Show that $\mathbb Q[x]/(x^2+2)$ and $\mathbb Q[x]/(x^2-2)$ are not isomorphic. I have a proof that says $\mathbb Q[x]/(x^2+2)$ and $\mathbb Q[x]/(x^2-2)$ are not isomorphic.
However I feel that it is not good one...
First I see that $x^2+2$ and $x^2-2$ are irreducible in $\mathbb Q$. Then I notice that $x^2=2$ and $x^2=-2$.
Now, if we take $(a+bx)(c+dx)$ on both, we end up
$ac-2bd + (ad + bc)x$ and $ac+2bd + (ad + bc)x$.
Can I now say, that these fields do not have the same structure, so they cannot be isomorphic?
I do not want exact answer (homework :) ), but a hint would be nice to guide me into right direction.
 A: Hint: Any isomorphism between the two fields will send $\mathbb{Q}$ to itself because $1$ has to map to $1$. Knowing this, we can examine which elements have square roots.
A: No, simply observing that the multiplication rules look different will not prove that the rings are not isomorphic. For example consider $\mathbb R$ with the ordinary addition and the "multiplication" $a\otimes b=-ab$. This has a different multiplication than $\mathbb R$ with the usual operations, but it is nevertheless isomorphic, by the isomorphism $x\mapsto -x$.

First off, it may help your intuition to realize that your two quotient rings are isomorphic to subrings of $\mathbb C$, namely $\mathbb Q+\sqrt2i\mathbb Q$ and $\mathbb Q+\sqrt2 \mathbb Q$.
Once you see this, you should be able to prove directly that one ring satisfies the property "$-(1+1)$ has a square root" and the other doesn't. And this means there cannot even be a homomorphism that takes the square root (in the ring where it exists) anywhere in the other ring.
A: Try the following: Assume there is an isomorphism $\psi:\Bbb Q[x]/(x^2 - 2)\to \Bbb Q[y]/(y^2 + 2)$. Then, since both these are two-dimensional over $\Bbb Q$, we have that $\psi$ is completely determined by $\psi(1)$ (which must be $1$) and $\psi(x) = a + by$ for some rational $a$ and $b$. Use this to derive a contradiction.
A: $\mathbb{Q}[x]/\langle x^2 -2 \rangle \cong \mathbb{Q}(\sqrt{2}) \Rightarrow$ if such an iso did exist we would have $\mathbb{Q}[x]/\langle x^2+2 \rangle \cong \mathbb{Q}(\sqrt{2}). $ So consider in  $\mathbb{Q}[x]/\langle x^2+2 \rangle $ the equivalence of $[a+bx] = [a]+[b][x]$. 
Recall $\mathbb{Q}(\sqrt{2}) = \{a+b\sqrt{2} : a,b \in \mathbb{Q} \}$. Hence $[x] \mapsto \sqrt{2}$ by some iso "$\phi$". However, this implies $\phi([x^2]) = \phi([x][x]) = \phi([x])\phi([x])=(\phi([x]))^2 = 2$. But $[x]^2 =-2 \Rightarrow \phi([x^2])=\phi(-2)=-2 \not = 2.$
Or just notice that there is not iso from $\mathbb{Q}(\sqrt{2}i) \to\mathbb{Q}(\sqrt{2}) $ since $[a+bi]=[a]+[b][i] \Rightarrow [i] \mapsto \sqrt{2}$, but then $[i^4]=[1] \mapsto 4 \Rightarrow$ we don't have an iso. 
A: $
K=\mathbb{Q}[x]/\langle x^2 -2 \rangle \cong \mathbb{Q}(\sqrt{2}) \subseteq \mathbb R
$
$
L=\mathbb{Q}[x]/\langle x^2 +2 \rangle \cong \mathbb{Q}(\sqrt{-2}) \not\subseteq \mathbb R
$
In words: $K$ can be embedded into the real numbers but $L$ cannot because it contains an element whose square is negative. Hence, $K$ and $L$ cannot be isomorphic.
