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Background

This question is motivated by trying to answer this question. But before going into the question straight let me give some background.

Definition 1. Let $R$ be a ring and $A$ be an ideal of $R$. Let $\mathcal{T}_A$ denote the set of all semiprime ideals of $R$ containing $A$. Then, $$\beta(A):=\bigcap_{Q\in \mathcal{T}_A}A$$ called the prime radical of $A$.

Let us now make the following definition,

Definition 2. Let $R$ be a ring and $A\subseteq R$. Let $\mathcal{T}_A$ denote the set of all semiprime ideals of $R$ containing $A$. Then, $$\eta(A):=\bigcap_{Q\in \mathcal{T}_A}A$$ we call $\eta(A)$ to be the prime radical of $A$ in $R$.

We now note the following,

Theorem 1. Let $R$ be a ring and $A\subseteq R$. Let $\mathscr{P}(R)$ denote the power set of $R$. Define, $\Phi:\mathscr{P}(R)\to \mathscr{P}(R)$ by, $$\Phi(A)=\eta(A)$$Then,

  • $A\subseteq \Phi(A)$ for all $A\in \mathscr{P}(R)$

  • $A\subseteq B\implies \Phi(A)\subseteq \Phi(B)$ for all $A,B\in \mathscr{P}(R)$

  • $\Phi(\Phi(A))=\Phi(A)$ for all $A\in \mathscr{P}(R)$

Each of the properties follows immediately from Definition 2. Consequently we claim that the operator $\Phi$ is a closure operator on $R$. So, it generates a topology on $R$. Denote this topological space by $(R,\tau)$ and call this space (for the time being) the prime radical space of $R$, the topology being prime radical topology on $R$.

Question

Is $(R,+,\cdot)$ a topological ring with respect to the prime radical topology?

I am interested in this question because if the answer of this question is affirmative then the answer of MO post linked above seems to be affirmative for at least for this topology on $R$.

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Question

Is $(R,+,\cdot)$ a topological ring with respect to the prime radical topology?

Answer

It is not true that $(R,+,\cdot)$ is always a topological ring with respect to the prime radical topology. We don't give any specific counterexample but instead point out a way to construct such a counterexample. The construction of an explicit counterexample is left to the interested reader.

Theorem 1. Let $(R,+,\cdot)$ be a ring such that $x^n\ne x$ for all $n\in \mathbb{N}\setminus\{0,1\}$ and for all non-zero, non-identity $x\in R$. Equip $R$ with prime radical topology. If $(R,+,\cdot)$ is a topological ring with respect to the prime radical topology then the prime radical space of $R$ is Hausdorff.

Proof. Since $(R,+,\cdot)$ is a topological ring, it follows that $(R,+)$ is a topological group under the same topology. Since a $T_0$-topological group is Hausdorff, it is enough to show that the prime radical space of $R$ is $T_0$.

For this purpose let $x,y\in R$ such that $x\ne y$. Then consider the $m$-systems $\mathscr{N}(x):=\{x,x^2,\ldots,x^n\}$ and $\mathscr{N}(y):=\{y,y^2,\ldots,y^n\}$. Note that both $\mathscr{N}(x)$ and $\mathscr{N}(y)$ are open.

If either $x=0$ then $\mathscr{N}(y)$ is an open set which contains $y$ but not $0$ and we are done in this case. Similar will be the argument if $y=0$.

So we assume that $x,y\ne 0$. If $y=x^n$ for some $n\in \mathbb{N}$ then note that $n>1$ and that $y\ne 1$. Now consider $\mathscr{N}(y)$. Notice that $y\in\mathscr{N}(y)$ but $x\notin \mathscr{N}(y)$ by our hypothesis. Otherwise if $x^n\ne y$ for all $n\in \mathbb{N}$ then it follows that $x\in \mathscr{N}(x)$ but $y\notin \mathscr{N}(x)$.

So we are done.

Theorem 2. The prime radical space of a nontrivial $R$ is always non-$T_1$.

Proof. It is enough to prove that there exist a singleton set in $R$ which is not closed. But this is easy since

  • The closed sets of the prime radical space of $R$ are precisely the semiprime ideals of $R$.

  • $R$ is nontrivial.

  • No singleton subset of $R\setminus\{0\}$ is an ideal of $R$.

So we are done.