Is $\mathbb R[x]/(x^2-2)=\mathbb R$? 
Is $\mathbb R[x]/(x^2-2)=\mathbb R$?

I am reading Artin's Algebra and trying to analyze quotient rings. In one example he shows that $\mathbb Z[i]/(i-2) \cong \mathbb Z/5\mathbb Z$, example $11.4.5$. Very cool.
Now, I am trying to do a similar analyzing to see what $\mathbb R[x]/(x^2-2)$ is.
We know $x^2-2$ is reducible and $x^2-2=(x-\sqrt 2)(x+\sqrt 2)$. So, is it true I that I can write $\mathbb R[x]/(x^2-2)=\mathbb R[x]/(x-\sqrt 2)(x+\sqrt 2)$?
Also, I presume that since the canoncial map $\mathbb R[x] \to \mathbb R[x]/(x^2-2)$ where $x \mapsto \sqrt 2$, then we can write $\mathbb R[x]/(x^2-2)=\mathbb R[\sqrt 2]$. But since $\sqrt 2 \in \mathbb R$, then $\mathbb R[\sqrt 2]=\mathbb R$
 A: Take the homomorphism:
$$\mathbb R[x]\to \mathbb R\times\mathbb R\\
p(x)\mapsto (p(\sqrt{2}),p(-\sqrt{2}))$$
The kernel of this homomorphism is the set of $p$ such that $p(\sqrt{2})=p(-\sqrt{2})=0$, which means the $p$ which are divisible by $x^2-2$. 
So the factor map: $\mathbb R[x]/(x^2-2)\to\mathbb R\times\mathbb R.$ is one-to-one.
Now you just have to prove that this map is onto.
A: More generally, if you have a principal domain $A$ and $p,q$ two relatively prime elements, $A/(pq)$ is isomorphic to $A/p\times A/q$.
You have quotients map $f_p:A\rightarrow A/p$ and $f_q$. $(f_p,f_q):A\rightarrow A/p\times A/q$ 
suppose that $(f_p(l),f_q(l))=0$, you can write $up+vq=1,u,v\in A$ this implies that $l-upl+vql$ since $f_p(l)=0, l=ap$, $l=bq$ implies $l=upbq+vqap=pq(ub+av)$, this implies that $l\in (pq)$ this implies that the kernel of $(f_p,f_q)$ is $(pq)$.
Let $x,y\in A$, $x=upx+vqx, y=upy+vqy, f_p(vqx+upy)=f_p(x), f_q(upy+vqy)=f_q(y)$ this implies that $(f_p,f_q)$ is surjective and $(f_p,f_q)$ induces an isomorphism $A/(pq)\rightarrow A/p\times A/q$.
