into polynomials of lower degree with coefficients in \(Z_3\). Therefore, it is irreducible, and \(Z_3[x]/(x^3 + 2x^2 + 1)\) is a field. C. If \(c = 1\), the polynomial becomes \(x^3 + x^2 + 1\). This polynomial has a root in \(Z_3\), namely \(x = 1\), since \(1^3 + 1^2 + 1 = 3 = 0\) in \(Z_3\). Therefore, the polynomial is not irreducible, and \(Z_3[x]/(x^3 + x^2 + 1)\) is not a field. D. The option \(c = 3\) is not valid since \(3\) is not an element of \(Z_3\). In \(Z_3\), \(3\) is equivalent to \(0\), which we have already checked in option A. Therefore, the correct answer is B. 2. |
B |
| deepseekr1 | To determine for which \( c \in \mathbb{Z}_3 \) the ring \( \mathbb{Z}_3[x]/(x^3 + cx^2 + 1) \) is a field, we need the polynomial \( x^3 + cx^2 + 1 \) to be irreducible over \( \mathbb{Z}_3 \). A cubic polynomial is irreducible if it has no roots in the field (as it cannot factor into linear or quadratic terms without roots). 1. **For \( c = 0 \):** The polynomial becomes \( x^3 + 1 \). Testing roots: - \( x = 0 \): \( 0 + 1 = 1 \neq 0 \). - \( x = 1 \): \( 1 + 1 = 2 \neq 0 \). - \( x = 2 \): \( 8 + 1 \equiv 2 + 1 = 0 \mod 3 \). **Root at \( x = 2 \)** ⇒ Reducible. **Not a field.** 2. **For \( c = 1 \):** The polynomial becomes \( x^3 + x^2 + 1 \). Testing roots: - \( x = 0 \): \( 0 + 0 + 1 = 1 \neq 0 \). - \( x = 1 \): \( 1 + 1 + 1 = 3 \equiv 0 \mod 3 \). **Root at \( x = 1 \)** ⇒ Reducible. **Not a field.** 3. **For \( c = 2 \):** The polynomial becomes \( x^3 + 2x^2 + 1 \). Testing roots: - \( x = 0 \): \( 0 + 0 + 1 = 1 \neq 0 \). - \( x = 1 \): \( 1 + 2 + 1 = 4 \equiv 1 \neq 0 \mod 3 \). - \( x = 2 \): \( 8 + 8 + 1 \equiv 2 + 2 + 1 = 5 \equiv 2 \neq 0 \mod 3 \). **No roots** ⇒ Irreducible. **The quotient is a field.** Since \( c = 2 \) (option B) is the only value for which the polynomial is irreducible, the answer is **B**. |
B |
| gpto1 | The question asks us to determine all \( c \) in \( \mathbb{Z}_3 \) such that \( \mathbb{Z}_3[x]/(x^3 + c x^2 + 1) \) is a field. This happens precisely when the polynomial \( x^3 + c x^2 + 1 \) is irreducible in \( \mathbb{Z}_3[x] \). We test each possible value of \( c \) in \( \mathbb{Z}_3 \): 1. **For \( c = 0 \)**: The polynomial becomes \( x^3 + 1 \). We check for roots in \( \mathbb{Z}_3 \): - \( x = 0 \): \( f(0) = 0 + 1 = 1 \neq 0 \) - \( x = 1 \): \( f(1) = 1 + 1 = 2 \neq 0 \) - \( x = 2 \): \( f(2) = (2)^3 + 1 = 8 + 1 = 9 \equiv 0 \mod 3 \) Since \( x = 2 \) is a root, \( x^3 + 1 \) is reducible. Therefore, the corresponding quotient ring is not a field. 2. **For \( c = 1 \)**: The polynomial is \( x^3 + x^2 + 1 \). Testing for roots: - \( x = 0 \): \( f(0) = 0 + 0 + 1 = 1 \neq 0 \) - \( x = 1 \): \( f(1) = 1 + 1 + 1 = 3 \equiv 0 \mod 3 \) - \( x = 2 \): \( f(2) = 8 + 4 + 1 = 13 \equiv 1 \mod 3 \) Since \( x = 1 \) is a root, the polynomial is reducible, and the quotient ring is not a field. 3. **For \( c = 2 \)**: The polynomial is \( x^3 + 2 x^2 + 1 \). Testing for roots: - \( x = 0 \): \( f(0) = 0 + 0 + 1 = 1 \neq 0 \) - \( x = 1 \): \( f(1) = 1 + 2 + 1 = 4 \equiv 1 \mod 3 \) - \( x = 2 \): \( f(2) = 8 + 8 + 1 = 17 \equiv 2 \mod 3 \) There are no roots in \( \mathbb{Z}_3 \), so the polynomial is irreducible, and the quotient ring \( \mathbb{Z}_3[x]/(x^3 + 2 x^2 + 1) \) is a field. **Therefore, the only value of \( c \) for which the quotient ring is a field is \( c = 2 \).** **Answer:** B |
B |
| GPT-4o | To determine which value of \( c \) makes \( \mathbb{Z}_3[x]/(x^3 + cx^2 + 1) \) a field, we need to ensure that the polynomial \( x^3 + cx^2 + 1 \) is irreducible over \( \mathbb{Z}_3 \). Consider the polynomial \( f(x) = x^3 + cx^2 + 1 \) over \( \mathbb{Z}_3 \). For \( \mathbb{Z}_3[x]/(f(x)) \) to be a field, \( f(x) \) must not have any roots in \( \mathbb{Z}_3 \) and must not be factorizable into lower-degree polynomials in \( \mathbb{Z}_3[x] \). Let's check each value of \( c \) in \( \mathbb{Z}_3 \) (which includes 0, 1, and 2): 1. **For \( c = 0 \)**: \( f(x) = x^3 + 1 \). Examine the roots: - \( f(0) = 0^3 + 1 = 1 \neq 0 \) - \( f(1) = 1^3 + 1 = 2 \neq 0 \) - \( f(2) = 2^3 + 1 = 8 + 1 = 9 \equiv 0 \pmod{3} \) So, \( x = 2 \) is a root, thus \( x^3 + 1 \) is reducible. 2. **For \( c = 1 \)**: \( f(x) = x^3 + x^2 + 1 \). Examine the roots: - \( f(0) = 0^3 + 0^2 + 1 = 1 \neq 0 \) - \( f(1) = 1^3 + 1^2 + 1 = 3 \equiv 0 \pmod{3} \) - \( f(2) = 2^3 + 2^2 + 1 = 8 + 4 + 1 = 13 \equiv 1 \pmod{3} \neq 0 \) So, \( x = 1 \) is a root, thus \( x^3 + x^2 + 1 \) is reducible. 3. **For \( c = 2 \)**: \( f(x) = x^3 + 2x^2 + 1 \). Examine the roots: - \( f(0) = 0^3 + 2 \cdot 0^2 + 1 = 1 \neq 0 \) - \( f(1) = 1^3 + 2 \cdot 1^2 + 1 = 1 + 2 + 1 = 4 \equiv 1 \pmod{3} \neq 0 \) - \( f(2) = 2^3 + 2 \cdot 2^2 + 1 = 8 + 8 + 1 = 17 \equiv 2 \pmod{3} \neq 0 \) Since \( f(x) \) has no roots in \( \mathbb{Z}_3 \), we need to check if it is irreducible. The degrees of possible factors must be 1 and 2. Given no linear factors exist (as there are no roots), the polynomial cannot be factored into polynomials of lower degree. Thus, \( x^3 + 2x^2 + 1 \) is irreducible over \( \mathbb{Z}_3 \), making \( \mathbb{Z}_3[x]/(x^3 + 2x^2 + 1) \) a field. The correct answer is: B. 2 |
B |
| llama3.1-70B-Instruct | To determine if Z_3[x]/(x^3 + cx^2 + 1) is a field, we need to check if the polynomial x^3 + cx^2 + 1 is irreducible over Z_3. If the polynomial is irreducible, then the quotient ring is a field. Let's check the possible values of c: * If c = 0, then the polynomial becomes x^3 + 1, which is reducible (x + 1)(x^2 - x + 1). * If c = 1, then the polynomial becomes x^3 + x^2 + 1, which is irreducible. * If c = 2, then the polynomial becomes x^3 + 2x^2 + 1, which is reducible (x + 2)(x^2 - 2x + 1). Therefore, the only value of c that makes Z_3[x]/(x^3 + cx^2 + 1) a field is c = 1. The correct answer is C. |
B |
| llama3.1-405B-Instruct | To determine when $\mathbb{Z}_3[x]/(x^3 + cx^2 + 1)$ is a field, we need to find conditions on $c$ such that the polynomial $x^3 + cx^2 + 1$ is irreducible over $\mathbb{Z}_3$. If the polynomial is irreducible, then the quotient ring is a field. Let's check each option: A. $c = 0$: The polynomial becomes $x^3 + 1$. This polynomial is reducible over $\mathbb{Z}_3$ because $x^3 + 1 = (x + 1)(x^2 - x + 1)$, so this option does not yield a field. B. $c = 2$: The polynomial becomes $x^3 + 2x^2 + 1$. We need to check if this polynomial is irreducible over $\mathbb{Z}_3$. By trying out all possible roots (0, 1, 2) in $\mathbb{Z}_3$, we find that none of them are roots of the polynomial, suggesting it is irreducible, and thus this option yields a field. C. $c = 1$: The polynomial becomes $x^3 + x^2 + 1$. Similar to option B, checking all possible roots in $\mathbb{Z}_3$ reveals that this polynomial is also irreducible over $\mathbb{Z}_3$, making this option yield a field as well. D. $c = 3$: Since we are working in $\mathbb{Z}_3$, $c = 3$ is equivalent to $c = 0$, which we already determined does not yield a field. Therefore, the values of $c$ in $\mathbb{Z}_3$ such that $\mathbb{Z}_3[x]/(x^3 + cx^2 + 1)$ is a field are $c = 1$ and $c = 2$. The best answer is C. |
B |
| DeepSeek-Coder-V2 | B | B |
| Qwen2-72B-Instruct | B. 2 | B |
| baichuan4 | B | B |
| doubao | B In Z_3, the possible values for c are 0, 1, and 2. For c = 0, the polynomial becomes x^3 + 1 = (x + 1)(x^2 - x + 1), which does not result in a field. For c = 1, the polynomial is x^3 + x^2 + 1, which is irreducible over Z_3, so Z_3[x]/(x^3 + x^2 + 1) is a field. For c = 2, the polynomial is x^3 + 2x^2 + 1 = (x + 1)(x^2 + x + 1), which does not result in a field. So the answer is B. |
B |
| minimax | To determine which values of \( c \) in \( \mathbb{Z}_3 \) make \( \mathbb{Z}_3[x]/(x^3 + cx^2 + 1) \) a field, we need to ensure that the polynomial \( x^3 + cx^2 + 1 \) is irreducible over \( \mathbb{Z}_3 \). A polynomial is irreducible over a field if it cannot be factored into non-trivial polynomials with coefficients in that field. Let's check each option: A. \( c = 0 \): The polynomial becomes \( x^3 + 1 \). This can be factored as \( (x + 1)(x^2 + x + 1) \) in \( \mathbb{Z}_3 \), so it is not irreducible. B. \( c = 2 \): The polynomial becomes \( x^3 + 2x^2 + 1 \). We need to check if there are any roots in \( \mathbb{Z}_3 \). By substituting \( x = 0, 1, 2 \), we get \( 1, 3, 1 \) respectively, which means there are no roots in \( \mathbb{Z}_3 \). We also need to check if it can be factored into lower-degree polynomials. This is not immediately obvious, so we would need to perform a more detailed check, which might involve trying to factor it or using the Eisenstein's criterion or other irreducibility tests. For the sake of this question, let's assume that \( x^3 + 2x^2 + 1 \) is irreducible over \( \mathbb{Z}_3 \). C. \( c = 1 \): The polynomial becomes \( x^3 + x^2 + 1 \). This polynomial also has no roots in \( \mathbb{Z}_3 \) since substituting \( x = 0, 1, 2 \) gives \( 1, 3, 1 \) respectively. We would also need to check if it can be factored into lower-degree polynomials. Again, let's assume that \( x^3 + x^2 + 1 \) is irreducible over \( \mathbb{Z}_3 \) for the sake of this question. D. \( c = 3 \): Since we are in \( \mathbb{Z}_3 \), \( c = 3 \) is equivalent to \( c = 0 \), which we have already determined is not irreducible. Given the above, the only options that might be irreducible are B and C. However, without further detailed checking, we cannot be certain. In an actual test setting, you would need to apply a more rigorous test for irreducibility. Assuming we have done that and found \( x^3 + 2x^2 + 1 \) to be irreducible, the answer would be: B. 2 Please note that this is a simplification and in a real test scenario, you would need to rigorously prove the irreducibility of the polynomial for the given value of \( c \). |
B |
| kimi | To determine the value of \( c \) in \( \mathbb{Z}_3 \) such that \( \mathbb{Z}_3[x]/(x^3 + cx^2 + 1) \) is a field, we need to ensure that the polynomial \( x^3 + cx^2 + 1 \) is irreducible over \( \mathbb{Z}_3 \). A polynomial is irreducible over a field if it cannot be factored into non-constant polynomials with coefficients in that field. For \( \mathbb{Z}_3 \), the coefficients can only be 0, 1, or 2. We need to check each possible value of \( c \) to see if \( x^3 + cx^2 + 1 \) has any roots in \( \mathbb{Z}_3 \). If it does not have any roots, then it is irreducible, and the quotient ring will be a field. Let's check each value of \( c \) in \( \mathbb{Z}_3 \): 1. For \( c = 0 \), the polynomial is \( x^3 + 1 \). This factors as \( (x + 1)(x^2 - x + 1) \), which is reducible over \( \mathbb{Z}_3 \). So, \( c = 0 \) is not a solution. 2. For \( c = 1 \), the polynomial is \( x^3 + x^2 + 1 \). We check for roots in \( \mathbb{Z}_3 \): - \( 0^3 + 0^2 + 1 = 1 \neq 0 \) - \( 1^3 + 1^2 + 1 = 3 \equiv 0 \) (mod 3), so \( x = 1 \) is a root. - \( 2^3 + 2^2 + 1 = 8 + 4 + 1 = 13 \equiv 1 \) (mod 3), so \( x = 2 \) is not a root. Since \( x = 1 \) is a root, \( x^3 + x^2 + 1 \) is reducible over \( \mathbb{Z}_3 \). So, \( c = 1 \) is not a solution. 3. For \( c = 2 \), the polynomial is \( x^3 + 2x^2 + 1 \). We check for roots in \( \mathbb{Z}_3 \): - \( 0^3 + 2 \cdot 0^2 + 1 = 1 \neq 0 \) - \( 1^3 + 2 \cdot 1^2 + 1 = 1 + 2 + 1 = 4 \equiv 1 \) (mod 3), so \( x = 1 \) is not a root. - \( 2^3 + 2 \cdot 2^2 + 1 = 8 + 8 + 1 = 17 \equiv 2 \) (mod 3), so \( x = 2 \) is not a root. Since \( x^3 + 2x^2 + 1 \) has no roots in \( \mathbb{Z}_3 \), it is irreducible over \( \mathbb{Z}_3 \). Therefore, \( c = 2 \) is a solution. 4. For \( c = 3 \), since \( \mathbb{Z}_3 \) only has elements 0, 1, and 2, \( c = 3 \) is equivalent to \( c = 0 \) in \( \mathbb{Z}_3 \), and we already determined that \( c = 0 \) is not a solution. Thus, the only value of \( c \) that makes \( \mathbb{Z}_3[x]/(x^3 + cx^2 + 1) \) a field is \( c = 2 \). Answer: B. 2 |
B |
| wenxinyiyan | B Explanation: For Z_3[x]/(x^3 + cx^2 + 1) to be a field, the polynomial x^3 + cx^2 + 1 must be irreducible over Z_3. We can check each value of c in Z_3 (0, 1, 2) to see which ones make the polynomial irreducible. 1. When c = 0, the polynomial becomes x^3 + 1, which is reducible since it has a root at x = -1 (mod 3). 2. When c = 1, the polynomial becomes x^3 + x^2 + 1. This is also reducible, as it can be factored as (x + 2)(x^2 + 2x + 2) (mod 3). Note that the quadratic factor is irreducible, but the presence of a linear factor makes the original polynomial reducible. 3. When c = 2, the polynomial becomes x^3 + 2x^2 + 1. This polynomial is irreducible over Z_3. One way to check this is to try all possible roots (0, 1, 2) and verify that none of them satisfy the equation x^3 + 2x^2 + 1 = 0 (mod 3). Since the polynomial is irreducible only when c = 2, the answer is B. |
B |
| zhipu | To determine the value of \( c \) in \(\mathbb{Z}_3\) such that \(\mathbb{Z}_3[x]/(x^3 + cx^2 + 1)\) is a field, we need to ensure that the polynomial \( x^3 + cx^2 + 1 \) is irreducible over \(\mathbb{Z}_3\). A polynomial over a finite field is irreducible if it cannot be factored into the product of two non-constant polynomials. Let's check each value of \( c \): 1. **\( c = 0 \):** \[ x^3 + 1 \] We check if \( x^3 + 1 \) has roots in \(\mathbb{Z}_3\): \[ x^3 + 1 = 0 \implies x^3 = -1 \equiv 2 \ (\text{mod } 3) \] Checking all elements in \(\mathbb{Z}_3\): - \( x = 0 \): \( 0^3 = 0 \) - \( x = 1 \): \( 1^3 = 1 \) - \( x = 2 \): \( 2^3 = 8 \equiv 2 \ (\text{mod } 3)\) \( x = 2 \) is a root, so \( x^3 + 1 \) is reducible. 2. **\( c = 1 \):** \[ x^3 + x^2 + 1 \] We check if \( x^3 + x^2 + 1 \) has roots in \(\mathbb{Z}_3\): \[ x^3 + x^2 + 1 = 0 \] Checking all elements in \(\mathbb{Z}_3\): - \( x = 0 \): \( 0^3 + 0^2 + 1 = 1 \) - \( x = 1 \): \( 1^3 + 1^2 + 1 = 3 \equiv 0 \ (\text{mod } 3)\) - \( x = 2 \): \( 2^3 + 2^2 + 1 = 8 + 4 + 1 = 13 \equiv 1 \ (\text{mod } 3)\) \( x = 1 \) is a root, so \( x^3 + x^2 + 1 \) is reducible. 3. **\( c = 2 \):** \[ x^3 + 2x^2 + 1 \] We check if \( x^3 + 2x^2 + 1 \) has roots in \(\mathbb{Z}_3\): \[ x^3 + 2x^2 + 1 = 0 \] Checking all elements in \(\mathbb{Z}_3\): - \( x = 0 \): \( 0^3 + 2 \cdot 0^2 + 1 = 1 \) - \( x = 1 \): \( 1^3 + 2 \cdot 1^2 + 1 = 1 + 2 + 1 = 4 \equiv 1 \ (\text{mod } 3)\) - \( x = 2 \): \( 2^3 + 2 \cdot 2^2 + 1 = 8 + 8 + 1 = 17 \equiv 2 \ (\text{mod } 3)\) No roots in \(\mathbb{Z}_3\). To confirm irreducibility, we check if it can be factored into polynomials of degree 1 and 2. Testing possible factorizations: \[ (x + a)(x^2 + bx + c) \] Expanding and comparing coefficients leads to contradictions. Hence, \( x^3 + 2x^2 + 1 \) is irreducible. Therefore, the value of \( c \) that makes \(\mathbb{Z}_3[x]/(x^3 + cx^2 + 1)\) a field is \( c = 2 \). **Answer: B** |
B |
| gpto1 | The question asks us to determine all \( c \) in \( \mathbb{Z}_3 \) such that \( \mathbb{Z}_3[x]/(x^3 + c x^2 + 1) \) is a field. This happens precisely when the polynomial \( x^3 + c x^2 + 1 \) is irreducible in \( \mathbb{Z}_3[x] \). We test each possible value of \( c \) in \( \mathbb{Z}_3 \): 1. **For \( c = 0 \)**: The polynomial becomes \( x^3 + 1 \). We check for roots in \( \mathbb{Z}_3 \): - \( x = 0 \): \( f(0) = 0 + 1 = 1 \neq 0 \) - \( x = 1 \): \( f(1) = 1 + 1 = 2 \neq 0 \) - \( x = 2 \): \( f(2) = (2)^3 + 1 = 8 + 1 = 9 \equiv 0 \mod 3 \) Since \( x = 2 \) is a root, \( x^3 + 1 \) is reducible. Therefore, the corresponding quotient ring is not a field. 2. **For \( c = 1 \)**: The polynomial is \( x^3 + x^2 + 1 \). Testing for roots: - \( x = 0 \): \( f(0) = 0 + 0 + 1 = 1 \neq 0 \) - \( x = 1 \): \( f(1) = 1 + 1 + 1 = 3 \equiv 0 \mod 3 \) - \( x = 2 \): \( f(2) = 8 + 4 + 1 = 13 \equiv 1 \mod 3 \) Since \( x = 1 \) is a root, the polynomial is reducible, and the quotient ring is not a field. 3. **For \( c = 2 \)**: The polynomial is \( x^3 + 2 x^2 + 1 \). Testing for roots: - \( x = 0 \): \( f(0) = 0 + 0 + 1 = 1 \neq 0 \) - \( x = 1 \): \( f(1) = 1 + 2 + 1 = 4 \equiv 1 \mod 3 \) - \( x = 2 \): \( f(2) = 8 + 8 + 1 = 17 \equiv 2 \mod 3 \) There are no roots in \( \mathbb{Z}_3 \), so the polynomial is irreducible, and the quotient ring \( \mathbb{Z}_3[x]/(x^3 + 2 x^2 + 1) \) is a field. **Therefore, the only value of \( c \) for which the quotient ring is a field is \( c = 2 \).** **Answer:** B |
B |
| deepseekv3 | To determine all \( c \in \mathbb{Z}_3 \) such that \( \mathbb{Z}_3[x]/(x^3 + cx^2 + 1) \) is a field, we need to check whether the polynomial \( x^3 + cx^2 + 1 \) is irreducible over \( \mathbb{Z}_3 \). If the polynomial is irreducible, then the quotient ring \( \mathbb{Z}_3[x]/(x^3 + cx^2 + 1) \) is a field. We will check the irreducibility of \( x^3 + cx^2 + 1 \) for each \( c \in \mathbb{Z}_3 \): 1. **For \( c = 0 \):** The polynomial is \( x^3 + 1 \). We check if it has any roots in \( \mathbb{Z}_3 \): - \( x = 0 \): \( 0^3 + 1 = 1 \neq 0 \) - \( x = 1 \): \( 1^3 + 1 = 2 \neq 0 \) - \( x = 2 \): \( 2^3 + 1 = 8 + 1 = 9 \equiv 0 \pmod{3} \) Since \( x = 2 \) is a root, \( x^3 + 1 \) is reducible. 2. **For \( c = 1 \):** The polynomial is \( x^3 + x^2 + 1 \). We check if it has any roots in \( \mathbb{Z}_3 \): - \( x = 0 \): \( 0^3 + 0^2 + 1 = 1 \neq 0 \) - \( x = 1 \): \( 1^3 + 1^2 + 1 = 1 + 1 + 1 = 3 \equiv 0 \pmod{3} \) Since \( x = 1 \) is a root, \( x^3 + x^2 + 1 \) is reducible. 3. **For \( c = 2 \):** The polynomial is \( x^3 + 2x^2 + 1 \). We check if it has any roots in \( \mathbb{Z}_3 \): - \( x = 0 \): \( 0^3 + 2 \cdot 0^2 + 1 = 1 \neq 0 \) - \( x = 1 \): \( 1^3 + 2 \cdot 1^2 + 1 = 1 + 2 + 1 = 4 \equiv 1 \pmod{3} \neq 0 \) - \( x = 2 \): \( 2^3 + 2 \cdot 2^2 + 1 = 8 + 8 + 1 = 17 \equiv 2 \pmod{3} \neq 0 \) Since there are no roots in \( \mathbb{Z}_3 \), \( x^3 + 2x^2 + 1 \) is irreducible. Therefore, the only value of \( c \) for which \( \mathbb{Z}_3[x]/(x^3 + cx^2 + 1) \) is a field is \( c = 2 \). Answer: **B** |
B |