# Spec($A$) is connected if $A$ is local

Another exercise from Balwant-Singh:

Show that if $A$ is local then Spec($A$) is connected in the Zariski topology.

Any hint ?

-

Another possible hint: a local ring $R$ have a unique maximal ideal $m$. For every proper ideal $I \lhd R$ we must have $m \in V(I)$ hence every openset $m \not \in D(I)$, for $I$ proper ideal.
I have understood your argument, but what do you mean by $D(I)$ ? –  WLOG Mar 31 '14 at 13:36
@WLOG Right I supposed that it was well known notation: $D(I) = Spec(R) \setminus V(I)$ a standard open set. –  Giorgio Mossa Mar 31 '14 at 13:37
HINT: $\operatorname{Spec}(A)$ is connected if and only if $A$ is connected.