Confusion about Spec of quotient ring Consider the ring $A := \dfrac{\mathbb C[x]}{(x(x-1)(x-2))}$. According to some sources (cf. Vakil) $sp(Spec (A))$ should be just the three points $\{0,1,2\}.$ It seems right, because $A$ is the ring of regular functions on these three points. However, I have the impression that there are many more prime ideals in $A$. Take for example $(x-0.1).$ Why is it not in the spec?
 A: Since $\mathbb{C}$ is a field, $\mathbb{C}[x]$ is a Euclidean domain, which implies that it is a PID. Therefore every prime ideal of $\mathbb{C}[x]$ is either $(0)$ or $(p(x))$ where $p(x)$ is an irreducible polynomial. By the correspondence theorem, the prime ideals of $A$ naturally correspond to the prime ideals of $\mathbb{C}[x]$ containing $(x(x-1)(x-2)).$ Those are of the form $(p(x))$ where $p(x)$ is irreducible and divides $x(x-1)(x-2).$ The only such ideals are $(x), (x-1), (x-2).$
A: To go with your example, if we consider $x-0.1$ in $\mathbb C [x]$ then we can multiply it by $x^2-2.9x+1.71$, which will give $x^3-3x^2+2x-0.171$, which is just $-0.171$ in our ring, since $x(x-1)(x-2)=x^3-3x^2+2x$. Hence it is invertible, since 0.171 is clearly invertible, and so is any other element.
I think that if you check that $(x-a)$ is the only prime ideal in $\frac{\mathbb C [x]}{(x-a)}$ then it will be simpler to compute and still give you the idea (indeed, if $a=0$ it is trivial that the only non-invertible ideal is $(x)$).
A: Just another viewpoint: the ideals $(x)$, $(x - 1)$, and $(x - 2)$ of $\mathbb{C}[x]$ are pairwise coprime, so by the Chinese remainder theorem your quotient is isomorphic to
\[
\frac{\mathbb{C}[x]}{(x)} \times \frac{\mathbb{C}[x]}{(x - 1)} \times \frac{\mathbb{C}[x]}{(x - 2)}.
\]
Each factor here is isomorphic to $\mathbb{C}$. [What's the map?] Now, check that in general the prime ideals of the product ring $A \times B$ are of the form $\mathfrak{p} \times B$ and $A \times \mathfrak{q}$.
