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Find the degree for the given field extension $Q\left(\operatorname{sqrt}(2)^{\star} \operatorname{sqrt}(3)\right.$ ) over $Q$. Which of the following is the right choice? Explain your answer.

(A) 0 (B) 4 (C) 2 (D) 6
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(B) 4
The field extension in question is $Q(\sqrt{2}, \sqrt{3})$ over $Q$. To find the degree of the extension, we consider the minimal polynomial of the elements being adjoined.
Both $\sqrt{2}$ and $\sqrt{3}$ are algebraic numbers, and their minimal polynomials over $Q$ are $x^2-2$ and $x^2-3$, respectively. These polynomials are irreducible over $Q$, so the degrees of the extensions $Q(\sqrt{2})$ and $Q(\sqrt{3})$ over $Q$ are both 2.
Now, consider the extension $Q(\sqrt{2})(\sqrt{3})$. Since $\sqrt{3}$ is not an element of $Q(\sqrt{2})$, the degree of the extension $Q(\sqrt{2}, \sqrt{3})$ over $Q(\sqrt{2})$ is 2 .

By the tower law, the degree of the extension $Q(\sqrt{2}, \sqrt{3})$ over $Q$ is the product of the degrees of the extensions $Q(\sqrt{2})$ over $Q$ and $Q(\sqrt{2}, \sqrt{3})$ over $Q(\sqrt{2})$. Hence, the degree is $2 \times 2=4$
by Diamond (89,043 points)

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