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Thread: dimension

  1. #1
    Senior Member Sampras's Avatar
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    dimension

    Consider $\displaystyle \mathbb{Q}(\sqrt{2}) $. Is the dimension of $\displaystyle \mathbb{Q}(\sqrt{2}) $ equaled to $\displaystyle 3 $? Because $\displaystyle \{1, \sqrt{2} \} $ is a basis for $\displaystyle \mathbb{Q}(\sqrt{2}) $ over $\displaystyle \mathbb{Q} $.

    Also why does $\displaystyle [\mathbb{Q}(\sqrt{2}+\sqrt{3}): \mathbb{Q}] = 4 $. Is it because $\displaystyle \text{irr}(\sqrt{2}+\sqrt{3}, \mathbb{Q}) = x^4-10x^{2}+1 $? And the degree of this polynomial is 4?
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  2. #2
    Senior Member Shanks's Avatar
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    the dimension of $\displaystyle Q(\sqrt{2})$ over Q is 2(the degree of the minimal polynomial of $\displaystyle \sqrt{2}$ over Q).
    The second probelm is similar to the aboved mentioned!
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