Find isomorphism between $\mathbb{Q}[T]/(T^2+3)$ and $\mathbb{Q}[T]/(T^2+T+1)$ The books states that the isomorphsim is $g(T)=2T+1$ and the identity when restricted to $\mathbb{Q}$.
I would like some help to understand what the process is to find $g$.
 A: The first step is to look at the roots of the polynomials (in $\mathbb C$ or just $\overline{\mathbb Q}$). Those are $\pm i\sqrt 3$ and $\frac{-1\pm i\sqrt 3}{2}$. So if $t$ is a root of the latter, then $2t+1$ is a root of the former!
A: One way to think about this is that substitution $2T+1$ for $T$ results in
$$T^2+3 \mapsto (4T^2+4T+1)+3=4(T^2+T+1).$$
If we let $I=\langle T^2+3\rangle $ and $J = \langle T^2+T+1 \rangle$, this shows that the $\mathbb Q$-algebra isomorphism $\varphi: \mathbb Q[T] \to \mathbb Q[T]$ given by $$T \mapsto 2T+1$$ satisfies 
$$\varphi(I)=J,$$ and hence we have an induced isomorphism
$$\mathbb Q[T]/I \simeq \mathbb Q[T]/J.$$
To summarize, one method for finding such an isomorphism of quotients by principal ideals is to look for "change of variables" that takes the generator of one ideal to a unit multiple of the generator of the other ideal.
A: The first thing to note is that as $1$ must be mapped to $1$ under an isomorphism we get that all of $\mathbf{Z}$ must be preserved by the isomorphism, from there we can see that as $g(a/b) = g(ab^{-1}) = g(a)(g(b))^{-1}$ we have that all of $\mathbf{Q}$ must be preserved too.
Now we have to find where $T$ can go and the rest of the isomorphism will be determined, as $T^2 + 3 = 0$ in $\mathbf{Q}[T]/(T^2 + 3)$ we have that $g(T^2 + 3) = g(0) = 0$ and so $g(T)^2 + 3 = 0$, now any representative of $g(T)$ in $\mathbf{Q}[T]/(T^2 + T + 1)$ can be written as $aT + b$ with $a,b\in\mathbf{Q}$ so solving for $a$ and $b$ in the above we get $a^2T^2 + 2abT + b^2 + 3 = 0$ and so $(2ab - a^2) T + (b^2 + 3 - a^2) = 0$ so we need $2ab = a^2$ and $b^2 + 3 = a^2$, as $a$ cannot be zero (we are looking for an isomorphism and $g(b)$ already maps to $b$ so $g(T)$ can't) we get $2b = a$ and then $b^2 + 3 = 4b^2$ so $b^2 = 1$. Putting this all together gives our possible isomorphisms, $g(T) = 2T + 1$ or $-2T - 1$.
