dimension

**Sampras** · November 30th 2009, 03:51 AM

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?

**Shanks** · November 30th 2009, 04:14 AM

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!

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