Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question
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

4 Answers 4


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

share|cite|improve this answer

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 $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$.

share|cite|improve this answer

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.)

share|cite|improve this answer

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.

share|cite|improve this answer

protected by user26857 Oct 19 at 20:01

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.