Topological Rings and Homothecy

Question

I am trying to show that in a topological ring $A$, if left homothecy $x \mapsto ax$ is continuous at $x=0$ for all $a \in A$ and multiplication $\mu$ is continuous at $(0,0)$, then multiplication is continuous. The key identity is $$x_{1}x_{2} - y_{1}y_{2} = x_{1}(x_{2} - y_{2}) + x_{2}(x_{1} - y_{1}) + (x_{1} - y_{1})(x_{2} - y_{2}).$$ Everything works out fine when the ring is a metric space, but I am having some trouble in the general case. If commutativity simplifies things, then assume commutative.

Also, if the ring is a quotient of a topological ring, does it suffice to just show that left homothecy is continuous for all $a$?

Answer 1

As it stands, your key equation is false for noncommutative rings.

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With respect to your second question, the quotient of a topological ring is always a topological ring, no further assumptions needed. The fundamental observation is the following:

Suppose $X,Y$ are topological spaces with equivalence relations $\mathcal{R},\mathcal{S}$ such that the canonical projections $X\to X/\mathcal{R}$ and $Y\to Y/\mathcal{S}$ are open maps. Then the canonical continuous bijection $$\phi:\frac{X\times Y}{\mathcal{R}\times\mathcal{S}}\longrightarrow X/\mathcal{R}\times Y/\mathcal{S}$$ is a homeomorphism.

($\mathcal{R}\times\mathcal{S}$ stands for the equivalence relation on $X\times Y$ whereby $(x,y)\sim(x',y')$ iff $x\mathcal{R}x'$ and $y\mathcal{S}y'$, and $\phi$ is defined as $\phi([x,y])=([x],[y])$) The map $\phi$ is always a continuous bijection, but in general may fail to be a homeomorphism.

If $(A,+,\times)$ is a topological ring, and $I\subset A$ is a two-sided ideal, then the quotient map $A\to A/I$ is open since it is the quotient map of a group action on a space. It then follows very smoothly that addition and multiplication on $A/I$ are continuous maps, since they may be viewed as maps emanating from a quotient space (a colimit) rather than maps emanating from a product of spaces (a limit).

Indeed, if $c$ is addition or multiplication on $A$, and $\tilde{c}$ the quotient operation on $A/I$, then the diagram $$ \begin{array}{ccccc} A\times A &&\xrightarrow{\,c\,}&& A\\ \downarrow &&&&\downarrow\\ \frac{A\times A}{\sim}&\xrightarrow{\sim}&A/I\times A/I&\xrightarrow{\, \tilde{c}\,}&A/I \end{array} $$ proves continuity of $\tilde{c}$.

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I assume $(A,+)$ is a topological group, i.e. addition $+:A\times A\to A, (a,b)\mapsto a+b$ and inverse $A\to A, a\mapsto -a$ are known to be continuous.

Let $a\in A$, and let $V$ be a neighborhood of $0$. Because $+$ is continuous, there is a neighborhood $0\in V'\subset V$ such that $V'+V'\subset V$. Since left multiplication by $a$ is continuous at $0$, there exists a neighborhood $W'$ of $0$ such that $aW'\subset V'$. Since multiplication is continuous at $(0,0)$, there exists a neighborhood $W''$ of $0$ such that $W''W''\subset V'$. Now put $W=W'\cap W''$. Since $(A,+)$ is a topological group, the map $A\to A,x\mapsto a+x$ is a homeomorphism, and $a+W$ is an open neighborhood of $a$. Let $a'=a+x\in a+W$. Then $$a'W\subset aW'+W''W''\subset V'+V'\subset V.$$ We have thus proved that under your assumption (no commutativity as of yet),

For any $a\in A$ and any open neighborhood $V$ of $0$, there exists $W_a$ an open neighborhood of $a$ and $W_0$, an open neighborhood of $0$ such that $$W_aW_0\subset V$$

(In other words, the product map $\mu:A\times A\to A$ is actually continuous at all points of $A\times\lbrace 0\rbrace$.)

Now to prove continuity of multiplication when $A$ is commutative. Let $x_0,y_0\in A$. For all $x,y\in A$, $$\begin{matrix} xy-x_0y_0&=&x(y-y_0)+(x-x_0)y_0\\ &=&x(y-y_0)+y_0(x-x_0) \end{matrix}$$ Let $V_{x_0y_0}=x_0y_0+V$ be a neighborhood of $x_0y_0$, and $0\in V'\subset V$ a neighborhood of $0$ such that $V'+V'\subset V$. From the above, it follows that there are neighborhoods $W_{x_0}',W_{y_0}',W_0$ of $x_0,y_0,0$ respectively such that $W_{x_0}'W_0\subset V'$ and $W_{y_0}'W_0\subset V'$. Now define $W_{x_0}=W_{x_0}'\cap\big(x_0+W_{0}\big)$ and $W_{y_0}=W_{y_0}'\cap\big(y_0+W_{0}\big)$. If $\mu:A\times A\to A$ denotes multiplication, then $$\mu(W_{x_0}\times W_{y_0})\subset V_{x_0y_0}$$ so that $\mu$ is continuous at $(x_0,y_0)$.

Answer 2

We need to start also with the hypothesis that addition is continuous.

Assume that left and right multiplications are continuous and multiplication is continuous at $(0,0)$. We want to see that multiplication is continuous at $(a,b)$, for all $a,b\in R$.

Let $V$ be a neighborhood of $0$; we need to find a neighborhood $U$ of zero, such that $$ \mu((a+U)\times(b+U))\subseteq ab+V $$ that is, $$ \mu(a+x,b+y)\in ab+V $$ for all $x,y\in U$. This becomes $$ xb+ay+xy\in V $$ for all $x,y\in V$. So, let $V'$ be such that $V'+V'+V'\subseteq V$, which exists by continuity of addition.

Now, right multiplication by $b$ is continuous at $0$, so there exists a neighborhood $U_1$ of zero such that $xb\in V'$, for all $x\in U_1$.

Since left multiplication by $a$ is continuous at $0$, there exists a neighborhood $U_2$ of zero such that $ay\in U_2$ for all $y\in U_2$.

Since multiplication is continuous at $0$, there exists a neighborhood $U_3$ of zero such that $xy\in V'$, for all $x,y\in U_3$.

Setting $U=U_1\cap U_2\cap U_3$ we get the neighborhood of $0$ we need.

I don't think one can get away with continuity of right and left multiplications in the case of noncommutative rings. For instance, suppose that $R$ has a basis of neighborhoods of zero consisting of right ideals. Then right multiplications are continuous, because if $I$ is a member of the basis and $b\in R$, we have $Ib\subseteq I$. Addition is obviously continuous, because $I+I=I$ for all right ideals.

Also multiplication is continuous at $(0,0)$, because $I\cdot I\subseteq I$.

However, continuity of multiplication requires that, for any $a\in R$ and any $I$ in the basis, the right ideal $$ I:a=\{x\in R:ax\in I\} $$ is a neighborhood of zero (you see that this is just continuity of the left multiplication by $a$). Just find a filter of right ideals on some ring $R$ which hasn't this property and you'll have the counterexample.

Note that continuity of multiplication is automatic when you take as a basis of neighborhoods of zero a filter of ideals in a commutative ring.