Consider the ideal $I=(x^2+1,y)$ in the polynomial ring $\mathbb{C}[x,y]$.Then which of the following is true?
a) $I$ is a maximal ideal
b) $I$ is a prime ideal but not a maximal ideal
c) $I$ is neither a maximal ideal nor a prime ideal
d) $I$ is a maximal ideal but not a prime ideal
My attempt: since $x^2+1$ is reducible in $\mathbb{C}(x,y)$, $I$ is not a maximal ideal and $I$ is also not a prime ideal. So option c is correct.
The argument is incomplete or incorrect, since maximal ideals can have reducible generators, e.g. $\,(x^2,x,y).\,$ But we can use reducibility of $\,x^2+1\,$ to show that $I = (x^2+1,y)\,$ is not prime, hence not maximal, by showing that $\,(x-i)(x+i) = x^2+1\in I,\,$ but $\,x\pm i\not\in I\,$ (since otherwise $\,x+i\, =\, (x^2+1) f + y\, g\,$ $\,\Rightarrow\, 2i = 0\,$ by evaluating at $\,x=i,y=0,\,$ and similarly for $\,x-i)$
Or, if quotient rings are known then $I$ is not prime by $\,\Bbb C[x,y]/I = \Bbb C[x]/(x^2+1)$ not a domain, because $\,x^2+1 = (x-i)(x+i)\,$ is reducible, yielding zero-divisors $\,x\pm i$ in the quotient ring (perhaps that's the type of argument you had in mind).
