# dimension

• Nov 30th 2009, 03:51 AM
Sampras
dimension
Consider $\mathbb{Q}(\sqrt{2})$. Is the dimension of $\mathbb{Q}(\sqrt{2})$ equaled to $3$? Because $\{1, \sqrt{2} \}$ is a basis for $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$.

Also why does $[\mathbb{Q}(\sqrt{2}+\sqrt{3}): \mathbb{Q}] = 4$. Is it because $\text{irr}(\sqrt{2}+\sqrt{3}, \mathbb{Q}) = x^4-10x^{2}+1$? And the degree of this polynomial is 4?
• Nov 30th 2009, 04:14 AM
Shanks
the dimension of $Q(\sqrt{2})$ over Q is 2(the degree of the minimal polynomial of $\sqrt{2}$ over Q).
The second probelm is similar to the aboved mentioned!