# The maximal ideal of ring of germs of regular functions

Today I am reading Hartshorne Algebraic geometry section 1.3

For $$P\in Y,$$ we define the local ring of $$P$$ on $$Y, O_P,$$ to be the ring of germs of regular functions on $$Y$$ near $$P$$. Its maximal ideal $$m$$ is the set of germs of regular functions which vanish at $$P.$$ For if $$f(P) \neq 0,$$ then $$1/f$$ is regular in some neighborhood of $$P$$.

I have questions about the last sentence.

My understanding is: it is (generally) not true that

For any topological spaces $$X,Y$$, continuous $$f:X\rightarrow Y$$, any element $$0\in Y$$, $$f(P)\neq 0\implies f(U)\neq 0$$ for some neighborhood $$U$$ of $$P$$. （*）

Here is my reasoning for (*):

Consider the contrapositive, the statement becomes: if $$x_{i}\rightarrow P$$, $$f(x_{i})=0$$, then $$f(P)=0.$$ For all topological spaces continuity implies sequential continuity, so we only need to show the sequence $$0,0,\cdots$$ converges uniquely to $$0.$$ This is only true when the topology of $$Y$$ is $$T_1.$$ Zariski topology (on algebraic varieties) is $$T_{1}$$ so the Hartshorne statament is true.

My question is: is my reasoning correct? If it is, is there any better way to understand (internalize) this statement, i.e., "For if $$f(P) \neq 0,$$ then $$1/f$$ is regular in some neighborhood of $$P$$''? I feel like I took too many steps to understand it.

• @MarianoSuárez-Álvarez I mean $f: X\rightarrow Y$ when $X,Y$ are topological spaces. The topology is any topology. I think you misunderstand my question. Commented Dec 22, 2022 at 9:36
• @MarianoSuárez-Álvarez Oh yes! This is indeed the faster way I am looking for. If $T_{1}$ then every point is closed so my statement is true for every point. No need to consider the contrapositive. Commented Dec 22, 2022 at 9:48
• @MarianoSuárez-Álvarez The statement is not true when $Y$ is not $T_{1}$ Commented Dec 22, 2022 at 10:26

As Mariano Suárez-Álvarez points out in the comment, I will write an answer for this.

My reasoning is correct, while there is a faster way to prove (*) when $$Y$$ is $$T_{1}$$.

If $$Y$$ is $$T_{1}$$, then any point set $$\{0\}\subseteq Y$$ is closed, so we can take $$U=f^{-1}(Y-\{0\})$$ as the neighborhood of $$P$$.