Define an explicit isomorphism between $\Bbb Q[x]/\langle x^2+1\rangle$ and $\Bbb Q[x]/\langle x^2+4 \rangle$.

Question

I know that $\Bbb Q[x]/\langle x^2+1\rangle$ is isomorphic to $\Bbb Q(i)$. I also know that $\Bbb Q[x]/\langle x^2+4 \rangle$ is isomorphic to $\Bbb Q(2i)$. Since $\Bbb Q(i)$=$\Bbb Q(2i)$, it should be the case that $\Bbb Q[x]/\langle x^2+1\rangle$ is isomorphic to $\Bbb Q[x]/\langle x^2+4 \rangle$. So may I ask for an explicit isomorphism between $\Bbb Q[x]/\langle x^2+1\rangle$ and $\Bbb Q[x]/\langle x^2+4 \rangle$?

Answer 1

Define $\phi:\Bbb Q[x]/(x^2 + 1) \to \Bbb Q[x]/(x^2 + 4)$ by $$ \phi(a + bx) = a + (b/2)x $$ Per the comment, it's easier to think of this as $\phi:\Bbb Q[x]/(x^2 + 1) \to \Bbb Q[y]/(y^2 + 4)$, defined by $\phi(a + bx) = a + (b/2)y$. Or, more simply, $\phi(x) = y/2$.

Answer 2

Define a map $f:\Bbb Q[x]\to\Bbb Q[x]$ by $f(x)=\frac12x$. The consider the composition

$$\Bbb Q[x]\overset{f}{\to}\Bbb Q[x]\to\Bbb Q[x]/(x^2+4).$$

This is surjective because $f$ is, and you can check that the kernel of this composition is $(x^2+1)$, giving the desired isomorphism.