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Please help me finding the basis and dimension of the splitting field of the polynomial $x^4+5x^2+6$ in the rational field.


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Step one: factor that polynomial. I'm sure you can do that. – Gerry Myerson Feb 6 '13 at 5:52

Hint: You can factor this as $(x^2+3)(x^2+2)$ so the splitting field is $\mathbb{Q}(\sqrt{-2},\sqrt{-3})$. You should then apply the tower rule to get that $[\mathbb{Q}(\sqrt{-2},\sqrt{-3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{-2},\sqrt{-3}][\mathbb{Q}(\sqrt{-2}):\mathbb{Q}]$. To find each of the degrees in the product try and find the degree of the minimal polynomial. The hint for this is that a quadratic annihilates things and so you'r each degree 1 or 2 at each step--then recall that you're degree one only if you were in the field all along!

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Now we'll never know whether OP can factor polynomials. – Gerry Myerson Feb 6 '13 at 6:03
@GerryMyerson I couldn't either--I used wolfram :) – Alex Youcis Feb 6 '13 at 6:04
Wolfram? to factor a quadratic?? – Gerry Myerson Feb 6 '13 at 6:29
@GerryMyerson You can restore your faith in mathematical humanity--it was a joke :) – Alex Youcis Feb 6 '13 at 6:30
You don't know how much better I feel now. – Gerry Myerson Feb 6 '13 at 6:32

Here, x=±√2 i,±√3 i. So splitting field is Q(√2 i,√3 i) whose elements are of the form Q((√2 i) )(√3 i)=Q_1 (√3 i), in particular, {q_0+q_1 (√2 i);q_i ϵQ_1} that is {(a_0+a_1 √2 i)+(a_2+a_3 √2 i)√3 i;a_i ϵQ}. Therefore, basis is {1,√2 i,√3 i,-√6} and dimension is 4

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