金声玉亮2.0模型评测报告

There is a single choice question about abstract algebra. Answer the question by replying A, B, C or D.
Question: Find the degree for the given field extension Q(sqrt(2)*sqrt(3)) over Q.
A. 0
B. 4
C. 2
D. 6
Answer:

答案

模型	模型答案	正确答案
金声玉亮2.0	C	C
deepseekr1	The field extension is \( \mathbb{Q}(\sqrt{2} \cdot \sqrt{3}) = \mathbb{Q}(\sqrt{6}) \). Since \( \sqrt{6} \) is irrational and square-free, the minimal polynomial over \( \mathbb{Q} \) is \( x^2 - 6 \), which is irreducible. Thus, the degree of the extension is the degree of this polynomial, which is 2. Answer: C. 2	C
gpto1	C	C
GPT-4o	C. 2	C
llama3.1-70B-Instruct	The correct answer is B.	C
llama3.1-405B-Instruct	B	C
DeepSeek-Coder-V2	C. 2	C
Qwen2-72B-Instruct	B. 4	C
baichuan4	C	C
doubao	C. 2. The degree of the extension Q(sqrt(2)sqrt(3)) over Q is 2 because the minimal polynomial of sqrt(2)sqrt(3) = sqrt(6) over Q is x^2 - 6, which has degree 2.	C
minimax	C. 2 The field extension $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}$ is generated by the two elements $\sqrt{2}$ and $\sqrt{3}$. Since $\sqrt{2}$ and $\sqrt{3}$ are not rational numbers, they are not in $\mathbb{Q}$, but their product $\sqrt{2} \cdot \sqrt{3} = \sqrt{6}$ is also not in $\mathbb{Q}$, which means that the two elements are linearly independent over $\mathbb{Q}$. The degree of the field extension is the dimension of the larger field as a vector space over the smaller field. Since $\sqrt{2}$ and $\sqrt{3}$ are linearly independent, the degree of the extension is the product of the degrees of the extensions $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ over $\mathbb{Q}$. Each of these extensions is of degree 2 because $\sqrt{2}$ and $\sqrt{3}$ are roots of the irreducible quadratic polynomials $x^2 - 2$ and $x^2 - 3$, respectively. Therefore, the degree of the extension $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}$ is $2 \times 2 = 4$. However, the question asks for the degree of the extension $\mathbb{Q}(\sqrt{2} \cdot \sqrt{3})$ over $\mathbb{Q}$, which is the same as the degree of the extension $\mathbb{Q}(\sqrt{6})$ over $\mathbb{Q}$. Since $\sqrt{6}$ is a root of the irreducible quadratic polynomial $x^2 - 6$, the degree of this extension is also 2. So the correct answer is C. 2.	C
kimi	B. 4	C
wenxinyiyan	C. 2 Explanation: The field extension Q(sqrt(2)*sqrt(3)) can be simplified as Q(sqrt(6)). Now, sqrt(6) is an algebraic number of degree 2 over Q, because it satisfies the polynomial equation x^2 - 6 = 0, which is irreducible over Q. Therefore, the degree of the field extension Q(sqrt(6)) over Q is 2.	C
zhipu	To determine the degree of the field extension \(\mathbb{Q}(\sqrt{2} \cdot \sqrt{3})\) over \(\mathbb{Q}\), we need to understand the structure of the field \(\mathbb{Q}(\sqrt{2} \cdot \sqrt{3})\). First, observe that \(\sqrt{2} \cdot \sqrt{3} = \sqrt{6}\). Therefore, the field \(\mathbb{Q}(\sqrt{2} \cdot \sqrt{3})\) is the same as \(\mathbb{Q}(\sqrt{6})\). Next, we need to find the degree of the extension \(\mathbb{Q}(\sqrt{6})\) over \(\mathbb{Q}\). This degree is the degree of the minimal polynomial of \(\sqrt{6}\) over \(\mathbb{Q}\). The minimal polynomial of \(\sqrt{6}\) over \(\mathbb{Q}\) is \(x^2 - 6\), which is a quadratic polynomial. The degree of this polynomial is 2. Therefore, the degree of the field extension \(\mathbb{Q}(\sqrt{6})\) over \(\mathbb{Q}\) is 2. So, the correct answer is: C. 2	C
gpto1	C	C
deepseekv3	To determine the degree of the field extension \( \mathbb{Q}(\sqrt{2}, \sqrt{3}) \) over \( \mathbb{Q} \), we can proceed as follows: 1. Identify the Minimal Polynomials: - The minimal polynomial of \( \sqrt{2} \) over \( \mathbb{Q} \) is \( x^2 - 2 \). - The minimal polynomial of \( \sqrt{3} \) over \( \mathbb{Q} \) is \( x^2 - 3 \). 2. Compute the Degree of the Extension: - Since \( \sqrt{2} \) and \( \sqrt{3} \) are linearly independent over \( \mathbb{Q} \), the degree of the extension \( \mathbb{Q}(\sqrt{2}, \sqrt{3}) \) over \( \mathbb{Q} \) is the product of the degrees of the minimal polynomials of \( \sqrt{2} \) and \( \sqrt{3} \). - Therefore, the degree is \( 2 \times 2 = 4 \). Thus, the correct answer is: B. 4	C

模型

模型答案

正确答案

金声玉亮2.0

deepseekr1

The field extension is $ \mathbb{Q}(\sqrt{2} \cdot \sqrt{3}) = \mathbb{Q}(\sqrt{6}) $. Since $ \sqrt{6} $ is irrational and square-free, the minimal polynomial over $ \mathbb{Q} $ is $ x^2 - 6 $, which is irreducible. Thus, the degree of the extension is the degree of this polynomial, which is 2.

**Answer:** C. 2

gpto1

GPT-4o

C. 2

llama3.1-70B-Instruct

The correct answer is B.

llama3.1-405B-Instruct

DeepSeek-Coder-V2

C. 2

Qwen2-72B-Instruct

B. 4

baichuan4

doubao

C. 2. The degree of the extension Q(sqrt(2)*sqrt(3)) over Q is 2 because the minimal polynomial of sqrt(2)*sqrt(3) = sqrt(6) over Q is x^2 - 6, which has degree 2.

minimax

C. 2

The field extension $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}$ is generated by the two elements $\sqrt{2}$ and $\sqrt{3}$. Since $\sqrt{2}$ and $\sqrt{3}$ are not rational numbers, they are not in $\mathbb{Q}$, but their product $\sqrt{2} \cdot \sqrt{3} = \sqrt{6}$ is also not in $\mathbb{Q}$, which means that the two elements are linearly independent over $\mathbb{Q}$.

The degree of the field extension is the dimension of the larger field as a vector space over the smaller field. Since $\sqrt{2}$ and $\sqrt{3}$ are linearly independent, the degree of the extension is the product of the degrees of the extensions $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ over $\mathbb{Q}$.

Each of these extensions is of degree 2 because $\sqrt{2}$ and $\sqrt{3}$ are roots of the irreducible quadratic polynomials $x^2 - 2$ and $x^2 - 3$, respectively. Therefore, the degree of the extension $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}$ is $2 \times 2 = 4$.

However, the question asks for the degree of the extension $\mathbb{Q}(\sqrt{2} \cdot \sqrt{3})$ over $\mathbb{Q}$, which is the same as the degree of the extension $\mathbb{Q}(\sqrt{6})$ over $\mathbb{Q}$. Since $\sqrt{6}$ is a root of the irreducible quadratic polynomial $x^2 - 6$, the degree of this extension is also 2.

So the correct answer is C. 2.

kimi

B. 4

wenxinyiyan

C. 2

Explanation:
The field extension Q(sqrt(2)*sqrt(3)) can be simplified as Q(sqrt(6)). Now, sqrt(6) is an algebraic number of degree 2 over Q, because it satisfies the polynomial equation x^2 - 6 = 0, which is irreducible over Q. Therefore, the degree of the field extension Q(sqrt(6)) over Q is 2.

zhipu

To determine the degree of the field extension $\mathbb{Q}(\sqrt{2} \cdot \sqrt{3})$ over $\mathbb{Q}$, we need to understand the structure of the field $\mathbb{Q}(\sqrt{2} \cdot \sqrt{3})$.

First, observe that $\sqrt{2} \cdot \sqrt{3} = \sqrt{6}$. Therefore, the field $\mathbb{Q}(\sqrt{2} \cdot \sqrt{3})$ is the same as $\mathbb{Q}(\sqrt{6})$.

Next, we need to find the degree of the extension $\mathbb{Q}(\sqrt{6})$ over $\mathbb{Q}$. This degree is the degree of the minimal polynomial of $\sqrt{6}$ over $\mathbb{Q}$.

The minimal polynomial of $\sqrt{6}$ over $\mathbb{Q}$ is $x^2 - 6$, which is a quadratic polynomial. The degree of this polynomial is 2.

Therefore, the degree of the field extension $\mathbb{Q}(\sqrt{6})$ over $\mathbb{Q}$ is 2.

So, the correct answer is:
C. 2

gpto1

deepseekv3

To determine the degree of the field extension $ \mathbb{Q}(\sqrt{2}, \sqrt{3}) $ over $ \mathbb{Q} $, we can proceed as follows:

1. **Identify the Minimal Polynomials:**
- The minimal polynomial of $ \sqrt{2} $ over $ \mathbb{Q} $ is $ x^2 - 2 $.
- The minimal polynomial of $ \sqrt{3} $ over $ \mathbb{Q} $ is $ x^2 - 3 $.

2. **Compute the Degree of the Extension:**
- Since $ \sqrt{2} $ and $ \sqrt{3} $ are linearly independent over $ \mathbb{Q} $, the degree of the extension $ \mathbb{Q}(\sqrt{2}, \sqrt{3}) $ over $ \mathbb{Q} $ is the product of the degrees of the minimal polynomials of $ \sqrt{2} $ and $ \sqrt{3} $.
- Therefore, the degree is $ 2 \times 2 = 4 $.

Thus, the correct answer is:

**B. 4**

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