Prove there isn't an isomorphism between quotient polynomial rings

Question

Prove there isn't an isomorphism $$\phi: {{\mathbb Q [x]} \over {I_1}} \to {{\mathbb Q [x]} \over {I_2}}$$

when $I_1=\langle x^2-2\rangle$, $I_2=\langle x^2+2\rangle$.

I want to assume there is an isomorphism, and say $\phi(1+I_1)=1+I_2$ so $\phi(x^2-2+I_1)=x^2-2+I_2$, and to get a contradiction from here, but why is that true that $\phi(x^2-2+I_1)=x^2-2+I_2$?

Answer 1

Remark That $Q[X]/I_1=\{a+xb, a,b\in Q, x^2=2\}$ $Q[X]/I_2=\{a+yb, a,b\in Q, y^2=-2\}$.

Suppose $\phi$ exists, write $\phi(x)=a+by$, $\phi(x^2)=2=(a+by)^2=a^2-2b^2+2aby=2$. This implies $ab=0$. If $a=0$, $-2b^2=2$ impossible in $Q$.

Suppose that $b=0$, $a^2=2$ impossible in $Q$.

Answer 2

One very simple reason is that, if they were isomorphic, their rings of integers would be too. Now the ring of integers $\mathbf Z[x]/(x^2-2)$ has only two units – $\pm 1$, while $\mathbf Z[x]/(x^2+2)$ has a group of units isomorphic to $\mathbf Z/2\mathbf Z\times\mathbf Z$, by Dirichlet's units theorem . In any case, even without using this theorem, it has two other units: $\;\pm(3+2x)$.