Which of the following fields are isomorphic? 
Let $\mathbb{Q},\mathbb{R}$ denote the fields of rational numbers and real numbers respectively. Which of the following fields are not isomorphic.
  (a) $\mathbb{Q}[x]/( x^2 + 1 )$ and $\mathbb{Q}[x]/( x^2 + x + 1 )$.
  (b) $\mathbb{R}[x]/( x^2 + 1 )$ and $\mathbb{R}[x]/( x^2 + x + 1 )$.      
Justify your answer.


My thoughts:
(a) $\mathbb{Q}[x]/( x^2 + 1 )$ is isomorphic to $\mathbb{Q}(i)$ and $\mathbb{Q}[x]/( x^2 + x + 1 )$ is isomorphic to $\mathbb{Q}(\sqrt{3},i)$. So they are not isomorphic.   
(b) $\mathbb{R}[x]/( x^2 + 1 )$ is isomorphic to $\mathbb{R}(i)$ that is $\mathbb{C}$ and $\mathbb{R}[x]/ ( x^2 + x + 1 )$ is isomorphic to $\mathbb{R}(\sqrt{3} , i)$ that is $\mathbb{C}$.
So they are isomorphic.
Are my arguments correct?
 A: One idea: In $\mathbb{Q}[x]/<x^2 + 1>$, all elements are of the form $a + bx$, where $a$ and $b$ are rational numbers, and $x^2 = -1$. 
Show that the polynomial $p(t) = t^2 + t + 1$ does not have any solutions in this field. Of course, $p(t)$ does have solutions in the field $\mathbb{Q}[x]/<x^2 + x + 1>$, so if it didn't have any solutions in the first field, you would show the fields are not isomorphic.
To show that $p(t)$ doesn't have solutions in the first field, suppose $t = a + bx$ were a solution. Expand $t^2 + t + 1$, and show that this places impossible restrictions on the rational numbers $a$ and $b$.
A: The notation I learned is $\mathbb{F}/(f)$, so I will use this here.
$\mathbb{Q}/(x^2+x+1)$ is not isomorphic to $\mathbb{Q}(\sqrt3,i)$ because neither of them can be generated. It should be $\mathbb{Q}(\sqrt{-3})$ that is isomorphic to $\mathbb{Q}/(x^2+x+1)$.
And there is a little problem with the other one. $\mathbb{R}/(x^2+x+1)$ is isomorphic to $\mathbb{R}(\sqrt3,i)$, which is of course true, is not so clear. The reason is the same as the $\mathbb{Q}$ case.
A: The isomorphism between R[x]/(x^2+1) and R[x]/(x^2+x+1)
Both are isomorphic to C. First isomorphism is f(i) and for the other one it is f((i*sqrt(3)-1)/2)
