A totally ordered ring with its order topology is not a topological ring.

Question

There is a theorem that, if $R$ is a totally ordered ring which is also a division ring, then $R$ is a topological division ring with respect to the order topology on $R$. I am certain that, in general, a totally ordered ring equipped with its order topology need not be a topological ring. Could anybody please provide an example of such a ring?

Note: A ring in this setting is not required to have a multiplicative identity. However, it would be great to have a unital example and a nonunital example. Also, a commutative and a noncommutative example for each case (unital or nonunital) will be helpful.

Answer 1

Since a totally ordered group is a topological group in the order topology, the thing that can stop an ordered ring from being a topological ring in the order topology is a discontinuous multiplication. Since addition is continuous, the multiplication is continuous if and only if

1. for each $x\in R$ the left multiplication $\lambda_x \colon y \mapsto xy$ is continuous at $0$,
2. for each $x\in R$ the right multiplication $\rho_x \colon y \mapsto yx$ is continuous at $0$, and
3. the multiplication $\mu \colon (x,y) \mapsto xy$ is continuous at $(0,0)$.

For a commutative ring, points 1. and 2. are of course equivalent.

We can make 1. or 2. fail if we have "infinitely large" elements but not "infinitely small" elements, and the order topology is not discrete.

And example of a ring with infinitely large elements but not infinitely small elements is a polynomial ring over an ordered ring with [big-endian] lexicographic order ($p > 0$ if and only if the leading coefficient of $p$ is positive).

So let $S$ be an ordered ring that is not discrete, and take $R = S[X]$, endowed with the lexicographic order. This gives a total order on $R$ that is compatible with the ring operations - the leading coefficient of a product of polynomials is the product of the leading coefficients of the factors, so the product of positive elements is positive, and the leading coefficient of the sum of two polynomials is either one of the leading coefficients (if the degrees are different) or the sum of the leading coefficients (if the degrees are equal), so the sum of positive polynomials is positive.

And in that ordered ring, the multiplication with $X$ (since $X$ is central, we have $\lambda_X = \rho_X$) is not continuous at $0$. Taking $0 < a \in S$, there is no neighbourhood $U$ of $0$ such that $\lambda_X(U) \subset (-a,a)$, since $p\cdot X > a$ for every $0 < p \in R$, and by non-discreteness of $S$, every neighbourhood of $0$ in $R$ contains positive elements.

$R$ is unital/commutative if and only if $S$ is, thus choosing appropriate⁽¹⁾ $S$ we get unital and non-unital, commutative and non-commutative examples.

Also, the ideal $X\cdot S[X]$ gives a non-unital example.

⁽¹⁾ Examples of noncommutative ordered rings are however not obvious.

Answer 2

I just realized that the theorem I referred to in my question has a sketchy proof. Therefore, I am supplying a proof for my own record.

Theorem. Let $D$ be a totally ordered division ring with respect to the total order $<$ on $D$. Then, equipped with its order topology, $D$ is a topological division ring.

Proof. Write $D^\times$ for the multiplicative group $D\setminus\{0\}$. Also, $a:D\times D\to D$, $m:D\times D\to D$, and $i:D^\times \to D^\times$ denote the addition, the multiplication, and the inverse maps, respectively. To distinguish an ordered pair from an open interval, we shall write $\langle x,y\rangle \in D\times D$ for an ordered pair of $x,y\in D$, and $(x,y)\subseteq D$ for the open interval $\big\{z\in D\,|\,x<z<y\big\}$ between $x,y\in D$ with $x<y$. From now on, let $x$ and $y$ be arbitrary elements of $D$ with $x<y$. Recall that $D$ is of characteristic $0$ (whence division by $2$ is well defined.)

We shall first verify that $a$ is continuous. Suppose $\langle u,v\rangle \in a^{-1}\big((x,y)\big)$. Then, we have that $x<u+v<y$, which leads to $u\in\left(\frac{x+u-v}{2},\frac{y+u-v}{2}\right)$ and $v\in\left(\frac{x-u+v}{2},\frac{y-u+v}{2}\right)$. Clearly, $$\left(\frac{x+u-v}{2},\frac{y+u-v}{2}\right)\times \left(\frac{x-u+v}{2},\frac{y-u+v}{2}\right)\subseteq a^{-1}\big((x,y)\big)\,.$$ Ergo, there exists an open neighborhood of $\langle u,v\rangle$ that is contained in $a^{-1}\big((x,y)\big)$. Thus, $a$ is continuous.

Next, we claim that $m$ is continuous. Suppose $\langle u,v\rangle \in m^{-1}\big((x,y)\big)$. By switching signs if necessary, we may assume without loss of generality that $u\geq 0$ and $v\geq 0$. If both $u>0$ and $v>0$ hold, then we can further suppose that $0<x<y$ (otherwise we can simply replace $x$ by $\frac{uv}{2}$). As $0<x<uv<y$, we have $u\in\left(\frac{u+xv^{-1}}{2},\frac{u+yv^{-1}}{2}\right)$ and $v\in \left(2\big(u+xv^{-1}\big)^{-1}x,2\big(u+yv^{-1}\big)^{-1}y\right)$. Hence, the open neighborhood $\left(\frac{u+xv^{-1}}{2},\frac{u+yv^{-1}}{2}\right)\times \left(2\big(u+xv^{-1}\big)^{-1}x,2\big(u+yv^{-1}\big)^{-1}y\right)$ of $\langle u,v\rangle$ is contained in $m^{-1}\big((x,y)\big)$. Next, we handle the case where exactly one of $u$ and $v$ is zero. We suppose $u>0$ and $v=0$. Then, $x<0<y$ and the open neighborhood $\left(\frac{u}{2},2u\right)\times\left(\frac{u^{-1}x}{2},\frac{u^{-1}y}{2}\right)$ of $\langle u,v\rangle=\langle u,0\rangle$ is contained in $m^{-1}\big((x,y)\big)$. Finally, suppose that $u=0$ and $v=0$. Then, $x<0<y$ and we can take $\left(-1,+1\right) \times \left(-t,+t\right)$, where $t:=\min\{-x,y\}$, to be a required open neighborhood of $\langle u,v\rangle =\langle0,0\rangle$.

Finally, to justify that $i$ is continuous, we can assume without loss of generality that $0<x<y$. Then, clearly, $i^{-1}\big((x,y)\big)=\left(y^{-1},x^{-1}\right)$ is an open set. The proof is now complete.