Two quadratic fields over $\mathbb{Q}$ I'm having a bit of trouble showing that the two quadratic fields $\mathbb{Q}[X]/(X^2+1)$ and $\mathbb{Q}[X]/(X^2+3)$ over $\mathbb{Q}$ are not isomorphic (as fields). Could someone help me? Perhaps the two fields are isomorphic to each other? Many thanks for your answers. 
 A: Hint: There is a polynomial $P$ with rational coefficients such that $P(t)=0$ has a solution in one of the fields but not in the other.
A: Show that there is no element $\alpha$ of $\Bbb Q[X]/(X^2+3)$ such that $-\alpha=\alpha^{-1}$. There is certainly such an element in $\Bbb Q[X]/(X^2+1)$--namely, the equivalence class of $X$. A field isomorphism would guarantee that such an element exists in the other, too.
Alternately (equivalently), show that $\Bbb Q[X]/(X^2+3)$ has no primitive $4$th root of unity.
A: Hint $\ $ Compare discriminants: $\rm\displaystyle\ \{\big(\alpha-\alpha'\big)^2:\ \alpha \in \mathbb Q(\sqrt{-1})\,\}\, =\, -\mathbb Q^{\:2}\, $ vs. 
$\rm\, -3\, \mathbb Q^{\:2}\, $ for $\rm\ \mathbb Q(\sqrt{-3})\ $
Note that if $\rm\ \alpha,\: \alpha'\ \not\in\mathbb Q\ $ are conjugate then they remain so under any field isomorphism since their minimal polynomial $\rm\ (x-\alpha)\ (x-\alpha')\:$ lies in $\rm\,\mathbb Q[x],\:$ so it is fixed by any isomorphism.
In fact quadratic fields are characterized uniquely be their discriminant.
