1
$\begingroup$

Show that $F(\alpha)=F(\beta)$ if $\alpha$ and $\beta$ are roots of the same irreducible polynomial $g$ over field $F$

I defined a map $\psi: F(\alpha) \to F(\beta)$ by sending $p(\alpha)$ to $p(\beta)$. This map is clearly onto. If $p_1(\alpha)=p_2(\alpha)$, then by minimality of the polynomial $g$, we must have that $g$ divides $p_1-p_2$ and hence $p_1(\beta)=p_2(\beta)$. The same argument backward gives that this map is one-one. It is also easy to see that $\psi$ is a homomorphism. Thus, $\psi$ is an isomorphism.

I showed that both these fields are isomorphic. Is this what the question wants me to show? Is there any other way to do it?

Thanks for the help!!

$\endgroup$
  • $\begingroup$ ''If p1(α)=p2(α), then by minimality of the polynomial g, we must have that g divides p1−p2 and hence p1(α)=p2(α).'' Assumption and conclusion are the same. $\endgroup$ – Wuestenfux Jul 23 '17 at 14:28
4
$\begingroup$

The statement in the question isn't true. Consider the polynomial $x^3 - 2$ (over $\mathbb Q$). Two of its roots are $\alpha = \sqrt[3]{2}$ and $\beta = \frac{-1+i\sqrt{3}}{2}\sqrt[3]{2}$. It is easy to see that $\mathbb Q(\alpha) \neq \mathbb Q(\beta)$ as on contains only reals and one contains non-real numbers.

The best you can get is $F(\alpha)$ is isomorphic to $F(\beta)$ for which your proof works.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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