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I am looking to prove that a polynomial function $f: \mathbb{R} \rightarrow \mathbb{R} $ is continuous for the Zariski topology on $\mathbb{R} $(both domain and range)

How do I do this?

Bernard
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Mary
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  • Well, have you tried taking an open set and seeing whether its inverse image is open? – Eric Wofsey Mar 06 '18 at 21:31
  • How do you define the Zariski topology on a field? There aren't many points… – Bernard Mar 06 '18 at 22:12
  • @bernard I'm not sure how you define it I haven't heard of this type of topology before? – Mary Mar 07 '18 at 21:00
  • @EricWofsey I kmow that's what I have to do but I sure not sure how I actually write these open sets down? – Mary Mar 07 '18 at 21:01
  • It is defined on the set of prime ideals of a ring (the spectrum of the ring), and a basis of open sets is is $D(a)={\mathfrak p\in\operatorname{Spec R}\mid \mathfrak p\not\ni a}$. As a field has only one prime ideal… – Bernard Mar 07 '18 at 21:04

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