I am trying to understand why if $\alpha$ and $\beta$ are roots of an irreducible polynomial p on $\mathbb{Q}$, then $\mathbb{Q}(\alpha) \simeq \mathbb{Q}(\beta)$ without using the abstract-algebra arguments like $\mathbb{Q}(\alpha) \simeq \mathbb{Q}[X]/(p) \simeq \mathbb{Q}(\beta)$.
So let
$\begin{align} \phi : \mathbb{Q}(\alpha)\rightarrow \mathbb{Q}(\beta) \\ u+v \alpha \mapsto u+v \beta \end{align}$
Hence:
$\phi(u+\alpha v) \phi (x+ \alpha y) = ux +\beta(vx+uy) + \beta^2vy$
and :
$\phi((u+\alpha v)(x+\alpha y)) = ux+\beta(vx+uy)+\phi(\alpha^2)vy$
Now if $\phi(\alpha^2) = \beta^2$, we would have a homomorphism. But this is not true in general, since if we take the reductible polynomial $p(x)=(x-2)^2(x-3)^2$ and $\alpha=\sqrt{2}$ and $\beta=\sqrt{3}$, then we have $\phi(\alpha^2)=2 \neq 3 = \phi(\beta^2)$. So how can I see this is true when $p(x)$ is irreducible?
Edit :
It is true that we don't just have linear factors in $\alpha$, so I reformulate the question :
Why if p is not irreductibe, we can have for $p(\alpha)$ and $q(\alpha)$ in $\mathbb{Q}(\alpha)$ that $\phi(p(\alpha)q(\alpha)) \neq \phi(p(\alpha))\phi(q(\alpha))$?
There is a problem with your definition if the degree of $P>2$, you cannot write every element of $\mathbb{Q}(\alpha)$ $ u+\alpha v, u,v\in \mathbb{Q}$.