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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.2012-12-08
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    On $X$ we also have $xw=yz$ and $zw=x^2 y$.2012-12-08
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    @BabakSorouh yes.2012-12-08
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    @MartinBrandenburg Is it needed? $xz=y^2$, $yw=z^2$ multiplying both sides then $xyzw=y^2z^2$, but can't I cancel $yz$?2012-12-08
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    Well you can only cancel when $y,z \neq 0$. How do you derive $xw=yz$ when $yz=0$?2012-12-08
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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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