Sufficient condition for multiplication to be continuous

Question

Given a commutative ring $(R,+,\cdot)$, and a topology $\tau$ on $R$ such that for any $a\in R$ the maps

\begin{align} \cdot a: &R\to R \\ &x\mapsto a\cdot x \\\\ &\mbox{and} \\\\ +a: &R\to R \\ &x\mapsto x+a \end{align} Are continuous, is it necessarily true that $(R,+,\cdot,\tau)$ is a topological ring?

What I have tried: We need to show that the maps

\begin{align} \cdot : &R^2\to R \\ &(x,y)\mapsto x\cdot y \\\\ &\mbox{and} \\\\ +: &R^2\to R \\ &(x,y)\mapsto x+y \end{align}

are continuous. To do so, let $V \in \tau$, and let $(x,y) \in \cdot^{-1}(V)$, then since $\cdot y$ is continuous, there is some $U_x \in \tau$ such that

$$U_x \subseteq \cdot y ^{-1}(V) \subseteq \pi_1\left(\cdot^{-1}(V)\right)$$

and similarly there is some $U_y$ such that

$$ U_y \subseteq \cdot x^{-1}(V) \subseteq \pi_2\left(\cdot^{-1}(V)\right) $$

Where $\pi_1, \pi_2$ are the canonical projections, now one would be tempted to conclude that

$$U_x\times U_y \subseteq \cdot^{-1}(V)$$

But this is not obviously true, since it is not true in general that

$$\pi_1(X) \times \pi_2(X) \subseteq X$$

However my intuition from metric spaces says that we should still be able to choose $U_x$ and $U_y$ to be "small enough" to guarantee that their product is contained in $\cdot^{-1}(V)$, but this intuition might be leading me astray. I suspect that the claim is in fact false, since it would almost surely be listed on Wikipedia as an alternative characterization, but yet I cannot find a counterexample. I have found some examples showing that this is not true for general continuous functions, but it is more tricky to come up with examples where the functions also turn the underlying set into a ring.

Answer 1

With no further conditions there are indeed counterexamples - consider for instance $\Bbb Q$ with the usual operations and the cofinite topology.

Then $+$ and $\times$ are separately continuous, because for any $a, y \in \Bbb Q$, there is exactly one solution to $a + x = y$ in $\Bbb Q$, and there is either one solution to $ax = y$, or no solutions, or every element of $\Bbb Q$ is a solution. It follows from this that the preimage of a finite set under the maps $x \mapsto a + x$ and $x \mapsto ax$ is either finite or $\Bbb Q$.

However neither operation is jointly continuous. For example, the preimage of $\{1\}$ in $\Bbb Q^2$ under either operation is not closed. One reason is that if $U$ is a nonempty basic open set in $\Bbb Q^2$ (with the product-cofinite topology), then $U$ must intersect both preimages, since there are finite sets $F_1$ and $F_2$ such that $U$ is exactly the set of points $(a, b)$ with $a \notin F_1$ and $b \notin F_2$, and both preimages contain infinitely many points with distinct coordinates.

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The cofinite topology is a standard example to show that "a quasitopological group is not necessarily a topological group". You may find more relevant material by searching for "quasitopological group", "semitopological group", and possibly "semitopological ring".