Are rings $\mathbb{R}[x]/(x^2+2)$ and $\mathbb{R}[x]/(x^2+3)$ isomorphic? 
We have two rings:
  $\mathbb{R}[x]/(x^2+2)$ and $\mathbb{R}[x]/(x^2+3)$. Find out if they are isomorphic.

I have several similar exercises to do and I do not have an idea how to proceed. I suspect that the first one is isomorphic with $\mathbb{R}[i\sqrt{2}]$ and second one with $\mathbb{R}[i\sqrt{3}]$.
 A: Hint: they are both field extensions of $\mathbb R$ of degree $2$.
A: Any homomorphism $h:\mathbb{R}[x]/(x^2+2)\longrightarrow\mathbb{R}[x]/(x^2+3)$ is defined entirely by $h(1)$ and $h(x)$. Of course, we must have that $h(1)=1$, so $h$ is entirely defined by $h(x)$.
On the other hand, we must have ${h(x)}^2=h(x)\cdot h(x)=h(x^2)=h(-2)=-2$. Can you take it from here?
EDIT: Adding from the comments. $i\sqrt{2}$ is not an element of $\mathbb{R}[x]/(x^2+3)$ per se, although there is an isomorphism between $\mathbb{R}[x]/(x^2+3)$ and $\mathbb{C}$. Explicitly, we have that $h(x)=ax+b$ for some $a,b\in\mathbb{R}$. To find them, we evaluate ${h(x)}^2$:
$${(ax+b)}^2=a^2x^2+2abx+b^2=2abx+(b^2-3a^2)$$
Hence, we must have $2ab=0$ and $b^2-3a^2=-2$. Since $a$ and $b$ are both real, it's easy to see the only solution is $b=0$ and $a=\pm\sqrt{2/3}=\pm\sqrt{6}/3$. In other words,
$$h(x)=\pm\frac{x\,\sqrt{6}}3$$
A: Show that they're both isomorphic to $\mathbb{C}$. You can do this by considering homomorphisms $\mathbb{R}[x] \to \mathbb{C}$ defined by evaultion at $\sqrt{2}i$ and $\sqrt{3}i$ respectively. Are these maps surjective? What are their kernels? What does this tell you?
