Quotients of topological rings Let $\varphi\colon R\to S$ be a surjective ring homomorphism and let $R$ be a topological ring.

Is there some nice characterization of the finest topology on $S$ for with both $S$ becomes a topological ring and $\varphi$ becomes continuous?

If needs be, further conditions can be imposed on $R$, $S$ and/or $\varphi$.
Thanks for any help!
 A: I think that hardmath's conjecture in comments that the answer is the quotient topology is correct.
Give $S$ the quotient topology (i.e., a subset $U\subseteq S$ is open if and only if $\varphi^{-1}(U)\subseteq R$ is open), and let $I=\ker(\varphi)$.
The quotient topology is the finest topology on $S$ for which $\varphi$ is continuous, so if we can show that it makes $S$ into a topological ring (i.e., all the ring operations are continuous) then we're done.
The key observation is that $\varphi$ is an open map: if $V\subseteq R$ is open, then 
$$\varphi^{-1}\left(\varphi\left(V\right)\right)=\bigcup_{x\in I}(V+x)$$
is a union of translates of $V$, and therefore open, and so $\varphi(V)\subseteq S$ is open.
It now easily follows that
$$\varphi\times\varphi:R\times R\to S\times S$$
is also an open map, since $(\varphi\times\varphi)(V_1\times V_2)=\varphi(V_1)\times\varphi(V_2)$ is open in $S\times S$ if $V_1$ and $V_2$ are open in $R$.
Consider the multiplication maps $\mu:R\times R\to R$ and $\overline{\mu}:S\times S\to S$, and let $U\subseteq S$ be open. Then $$\overline{\mu}^{-1}(U)=\left(\varphi\times\varphi\right)\left(\mu^{-1}\varphi^{-1}(U)\right)$$
is open. So $\overline{\mu}$ is continuous, and similar proofs show that the other ring operations on $S$ are continuous. 
