Recently, I proved the following nice characterization of the saturation of the multiplicative subset $ S_f: = \{1,f,f^2,\ldots\}$ for $f$ an element of a commutative unital ring $A$ (however, I only need to work with $A$ a f.g. $k$-algebra over a field $k$).
Let me denote by $\overline{S}$ to the saturation of a multiplicative set $S$. I proved that,
$$\overline{S_f}=A-\bigcup\{ m \mid m \text{ maximal and does not contain } f\}$$
My problem is this result will automatically imply the following —unexpected— equality.
$$\bigcup\{ m \mid m \text{ maximal ideal in A that does not contain } f\} = \bigcup\{ p \mid p \text{ prime ideal in A that does not contain } f\}$$
Which I simply do not believe. But I tried to produce a counter-example using schemes and it looks the examples are avoiding the difference between those two sets.
Best Answer
Take $R= k[[X]]$ for a field $k$ and $f=X$.
Then $R$ is local so has only one maximal ideal and it contains $X$. Therefore the LHS is $\emptyset$ : there is no maximal ideal that does not contain $f$
However $f$ is not nilpotent $(X^n \neq 0$ for all $n$), so $f$ is not in all prime ideals: the RHS is not empty, there is a prime ideal that does not contain $f$.
So you probably made a mistake in your proof