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Is it true that a morphism of affine algebraic varieties is continuous in Zariski topology? How should I proceed? thank you

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    Hint: a function is continuous if and only if the pre-image of a closed set under the function is closed. So you should try to show that the pre-image of an algebraic variety is itself an algebraic variety, by finding an ideal that it is the variety of.2012-04-05
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    thank you for the reply.But need an explicit answer.it would be nice if you elaborate by an example.2012-04-05
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    That might be nice, but it would be *best* if you tried to use Matt's hint, and then told us where you get stuck. Why do you *need* an explicit answer?2012-04-05
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    @MimMim: have you tried thinking of one for yourself?2012-04-05
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    What's your definition of a morphism?2012-04-06
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    Dear Sir, My Definition of morphism is a polynomial map: $$\mathbb{A}^n\rightarrow \mathbb{A}^m$$ $$x\mapsto (F_1(x),\dots,F_m(x))$$2012-04-06

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