Is the ideal generated by the polynomial $x^2-2$ maximal in the ring $\mathbb{C}[x]$?

Question

Personal question: Is the ideal generated by the polynomial $x^2-2$ maximal in the ring $\mathbb{C}[x]$?

I know the ideal in $\mathbb{Q}[x]$ generated by $x^2-2$ is maximal, considering $\mathbb{Q}[x]/(x^2-2) \cong \mathbb{Q}\sqrt{2}$.

Answer 1

Nope. You can factor it as $(x-\sqrt{2})(x+\sqrt{2})$ (and neither factor is a unit), so the ideal generated by $x-\sqrt{2}$ (or $x+\sqrt{2}$) is a larger ideal containing it. More generally, in a polynomial ring $K[x]$ over a field $K$, you can show that the ideal generated by a single polynomial is maximal iff the polynomial is irreducible over $K$.

Answer 2

The (nonzero) prime and hence maximal ideals of $\mathbf{C}[x]$ are precisely linear polynomials.

One can see directly that $(x^2-2)$ is not maximal in several ways though. One is to note that $x^2 - 2 = (x+ \sqrt{2}) (x-\sqrt{2})$. This shows that it is not irreducible and therefore, not prime (hence not maximal).