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I have to show that if $\alpha$ has the minimal polynomial $t^2-2$ over $\mathbb{Q}$ and $\beta$ has the minimal polynomial $t^2-4t+2$ over $\mathbb{Q}$ then the extensions $\mathbb{Q}(\alpha):\mathbb{Q}$ and $\mathbb{Q}(\beta):\mathbb{Q}$ are isomorphic.

However the only way that I know how to show that two extensions are isomorphic is if $\alpha$ and $\beta$ had the same minimal polynomial, which they don't?

Cheers Folks

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You mean isomorphic as rational vector spaces...? –  DonAntonio Jan 12 '13 at 17:44
$t^2-4t+3$ is reducible over $\Bbb{Q}$. It is impossible for it to be the minimal polynomial of anything. –  Chris Eagle Jan 12 '13 at 17:47
@ChrisEagle It was supposed to be $t^2-4t+2$ sorry –  Joe Cabel Jan 12 '13 at 17:51

2 Answers 2


If we imagine both fields embedded in $\Bbb C$, then clearly $\alpha$ must be $\pm\sqrt 2$ while by the quadratic formula $\beta = \frac{4\pm\sqrt{16-8}}{2}=2\pm\sqrt{2}$.

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Another solution: Choose a common extension, say $\mathbb{C}$. Then $0 = \beta^2-4\beta+2 = (\beta-2)^2-2$ implies $\beta-2=\pm \alpha$, hence $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$.

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