7
$\begingroup$

Can you tell me if this is correct?

Let $R$ be a ring and let it have the $I$-adic topology for some ideal $I$ in $R$. I want to show that $+: R \times R \to R$ is continuous at $(x_0, y_0)$.

Proof: Neighbourhoods of zero look like $I^nR$ so that neighbourhoods of $x$ in $R$ look like $x + I^nR$. What I want to show is that for a neighbourhood $x_0 + y_0 + I^{n_0}R$ of $x_0 + y_0$ there exists a neighbourhood $N = I^nR \times I^mR$ of $(x_0, y_0)$ such that for all $(x,y) \in I^nR \times I^mR$ we have $x+y \in x_0 + y_0 + I^{n_0}R$. Pick $n = m = n_0$. Now $+(x_0 + I^{n_0}R \times y_0 + I^{n_0}R) = x_0 + y_0 + I^{n_0}R$ which proves the claim.


Similarly, $\cdot : R \times R \to R$ is continuous at $(x_0, y_0)$. Let $N=I^{n_0}R$ be a neighbourhood of zero so that $x_0 y_0 + I^{n_0} R$ is a neighbourhood of $x_0 y_0$. Now we want to find $n,m$ such that $\cdot(x_0 + I^n R \times y_0 + I^m R ) \subset x_0y_0 + I^{n_0}R$. Again let $n=m=n_0$. Then $\cdot(x_0 + I^{n_0} R \times y_0 + I^{n_0} R ) = x_0 y_0 + I^{2n_0}R \subset x_0 y_0 + I^{n_0}R $. Which proves the claim.


Thanks for your help.

  • 1
    Seems correct to me2012-07-31

1 Answers 1

1

As Andrea Mori remarks, the proofs appear correct.

One minor technicality: Due to the appearance of cross terms in $(x_0+i)(y_0+j) = x_0y_0 + x_0j+iy_0+ij$ we can only infer that $(x_0+I^{n_0}R) \cdot (y_0+I^{n_0}R) \subseteq x_0y_0+I^{n_0}R$ Of course, this does not change the essence of the proof.