$\mathbb Q[x]/(x^2+1)$ is not isomorphic to $\mathbb Q[x]/(x^2+2)$ I found an argument online that the two fields are not isomorphic but I can't make sense of the argument. You find it here.
These are the things I'm confused about when it comes to the argument.  


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*It says that the field $\mathbb Q[x]/(x^2+1)$ contains a root of the polynomial $x^2+1$. I suppose this is the congruence class $[x]$? But this is not the same as the zero element of the field $\mathbb Q[x]/(x^2+1)$, right?  

*It says that every isomorfism takes $0$ to $0$. I interpret this as the zero-element in $\mathbb Q[x]/(x^2+1)$ (which I interpret to be $[0]$) is taken to the zero-element in $\mathbb Q[x]/(x^2+2)$ denoted $[0]_2$. But in the argument it says that this means that if $\mathbb Q[x]/(x^2+1)$ has a root of the polynomial, then also $\mathbb Q[x]/(x^2+2)$ must have a root of this same polynomial. Why is that?  
Greatful for any help!
 A: 1) Yes to both. Note that $[x]^2 + [1] = [x^2 + 1] = [0]$.
2) Not only must every isomorphism take 0 to 0, it must take every element $a$ of ${\mathbb Q}$ to itself (i.e., it must take $[a]$ to $[a]$). Even better, it must preserve polynomial expressions over ${\mathbb Q}$, in the sense that an element $a_0 + a_1 \beta + a_2 \beta^2 + \dots a_k \beta^k$ (with the $a_i \in {\mathbb Q}$ and $\beta \in {\mathbb Q}[x]/(x^2+1))$ maps to $a_0 + a_1 \beta' + a_2 \beta'^2 + \dots a_k \beta'^k$, where $\beta' \in {\mathbb Q}[x]/(x^2+2)$ is the image of $\beta$. In particular, if $1 + \beta^2 = 0$, then $1 + \beta'^2 = 0$.
Now because there indeed such an element $\beta \in {\mathbb Q}[x]/(x^2+1)$ with $1 + \beta^2 = 0$ (namely $[x]$), there should also be a $\beta' \in {\mathbb Q}[x]/(x^2+2)$ with $1 + \beta'^2 = 0$ (assuming that the fields are isomorphic).
What is left is to show that such an element of ${\mathbb Q}[x]/(x^2 + 2)$ does not exist. For this, write $\beta' = [a_0 + a_1 x]$ with $a_0, a_1 \in {\mathbb Q}$. Work out $1 + \beta'^2 = [1] + [a_0 + a_1 x]^2$ using $[x]^2 = [-2]$ in the form $[b_0 + b_1 x]$ and see that this can never be $[0]$.
A: First,we define an operation on $\mathbb{Q}[x]/(x^{2}+2)$ as follows:
Let $f(x) \in\mathbb{Q}[x]$, By Euclidean Alogrithm, we have:
$f(x)=g(x)(x^2+2)+ax+b$
$\Rightarrow \overline{f(x)}=ax+b\in \mathbb{Q}[x]/(x^{2}+2)$
For $\overline{f(x)}=ax+b;\overline{g(x)}=cx+d$, $f,g \in\mathbb{Q}[x]/(x^{2}+2)$ ,we have:
$\overline{f(x)g(x)}=(ad+bc)x+bd-2ac$
With the operation defined as above,we contruct a ring homomorphism as follows
$\mathbb{Q}[x]/(x^2+2)\overset{\varphi }{\rightarrow} Im\varphi$
$\varphi(ax+b) \rightarrow ai\sqrt{2}+b$
It is easy to check that  $\varphi$ is a ring homomorphism. Moreover, clearly $\varphi$ is an bijection.
Hence, $\varphi$ is a ring isomorphism and
$Im\varphi =\mathbb{Q}[\sqrt{-2}]=\mathbb{Q}[i\sqrt{2}]$
Therefore,we deduce $\mathbb{Q}[x]/(x^2+2)\cong \mathbb{Q}[i\sqrt{2}]$
Instead of doing as above, we can contruct a ring homomorphism 
$\mathbb{Q}[x]\rightarrow \mathbb{Q}[i\sqrt{2}]$
$\varphi(f(x))\rightarrow f(i\sqrt{2})$
We claim that : $Ker\varphi=(x^2+2)$
Indeed, Let $f(x)∈\mathbb{Q}[x]$, By Euclidean Alogrithm, we have: $f(x)=g(x)(x^2+2)+ax+b$
So, $f(i\sqrt{2})=0$ is equivalent to that $ai\sqrt{2}+b=0$
which gives $a=b=0$.
Consequently,$f(x)=g(x)(x^2+2)\in(x^2+2)$ (Ideal generated by $x^2+2$). 
So $Ker\varphi\subseteq (x^2+2)$ ($1$)
Conversely, if $f \in (x^2+2)$ then $f=(x^2+2)g(x)$ for $g(x)$ some polynomial.
Clearly, $\varphi(f)=0$ which implies $(x^2+2)\subseteq Ker\varphi$ ($2$)
Combining ($1$) and ($2$) we have $Ker\varphi= (x^2+2)$.
and $Im\varphi=\mathbb{Q}[i\sqrt{2}]$
By the first Isomorphism theorem, we deduce that :$\mathbb{Q}[x]/(x^2+2)\cong \mathbb{Q}[i\sqrt{2}]$.
By similar argument we can easily get $\mathbb{Q}[x]/(x^2+1)\cong \mathbb{Q}[i]$
It remains to show that $\mathbb{Q}[i\sqrt{2}]$ is not isomorphic  $\mathbb{Q}[i]$
Because one has an element whose square is $-2$ but no element whose square is $-1$, while the other has an element whose square is $-1$ but no element whose square is $-2$.
I should have made it clearer here,
Assume a contradiction that $\mathbb{Q}[i\sqrt{2}]\cong\mathbb{Q}[i]$ then there exists an isomorphism $\psi$ from $\mathbb{Q}[i\sqrt{2}]\rightarrow\mathbb{Q}[i]$
We have $\psi(i\sqrt{2})=a+bi$ for some $a,b \in \mathbb{Q}$
On the other hand $-2=-2\psi(1)=\psi(-2)=\psi ((i\sqrt{2})(i\sqrt{2}))= \psi^2(i\sqrt{2})=(a+bi)^2$
So $-2=a^2-b^2+2abi$ which gives $a=0$ or $b=0$
If $a=0$ then $b^2=2$ has no rational solutions
If $b=0$ then $a^2=-2$ has no rational solutions
Consequently we get a contradiction.
So $\mathbb{Q}[x]/(x^2+2)\cong \mathbb{Q}[i\sqrt{2}] \ncong \mathbb{Q}[i] \cong \mathbb{Q}[x]/(x^2+1)$
A: The other answers are so long! One field has a square root of negative one call it $i$ for obvious reasons. The other one doesn't. If $\phi$ is an isomorphism what is $\phi(i)^2$?
