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Let $\pi:X\rightarrow \mathbb A^{n}$ be a finite surjective morphism ,where $X$ is irreducible affine variety and dim$X=n$ and $K$ be a proper closed subset of $X$.Can we conclude that dimension of $\pi(K)$ is strictly less than $n$ ?

I think that the dimension should be less than n,since $K$ is proper closed in $X$ so dimension has to be strictly less than n and $\pi(K)$ has to be a proper in $\mathbb A^{n}$ but not getting convinced .

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Is your base-scheme Jacobson ? –  Damien L Feb 4 '13 at 16:32
    
I am sorry that i am not aware of base-scheme Jacobson?But I think now question is clear to you. –  Shraddha Srivastava Feb 4 '13 at 16:38
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up vote 1 down vote accepted

As $\pi$ is finite, $K$ and $\pi(K)$ have the same dimension.

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sorry,I did not get your idea completely.I can show that if $\pi $ is finite surjective morphism from irreducible to irreducible ,then dim$X=$dim$Y$.But why should restriction of finite morphism finite? –  Shraddha Srivastava Feb 4 '13 at 16:54
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Sorry I got your idea as restriction of a finite morphism on a closed set is finite. –  Shraddha Srivastava Feb 4 '13 at 17:00
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