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Another exercise from Balwant-Singh:

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

Any hint ?

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up vote 6 down vote accepted

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.

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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.

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