Proof Check:if $a+b\sqrt c$ is a root of a poly in $Q[x]$,so is $a-b\sqrt c$

Question

The question: If $a+b\sqrt c$ is a root of a poly in $Q[x]$,so is $a-b\sqrt c$,in which $a,b,c$ is rational number, and $\sqrt c$ is irrational.

My proof:Let $p(x)$ be the minimum polynomial of $a+b\sqrt c$ and $F$ be the splitting field of $p(x)$ over $Q$ and therefore normal . Since $a-b\sqrt c=2a-a-b\sqrt c$ ,the minimum polynomial of $a-b\sqrt c$,denoted by $q(x)$ ,splits in $F$.Thus,the splitting field of $p,q$ over Q is actually a subfield of $F$.However,$F$ is actually $Q(u_1,\dots,u_n)$,$u_i$ is root of $p$, which means it is an intersection of all fields containing $Q$ and $u_i$.So up to now we can conclude that $p(x)$ and $q(x)$ have the same roots .Therefore we finished our proof.

It is my first try in field theory.I wonder if my proof is correct.If it is incorrect,how could i amend it.

Appreciate!

Answer 1

I think that your proof is incorrect or at least inaccurate. Just because $q$ splits over $F$ we cannot conclude that it shares its roots with $p$. Here's an example:

Take the polynomial $x^8+6x^4+1$, which is irreducible over $\mathbb{Q}$ by

Irreducibility of $x^8+6x^4+1$

If $\alpha$ is a root of this polynomial in its splitting field, then it can be shown that

$$Q(\alpha)=\frac{\alpha^4+1}{\alpha^2}$$ is also in the splitting field, but is root of the polynomial $x^2+4$ over $\mathbb{Q}$.

Maybe I'm wrong, but this example seems to share all the assumptions you make in the proof.

I think it would be much easier to show that the minimal polynomial of $a+b\sqrt{c}$ is quadratic, i.e. is of the form $x^2+dx+f$ and then showing that $a-b\sqrt{c}$ is also root of this polynomial.