My problem is finding all maximal ideals of the commutative ring $\Bbb R[x,y]/(x^{2}+1,y^{2}-1)$.
I'm aware of the statement that maximal ideals of $\Bbb R[x,y]/(x^{2}+1,y^{2}-1)$ are the same as maximal ideals of $\Bbb R[x,y]$ containing $(x^{2}+1,y^{2}-1)$. It is standard that $\Bbb R[x,y]=(\Bbb R[x])[y]$. Since $\mathbb{R}[x,y]/(x^{2}+1,y^{2}-1) = \left(\mathbb{R}[y]\right)[x]/(x^{2}+1,y^{2}-1) $, we can apply the third isomorphism theorem to say that: $$ \left(\mathbb{R}[y]\right)[x]/(x^{2}+1,y^{2}-1) \cong \frac{\left(\mathbb{R}[y]\right)[x]/(x^{2}+1)}{(x^{2}+1,y^{2}-1)/(x^{2}+1)} $$ The numerator is isomorphic to some filed $F[y]$ since $x^2+1$ is irreducible (?) in $\left(\mathbb{R}[y]\right)[x]$, and the bottom is the ideal $(y^2-1)$ in that ring, so we indeed have $F[y]/(y^{2}-1)$. By Hilbert's Nullstellensatz's theorem, all ideals are of the form $(y-a)$ for real number $a$ such that $a^2-1=0$ which forces $a=\pm1$. We conclude that $\mathbb{R}[x,y]/(x^{2}+1,y^{2}-1)$ has $2$ maximal ideals: $(y\pm1)/(y^{2}-1)$.
I'm not sure whether the above solution is correct since $x^2+1$ is irreducible in $\mathbb{R}[x]$ but what about $\left(\mathbb{R}[y]\right)[x]$? Using Hilbert's Nullstellensatz's theorem may be an overkill however it is an useful tool in general. Please light up my mind by your comments.
