Prove $\frac{\mathbb{Q}[x]}{} \cong \frac{\mathbb{Q}[x]}{}$. It's obvious that $\frac{\mathbb{Q}[x]}{} \cong \mathbb{Q}[\sqrt{2}]$ but I can't get the other way, and it seems to be a logical way to prove this.
well, it should also be obvious that $\frac{\mathbb{Q}[x]}{\langle x^2+4x+2 \rangle} \cong \mathbb{Q}[\sqrt{2} - 2] = \mathbb{Q}[\sqrt{2}]$.