Prove that there is no irreducible polynomial in $\mathbb Q[x]$ which is zero at both $x=\sqrt 5$ and $x=\sqrt 7$ Prove that there is no irreducible polynomial in
    $\mathbb Q[x]$ which is zero at both $x=\sqrt 5$ and $x=\sqrt 7$
I must confess that I have no intuition about this problem. Unless it's just that we could factor out x minus each root individually. But this is in the context of learning about splitting fields. 
 A: The minimal polynomials of $\sqrt 5$ and $\sqrt 7$ are $f_1=x^2-5$ and $f_2=x^2-7$ so suppose there exists a polynomial $f(x)$ with these two roots, then we know that:
$f_1, f_2$ divides $f$ $\implies$ so $f$  can't be irreducible
A: If $p(x)\in\mathbb Q[x]$ is an irreducible polynomial with $\sqrt{5}, \sqrt{7}$ as roots, then if $F$ is the splitting field of $p$, there is an automorphism of $F$ that maps $\sqrt{5}\mapsto\sqrt{7}$. This automorphism thus maps $5\mapsto 7$, which is impossible. Thus, $p$ cannot exist. 
A: $f \,=\, \overbrace{(x^2\!-\!5)\, q + ax\!+\!b}^{\rm Division\ Algorithm},\ $ so $\ 0 = f(\sqrt 5) = a\sqrt 5 + b,\ $ so $\,a=0=b,\, $ else $\,\sqrt 5 = -\frac{b}a\in \Bbb Q $
Thus $\,f = (x^2\!-\!5)\,q,\ $ so $\ 0 = f(\sqrt 7)  = 2\, q(\sqrt 7).\,$ As above we deduce $\, q = (x^2\!-\!7)g,\ $   
thus  $\ f = (x^2\!-\!5)(x^2\!-\!7)g\ $ is reducible. 
Remark $\ $ Generally, $\,\Bbb Q[x]\,$ is Euclidean $\Rightarrow$ PID $\Rightarrow$  UFD, so $\ g,h\mid f\iff{\rm lcm}(g,h)\mid f,\ $ and $\,\gcd(g,h) = 1\iff{\rm lcm}(g,h) = gh,\,$ just as in $\,\Bbb Z.\,$ Above $\,g,h\,$ are nonassociate primes (being irreducible in a UFD), and it is true in any domain that lcm = product for nonassociate primes. 
Generally if $\,\alpha\,$ is an algebraic number with minimal polynomial $\,g\,$ then $\,f(\alpha) = 0\iff g\mid f,\,$ since the polynomials $\,\in \Bbb Q[x]\,$ having $\,\alpha\,$  as a root form an ideal $\,I,\,$ hence, $\,\Bbb Q[x]\,$ being Euclidean, $I$ is generated by any element $\,g\,$ of minimal degree - the "minimal" polynomial (for, if not, then $\,g\nmid f\,$ so $\,0\neq  f\ {\rm mod}\ g = f - qg\in I\,$ and has smaller degree than $\,g,\,$ contra minimality of $\,g)$
A: A helpful fact is the following:
If $h(x) \in \mathbb{Q}[x]$ has a root $\alpha$ in some extension $K$ of $\mathbb{Q}$, then $m(x) | h(x)$, where $m(x)$ is the minimal polynomial of $\alpha$.  

Alternative:
We know that, if $f(x) \in F[x]$ is irreducible, then $F[x] / \langle f(x) \rangle \cong F[\alpha]$ for any root $\alpha$ of $f(x)$.  So if there were such an irreducible polynomial, we'd have $\mathbb{Q}[\sqrt{2}] \cong \mathbb{Q}[\sqrt{5}]$.  Can you show that those two field extensions of $\mathbb{Q}$ are not isomorphic?
