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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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    Ok if $y=0$, then $x=0$ from $y^4=x^5$ and then $xw=0$. If $y \neq 0$ and $z=0$, then $yw=z^2$ implies $w=0$, then $xw=0$. So we don't need $xw=yz$. But remark that we need $xw-yz$ when we want to compute $I(X)$.2012-12-08

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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 $Z$.