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!!

  • $\begingroup$ ''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

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.


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