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

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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Let $\alpha$ have a minimal polynomial $t^2-2$ over $\mathbb{Q}$, and $\beta$ have a minimal polynomial $t^2+4t+2$ over $\mathbb{Q}$. We need to show that the extensions $\mathbb{Q}(\alpha): \mathbb{Q}$ and $\mathbb{Q}(\beta): \mathbb{Q}$ are isomorphic. It is enough to show that $\alpha$ and $\beta$ can be written as a linear combination of one another because of the definition of a simple extension.

If $\alpha$ have a minimal polynomial $t^2-2$ over $\mathbb{Q}$, then $\alpha=\pm \sqrt{2}$. Similarly, if $\beta$ have a minimal polynomial $t^2+4t+2$ over $\mathbb{Q}$, by the quadratic formula $$\beta=\frac{4\pm \sqrt{16-8}}{2}=\frac{4\pm 2\sqrt{2}}{2}=2\pm \sqrt{2},$$ then $\beta=2\pm 2$. Note that $$0=\beta^2-4\beta+2=\beta^2-4\beta+4+(-4+2)=(\beta-2)^2-2.$$ So $$(\beta-2)^2=2 \iff \beta-2=\pm \sqrt{2}=\beta-2=\pm \alpha.$$ Hence $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta).$ (i.e. The extensions $\mathbb{Q}(\alpha): \mathbb{Q}$ and $\mathbb{Q}(\beta): \mathbb{Q}$ are isomorphic.)

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Consider all elements you can obtain using addition, subtraction, multiplication, and division with the rationals and $\alpha$ and deduce that you can obtain all elements of $\mathbb{Q}(\beta)$. This shows that $\mathbb{Q}(\beta)$ is a subfield of $\mathbb{Q}(\alpha)$. The do the converse to show they are actually equal.

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