# If $A$ is reduced, Spec $A$ has no embedded points (How to use the hints)

I am struggling with the last part of the following exercise of Vakil's FOAG:

5.5.C. EXERCISE (ASSUMING (A)). Show that if $A$ is reduced, Spec $A$ has no embedded points. Hints: (i) first deal with the case where $A$ is integral, i.e., where Spec $A$ is irreducible. (ii) Then deal with the general case. If $f$ is a nonzero function on a reduced affine scheme, show that $\operatorname{Supp}f = \overline{D(f)}$: the support is the closure of the locus where $f$ doesn’t vanish. Show that $\overline{D(f)}$ is the union of the irreducible components meeting $D(f)$, using (i).

I wonder how to show that Spec$A$ has no embedded points based on his hint (ii). Suppose Spec$(A)$ does have an embedded point, say $P$. By his definition, $P$ is not an associated point w.r.t the support of the global section $1\in A$. But I don't know how to continue from here to find a contradiction.

Once you've shown that $\mathrm{Supp}(f) = \overline{D(f)}$ (which requires reducedness), then part (i) of the hint follows immediately because any non-empty open in an irreducible affine is dense.
In the general case then if $\overline{D(f)}$ intersects an irreducible component, then it contains the irreducible component by the same argument as above. So you will find that $\overline{D(f)}$ is the union of irreducible components intersecting $\overline{D(f)}$, for any $f\in A$. But this tells you that there are no embedded points, since this discussion tells you that the generic points of irreducible components of $\mathrm{Supp}(f)$ are all generic points of irreducible components of your scheme.