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In a topological ring $R$ (assume identity if necessary), is it true that if $I$ is an ideal then $\bar{I}$ , i.e. closure of $I$ is also an ideal?

It's easy to show that if $x, y \in \bar{I}$ then so is $-x$ and to show $x+y$ lies in closure, we take $x+U$ and $y+V$ as neighborhoods of $x$ and $y$ that intersect with $I$ where $U,V$ are neighborhoods of 0. Then $x+y+(U\cap V)$ works to show that $x+y$ lies in the closure.

But to show that $rx \in \bar{I}$, $r\in R$, not every neighborhood of $rx$ comes from a nbd of $0$ as $r$ doesn't have to be a unit and multiplication by $r$ doesn't have to be an homeomorphism, right?

So is the statement false or am I missing something?

Bhaskar Vashishth
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Let $(x_i)_{i\in E}$ be a net with values in $I$ converging towards $x$. Then since multiplication is continuous, $rx_i\to rx$, and $rx_i \in I$ so that $rx \in \overline{I}$.

If you don't know about nets and want to translate this in terms of opens, here's how you do it : let $V$ be a neighbourhood of $rx$; let $W$ be a neighbourhood of $x$ such that $rW\subset V$ (continuity of $y\mapsto ry$). Then there is $y\in I\cap W$ so $ry\in V$ and $ry\in I$.

Maxime Ramzi
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