# Show that $\mathbb Q[x]/(x^2+2)$ and $\mathbb Q[x]/(x^2-2)$ are not isomorphic. [duplicate]

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.

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.

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.

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.

$\mathbb{Q}[x]/\langle x^2 -2 \rangle \cong \mathbb{Q}(\sqrt{2}) \Rightarrow$ 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.

$K=\mathbb{Q}[x]/\langle x^2 -2 \rangle \cong \mathbb{Q}(\sqrt{2}) \subseteq \mathbb R$

$L=\mathbb{Q}[x]/\langle x^2 +2 \rangle \cong \mathbb{Q}(\sqrt{-2}) \not\subseteq \mathbb R$

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.