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