Isomorphism of Quotient Fields Are the two fields $\mathbb R[x]/\langle(x^2+1)\rangle $ and $\mathbb R[x]/\langle(x^2+x+1)\rangle $ isomorphic?
Solution
Now $\mathbb C$ is algebraic over $\mathbb R$ .Consider the evaluation homomorphism $\phi_i:\mathbb R[x]\rightarrow \mathbb C$.Now $i$ is irreducible over $\mathbb R$ and its irreducible polynomial is $x^2+1$.
We have $\mathbb R[x]/\langle(x^2+1)\rangle \cong \mathbb R[i]$ which is $\mathbb C$ .Similarly $\mathbb R[x]/\langle(x^2+x+1)\rangle \cong \mathbb R[\omega]$
  where $\omega ^2+\omega +1=0$ arguing the same as above 
Thus they are both isomorphic to $\mathbb C$ 
Is it correct?Please say if the reasoning is correct
Looking for your help
 A: Let's say we didn't know about $\Bbb C$. We can still exhibit an ring-isomorphism between $\Bbb R[x]/\langle x^2 + 1\rangle$ and $\Bbb R[x]/\langle x^2 + x + 1\rangle$.
For notational simplicity, call $\langle x^2 + 1\rangle = I$, and $\langle x^2 + x + 1\rangle = J$.
Define: $\phi: \Bbb R[x]/I \to \Bbb R[x]/J$ by:
$\phi(a + bx + I) = \left(a - \dfrac{b}{\sqrt{3}}\right) - \dfrac{2b}{\sqrt{3}}x + J$
This is clearly an bijective $\Bbb R$-linear map (regarding these as extension rings of $\Bbb R$), so we show $\phi$ is multiplicative.
$\phi((a + bx + I)(c + dx + I)) = \phi((ac - bd) + (ad + bc)x + I))$
$= \left(ac - bd - \dfrac{ad+bc}{\sqrt{3}}\right) - \dfrac{2(ad+bc)}{\sqrt{3}}x + J$
$= \left(ac + \dfrac{bd}{3} - \dfrac{ad}{\sqrt{3}} - \dfrac{bc}{\sqrt{3}} - \dfrac{4bd}{3}\right) + \left(\dfrac{-2ad}{\sqrt{3}} + \dfrac{2bd}{3} - \dfrac{2bc}{\sqrt{3}} + \dfrac{2bd}{3} - \dfrac{4bd}{3}\right)x + J$
$= \left[\left(a - \dfrac{b}{\sqrt{3}}\right)\left(c - \dfrac{d}{\sqrt{3}}\right) - \left(\dfrac{-2b}{\sqrt{3}}\right)\left(\dfrac{-2d}{\sqrt{3}}\right)\right]$ 
$+ \left[\left(a - \dfrac{b}{\sqrt{3}}\right)\left(\dfrac{-2d}{\sqrt{3}}\right) + \left(\dfrac{-2b}{\sqrt{3}}\right)\left(c - \dfrac{d}{\sqrt{3}}\right) - \left(\dfrac{-2b}{\sqrt{3}}\right)\left(\dfrac{-2d}{\sqrt{3}}\right)\right]x + J$
$= \left[\left(a - \dfrac{b}{\sqrt{3}}\right) - \dfrac{2b}{\sqrt{3}}x + J\right]\left[\left(c - \dfrac{d}{\sqrt{3}}\right)- \dfrac{2d}{\sqrt{3}}x + J\right]$
$= \phi(a + bx + I)\phi(c + dx + I)$
(since in $\Bbb R[x]/J$ we have:
$(a' + b'x + J)(c' + d'x + J) = (a'c' - b'd') + (a'd' + b'c' - b'd')x + J$, as is easily verified).
One might conjecture that this yields an "unusual" automorphism of $\Bbb C$, but this is not so-the multiplications in our two rings are two different rules.
