How to prove extension field question.

Question

$f(x) \in \mathbb{Q}[x]$ has $\sqrt7$ as a root if and only if $x^2-7$ divides $f(x)$ in $\mathbb{Q}[x]$.

I think both sides is True, but what theorem can i apply to make clean proof?

Answer 1

The converse is the easiest, if $x^2-7$ divides $f$ then there exists $g \in\mathbb{Q}[x]$ such that: $$(x^2-7)g(x) = f(x)$$

Therefore, $f(\sqrt{7}) = (7-7)g(\sqrt{7}) = 0 $ so $\sqrt{7}$ is root of $f$.

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Now if $\sqrt{7}$ is root of $f$. By the Euclidean division's theorem there exists $r,q \in \mathbb{Q}[x]$ such that $\deg(r)<\deg(x^2-2)=2$ verifying, $$f(x) = (x^2-7)q(x)+r(x)$$

As $\deg(r)<2$ there exists $\alpha, \beta \in \mathbb{Q}$ such that $r = \alpha x+\beta$.

Moreover $f(\sqrt{7})=r(\sqrt{7})$ and we know that $f(\sqrt{7})=0$ therefore $r(\sqrt{7})=0$, hence $\alpha \sqrt{7}+\beta=0$ so if $\alpha \neq 0$ we would have $\sqrt{7} = -\frac{\beta}{\alpha}\in \mathbb{Q}$ which is not true. Therefore $\alpha = 0$, then we deduce $\beta = 0$. Consequently $r=0$.

Finally we have,

$$f(x) = (x^2-7)q(x)$$

Hence, $x^2-7$ divides $f$.