Support of $f \in A$ in Spec $A$ for a reduced ring $A$

Question

Let $A$ be a reduced ring (i.e. no nilpotents). Take any $f \in A$ and I would like to show that $\operatorname{Supp} f = \overline{D(f)}$, where $D(f)$ denotes the distinguished open set of $\operatorname{Spec}A$ where $f$ does not vanish. I think that the fact that reducible is crucial here, but somehow my argument doesn't use this fact. I can't see where I am going wrong and I would appreciate some assistance. Thanks!

Answer 1

1) Given a scheme and a global section $f\in \Gamma(X,\mathcal O_X)$, the support of $f$ is the set of points $x\in X$ where the germ of $f$ is non-zero: $f_x\neq0$ .
Note carefully that this subset $\operatorname {supp} f\subset X$ is trivially closed (since its complement is open by the very definition of "germ" !) and that you need not take its closure.

2) In the case of an affine scheme $X=\operatorname {Spec} A$ the support of $f\in A=\Gamma(X,\mathcal O_X)$ is thus the set of prime ideals $\mathfrak p\subset A $ such that $\frac 01\neq \frac f1\in A_\mathfrak p$.
If for example $A$ is a domain the support of any $f\neq 0$ is $\operatorname {supp} f=\operatorname {Spec} A$,

3) The slightly counterintuitive example above is explained as follows:
Given a global section $f\in \Gamma(X,\mathcal O_X)$, it has a value $f(x)$ at every $x\in X$, namely: $$f(x)=\operatorname {class} f_x\in \kappa (x)=\mathcal O_{X,x}/\mathfrak m _x$$ We may then consider the subset $Z_X(f)\subset X$ of points $x\in X$ where $f(x)=0$ and its complement $$D_X(f)=X\setminus Z_X(f)=\{x\in X|f_x\notin \mathfrak m_x\}\subset X$$ Since obviously $f_x\notin \mathfrak m_x\implies f_x\neq 0$ we have $D_X(f)\subset \operatorname {supp} f$ and thus (since this last set is closed) $$\overline {D_X(f)}\subset \operatorname {supp} f$$[Thanks to user tracing for reminding me to mention this fact, which I had forgotten to state in my original answer]

In the important special case of an affine scheme $X=\operatorname {Spec} A$ we have $$\kappa (x)=\mathcal O_{X,x}/\mathfrak m _x=A_\mathfrak p/\mathfrak pA_\mathfrak p=\operatorname {Frac}(A/\mathfrak p) $$ and so $Z_X(f)=\{\mathfrak p\in \operatorname {Spec} A|f\in \mathfrak p\}=V(f)$.
This by the way proves that for a general scheme $X$ the subset $D_X(f)\subset X$ is always open (but not closed in general).

4) All of the above has nothing to do with reducedness!
Reducedness comes in when you wonder what it means for a function $f\in A$ to have value zero everywhere, in other words what does $Z(f)=\operatorname {Spec} A$ mean?
Answer: it means that $f\in \bigcap \mathfrak p= \operatorname { Nil} A$.
Hence if $A$ is reduced we have $Z(f)=\operatorname {Spec} A\iff f=0$ .
However given a field $k$ and the non-reduced ring $A=k[T]/(T^2)=k[\epsilon]$, the "function" $f=\epsilon $ is zero everywhere although its support is all of $\operatorname {Spec} A=\{(\epsilon)\}$.

Answer 2

Yes, reducedness is crucial for this. In general, all you can say is that $\operatorname{Supp} f$ contains $\overline{D(f)}$.

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You might want to consider the ring $k[x,y]/(xy,y^2)$, taking $f = y$. One has $D(y) = \emptyset,$ but $\operatorname{Supp} y$ is equal to the closed point $(x,y)$. (Georges's example with the dual numbers, taking $f = \epsilon$, is another good illustration of how your statement fails in the non-reduced case.)