Why is there a basic open, everywhere dense neighborhood of every closed point on a reduced affine scheme?

Question

Let $R$ be a reduced, noetherian ring with minimal prime ideals $P_1,\ldots,P_m$. Let $M$ be a maximal ideal of $R$. Does there exist a regular element of $R$, i.e. $a \in R \setminus \bigcup_{i=1}^m P_i$, such that $a\notin M$ and $a \notin R^\times$?

Or put more geometrically: For a closed point $x$ on $X = \operatorname{Spec}(R)$, is there a basic open neighborhood $U = D(a) \subsetneq X$ of $x$ which meets every irreducible component of $X$?

If this is not true, then this means that all regular elements of $R$ that vanish somewhere, vanish in some common closed point. This sounds bad, but I do not see an immediate contradiction.

Any help is appreciated!

Edit: Of course, $R$ should be non-local since then this obviously fails.

Answer 1

No, there does not always exist such an element $a$. Here is a counterexample:
Let $R$ be any discrete valuation ring, which is certainly noetherian and reduced, and let $M$ be its unique maximal ideal.
Then it is already impossible to have $a\notin M$ and $a\notin R^\times$, since $R=M\cup R^\times$ !

Answer 2

I agree with Georges that the question is confusing/misleading. But let me follow up by adding a non-local counterexample. Any Artinian ring $R$ that has more than one prime ideal (so it is certainly not local) should not satisfy the condition that there is an $a\in R$ not in any minimal prime ideal such that $a\not\in R^\times$. For these rings any prime ideal is both maximal and minimal. For instance take the product of two fields.