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Field extensions question

  1. Apr 20, 2010 #1
    how do I find all u in R such that Q(u)=Q(√2,³ √5 ) (square root of two and cubed root of 5) and prove they are the same. PLEASE help im desperate! I understand that one possible u is √2+³ √5 but i dont know hoe to prove it withou pages of algebra and i dont know what the others are.
  2. jcsd
  3. Apr 24, 2010 #2
    Well, the degree of [tex] \mathbb{Q}(\sqrt{2}, \sqrt[3]{5}) [/tex] over [tex] \mathbb{Q} [/tex] is 6. (Prove this.) Thus, we know that if [tex] \{ 1, \alpha \} [/tex] and [tex] \{ 1, \beta, \beta^2 \} [/tex] are bases for [tex] \mathbb{Q}(\sqrt{2}) [/tex] and [tex] \mathbb{Q}(\sqrt[3]{5}) [/tex], respectively, then [tex] \{ 1, \alpha, \beta, \alpha \beta, \beta^2, \alpha \beta^2 \} [/tex] is a basis for [tex] \mathbb{Q}(\sqrt{2}, \sqrt[3]{5}) [/tex] over [tex] \mathbb{Q} [/tex]. In particular, we can write
    u = a \alpha + b \beta + c \alpha \beta + d \beta^2 + e \alpha \beta^2
    (why can we assume that the [tex] 1 [/tex]-component of [tex] u [/tex] is [tex] 0 [/tex] WLOG?). At this point, since you know the degrees of all of the basis elements over [tex] \mathbb{Q} [/tex], you should be able to find conditions on the coefficients [tex] a,b,c,d,e [/tex] that guarantee that [tex] [ \mathbb{Q}(u) : \mathbb{Q} ] = 6 [/tex], which is enough to show that [tex] \mathbb{Q}(u) = \mathbb{Q}(\sqrt{2}, \sqrt[3]{5}) [/tex]. (HINT: If the degrees of [tex] x_1 [/tex] and [tex] x_2 [/tex] over [tex] \mathbb{Q} [/tex] are [tex] d_1, d_2 [/tex], and [tex] \gcd(d_1, d_2) = 1 [/tex], what is the degree of [tex] x_1 + x_2 [/tex]?)
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