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Let $X=\{(t^4,t^5,t^6,t^7)\in k^4, t \in k\}$, $Y=\{(t,t^2,t^3,t^4)\in k^4, t \in k\}$

Obviously $Y=V(y-x^2,z-x^3,w-x^4)$, and also $X$ seems to be $X=V(y^4-x^5,xz-y^2,yw-z^2)$.

  1. Is $X$ right? I'm quite not sure, since there can be similar things $z^4-x^6, w^4-x^7$ etc..

  2. What's the difference between $X$ and $Y$?

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Is $k$ a field? thanks. –  B. S. Dec 8 '12 at 9:12
    
On $X$ we also have $xw=yz$ and $zw=x^2 y$. –  Martin Brandenburg Dec 8 '12 at 9:18
    
@BabakSorouh yes. –  Gobi Dec 8 '12 at 9:20
    
@MartinBrandenburg Is it needed? $xz=y^2$, $yw=z^2$ multiplying both sides then $xyzw=y^2z^2$, but can't I cancel $yz$? –  Gobi Dec 8 '12 at 9:23
    
Well you can only cancel when $y,z \neq 0$. How do you derive $xw=yz$ when $yz=0$? –  Martin Brandenburg Dec 8 '12 at 9:25
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1 Answer 1

The parametrization $t\to (t, t^2, t^3, t^4)$ is a closed immersion of the affine line $\mathbb A^1$ to $\mathbb A^4$ (it is the restriction of the $4$-uple embedding of $\mathbb P^1$ to $\mathbb P^4$). So we have a morphism $Y\to Z$ which is clearly an isomorphism outside from $(0,0,0,0)$ because $t$ is invertible there.

At $0$, the second parametrization $f : \mathbb A^1\to \mathbb A^4$ is not a closed immersion. The corresponding ring homomorphism is
$$k[x,y, z, w] \to k[t], \quad x\mapsto t^4, y\mapsto t^5, z\mapsto t^6, w\mapsto t^7.$$ When localization at $(0,0,0,0)$ the image of the maximal ideal $(x,y,z,w)$ doesn't generate the maximal ideal $(t)$ of $k[t]_{tk[t]}$.

The morphism $Y\to Z$ is birational but not isomorphic. It is in fact the normalization of $Z$.

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