Let $\alpha = \sqrt2+\sqrt3$. Prove that $I =\{f(x) \in \mathbb{Q}[x] : f(\alpha) = 0\}$ is a nontrivial principal ideal in $\mathbb{Q}[x]$. Discuss if $\mathbb{Q}[x]/I$ is a field.
This is my solution:
If $\alpha = \sqrt2+\sqrt3$, then $\alpha^{4}-10\alpha^{2}+1 = 0$. Thus $\alpha$ is a root of the polynomial $p(x) = x^{4}-10x^{2}+1 \in \mathbb{Q}[x]$. We now consider the map $\phi_{\alpha}: \mathbb{Q}[x] \to \mathbb{Q}[\alpha]$ extending the identity in $\mathbb{Q}$ and sending $x \mapsto \alpha$ (ie, the evaluation map). Then $\phi$ is a ring homomorphism, and it induces an isomorphism $\mathbb{Q}[x]/Ker(\phi_{\alpha}) \cong Im (\phi_{\alpha})$. We now prove that $Ker(\phi_{\alpha}) = I$. First, for every $f(x) \in I$ we have $f(\alpha) = 0$, therefore $f(x) \in Ker(\phi_{\alpha})$. In other words, $I \subset Ker(\phi_{\alpha})$. To see the inclusion in the other direction, suppose that $f(x) \in Ker(\phi_{\alpha})$, then $f(\alpha) = 0$ and $f(x) \in I$. It follows that $Ker(\phi_{\alpha}) = I$. Hence, $I$ is an ideal of $\mathbb{Q}[x]$. Moreover, $\phi_{\alpha}$ is surjective, for, $\mathbb{Q}[\alpha]$ is a vector space over $\mathbb{Q}$ of dimension 4, a basis is $\{1, \alpha, \alpha^{2}, \alpha^{3}\}$. Thus, every element in $\mathbb{Q}[\alpha]$ can be written uniquely as $a + b\alpha + c\alpha^{2}+d\alpha^{3} = \phi_{\alpha} (a + bx + cx^{2}+dx^{3}) \in Im(\phi_{\alpha})$ (for some $a, b, c, d \in \mathbb{Q}$. Hence, we get an isomorphism $\mathbb{Q}[x]/I \cong \mathbb{Q}[\alpha]$. Since $\mathbb{Q}[\alpha]$ is a field, the quotient $\mathbb{Q}[x]/I$ is also a field and $I$ is a maximal ideal generated by a unique monic polynomial of minimal degree. This is: $I = (x^{4}-10x^{2}+1)$
I´ve also though about a more hands-on argument showing that $x^{4}-10x^{2}+1$ is irreducible in $\mathbb{Q}[x]$ and, since $\mathbb Q[x]$ is a PID, the ideal $(x^{4}-10x^{2}+1)$ is prime and hence maximal, which implies that $\mathbb{Q}[x]/I$ is a field. I´ve tried modig out by $3$ but it factors as $(x^{2}+1)^{2}$. However if I assume for the sake of contradiction that it factors as
$$x^{4}-10x^{2}+1 = (x^{2}+ax+1)(x^{2}+bx+1)$$
I get that $a=\sqrt12, b = -\sqrt12$ which implies that it is irreducible in $\mathbb{Q}[x]$.