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Let $\mathbb A^n$ be the n-dimensional affine space over a field K (algebraically closed if that makes it easier), so $\mathbb A^n= \text{Spec }K[x_1,...,x_n]$, and $\mathbb A^{n-1}$ the (n-1)-dimensional analog. The inclusion $K[x_1,...,x_{n-1}]\to K[x_1,...,x_n]$ induces a map $\mathbb A^n \to\mathbb A^{n-1}$. Is this a projective map?

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closed as off topic by Will Sawin, Qing Liu, Dan Petersen, Kevin Ventullo, Andy Putman Nov 18 '12 at 0:17

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For it to be projective (in particular proper), the image of a closed set should be a closed set. Look at the case $n = 2$. Then the map $\mathbb{A}^2 \rightarrow \mathbb{A}^1$ with coordinates $x,y$ and $x$ resp. The map on rings you wrote corresponds to the projection $(x,y) \mapsto x$. Now, $\mathbb{A}^2$ has the closed set $xy = 1$, the image of which is all of $\mathbb{A}^1$ except the origin. Since this isn't a closed set, the map isn't proper. Similarly, you can rewrite the standard counterexample for larger $n$. (Btw, maybe you want to tag this "algebraic-geometry"). –  LMN Nov 17 '12 at 21:45
    
that was very useful, thanks a lot –  kwkwkw Nov 17 '12 at 22:02
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maybe you meant to ask if this were an affine map? (it makes less sense to ask about projective maps when the spaces are all affine.) –  roy smith Nov 18 '12 at 6:48
    
It makes sense, it's just the same as talking about finite maps. –  Will Sawin Nov 18 '12 at 7:24
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