I am currently studying field extensions from Serge Lang's book Undergraduate Algebra. On page 269, he says this:

'Special case: Suppose that $\sigma$ is the identity on $F$ (any field), and let $\alpha,\beta$ be two roots of an irreducible polynomial in $F[t]$. Then there exists an isomorphism $\tau : F(a) \to F(b)$ which is the identity on $F$ and which maps $\alpha$ to $\beta$.'

So, I decided to test this theory out with the polynomial $t^4-5t^2+6$ whose roots are $\sqrt{2},-\sqrt{2},\sqrt{3},-\sqrt{3}$. So, by Lang's words there must be an isomorphism $\Bbb Q(\sqrt{2})$ onto $\Bbb Q(\sqrt{3})$ which is the identity on $F$ and maps $\alpha$ to $\beta$, so therefore $a+b\sqrt{2}$ must be mapped onto $a+b\sqrt{3}$ by this mapping. However, even though this mapping satisfies $\tau(v+w)=\tau(v)+\tau(w)$ where $v,w$ are in $\mathbb{Q}(\sqrt2)$, it does not satisfy $\tau(vw)=\tau(v)\tau(w)$ as $(ac+2bd)+(ad+bc)\sqrt3$ does not equal $(ac+3bd)+(ad+bc)\sqrt3$ - note the difference between $2bd$ and $3bd$. So, this mapping $\tau$ is clearly not an isomorphism. Where have I gone wrong with this, have I missed out a step or a hidden subtlety?

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    $\begingroup$ But its not irreducible so the result does not apply $\ (t^2-2)(t^2-3)\ \ $ $\endgroup$ – Bill Dubuque Jul 3 '17 at 16:06
  • $\begingroup$ $t^4-5t+6=(t^2-2)(t^2-3)$ $\endgroup$ – Francesco Polizzi Jul 3 '17 at 16:07
  • $\begingroup$ I've applied LaTeX formatting to part of your question, so you can see how it's done. To get greek letters, you type \tau or \beta; I didn't apply these changes because I wanted to mostly preserve what you wrote. $\endgroup$ – John Hughes Jul 3 '17 at 16:08
  • $\begingroup$ Oh god sorry. I thought accidentally for a second that irreducible just meant that the roots were not in Q because I was only considering linear factors. Oh well my mistake sorry $\endgroup$ – Daniele1234 Jul 3 '17 at 16:08
  • $\begingroup$ How do I close the question now? $\endgroup$ – Daniele1234 Jul 3 '17 at 16:09

$t^4-5t^2+6 = (t^2-3)(t^2-2)$ is reducible, and the theorem needs an irreducible polynomial.


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