Is $(X^2+1,Y^2-2)\subset\mathbb{Q}[X,Y]$ a prime ideal? Is it maximal?
I tried the following, and was wondering if my approach is correct. I appreciate if anyone can verify my solution or tell me where I went wrong.
First define $\phi:(\mathbb{Q}[Y])[X]\to(\mathbb{Q}[i])[Y]:\sum_{j=0}^np_jX^j\mapsto\sum_{j=0}^np_ji^j$, with $p_j\in\mathbb{Q}[Y]$. This is a surjective ring homomorphism with kernel $(X^2+1)$ and thus it follows from the first isomorphism theorem that $\mathbb{Q}[Y])[X]/(X^2+1)\cong(\mathbb{Q}[i])[Y]$. Then by the third isomorphism theorem, $$\mathbb{Q}[Y])[X]/(X^2+1,Y^2-2)\cong(\mathbb{Q}[Y])[X]/(X^2+1))/((X^2+1,Y^2-2)/(X^2+1))\cong(\mathbb{Q}[i])[Y]/(Y^2-2).$$ Now I want to show that if $(Y^2-2)\subset J\subsetneq(\mathbb{Q}[i])[Y]$, that $J=(Y^2-2)$. I first observe that since $\mathbb{Q}[i]$ is a field, $(\mathbb{Q}[i])[Y]$ is a principal ideal domain, and thus there exists a $g\in(\mathbb{Q}[i])[Y]$ such that $J=(g)$. Furthermore, $Y^2-2$ is irreducible over $(\mathbb{Q}[i])[Y]$, so $Y^2-2=fg$ for an $f\in(\mathbb{Q}[i])]Y]$ implies $\deg(g)=2$, and thus $\deg(f)=0$, so $f$ is invertible and $g=f^{-1}(Y^2-2)$. Thus $(Y^2-2)=J$, and thus $(Y^2-2)$ is maximal, $(\mathbb{Q}[i])[Y]/(Y^2-2)$ is a field. Because of the isomorphism shown above, $\mathbb{Q}[Y])[X]/(X^2+1,Y^2-2)$ is a field, and thus $(X^2+1,Y^2-2)$ is maximal.
Best Answer
Hint: Consider $\mathbb{Q}[X,Y] \to \mathbb{Q}[i,\sqrt2]$ given by $X \mapsto i$ and $Y \mapsto \sqrt 2$.