# Proof of equality of $2$ polynomials without using Galois Theory?

I have the following situation and ask myself whether an argument is available to prove the equality of the $2$ following polynomials.

Both polynomials are of degree $n$ and are monic.

The polynomial $p(x)\in\mathbb Z[x]$ is irreducible over $\mathbb Q$.

The other polynomial is $q(x)\in\mathbb R[x]$ according to the information I have. Both have a real number $\alpha\in\mathbb R$ as a common root. So $\alpha$ is algebraic over $\mathbb Q$ of degree $n$. What I know additionally is that $q(x)$ has $n$ different real roots.

If one could prove that $q(x)\in\mathbb Q[x]$ the equality of the polynomials would from my point of view follow as the $p(x)$ is the minimal polynomial of $\alpha$ and so $p(x)$ would divide in $\mathbb Q[x]$ the $q(x)$. The equal degree would then assure the equality of both.

But now I am confused whether one can talk about a splitting field $L=\mathbb Q(\text{list of the$n$real roots})$ of $q(x)$ over $\mathbb Q$ as I dont know whether the polynomial coefficients are in $\mathbb Q$ or e.g. in some subfield of $L$ defined by irrational radical expressions.

I dont see now whether one could somehow argue hopefully by using pure field theory and its Tower Theorem that the subfield of $L$ containing the coefficients of $q(x)$ must indeed be $\mathbb Q$.

This way I could avoid further search for application of the Galois Theory.

• counterexample: $p(x)=x^2-2$ and $q(x)=(x-\sqrt{2})(x-\sqrt{3})$ – Sfarla Jan 29 '16 at 17:31
• thats a fine and obvious counterexample. – Wolfgang Tintemann Jan 29 '16 at 19:30

(Edit) To risk error by simplifying: pick $n-1$ distinct random irrationals and use them and $\alpha$ as the roots of $q$. Then $q$ is in $\Bbb{R}[x]$, but it is overwhelmingly unlikely that the random roots are in $\Bbb{Q}[\alpha]$, much less that they are the conjugates of $\alpha$. (/Edit)
Let $t_1, \dots, t_{n-1}$ be $n-1$ $\Bbb{Q}[\alpha]$-independent real transcendentals. ("$\Bbb{Q}[\alpha]$-independent" means none of them are in the extension made by the others, e.g., $t_1 \not \in \Bbb{Q}[\alpha]$ and $t_j \not \in \Bbb{Q}[\alpha, t_1, ,t_2, \dots, t_{j-1}]$ for $j \in [2,n-1]$. There are infinite sequences of these, so no problem getting $n-1$ of them.) Then set $q(x) = (x-\alpha) \prod_{i=1}^{n-1} (x-t_i)$. It's hopeless that any of $q \in \Bbb{Z}[x]$, $q \in \Bbb{Q}[x]$, or $q \in \Bbb{Q}[\alpha,x]$.
(Edit) For $n=2$, try $q(x) = (x - \pi)(x - \alpha) = x^2 -(\pi + \alpha)x + \pi\alpha$. Since $\pi$ is transcendental, it is not in any algebraic extension of $\Bbb{Q}$. The existence of Hamel bases (via AC) allows us to conclude that there are large enough collections of transcendentals to satisfy our needs for any $n$.(/Edit)
All you can actually promise is that $(x-\alpha)$ divides $\gcd(p(x), q(x))$, not that $p(x)$ divides $q(x)$.