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Let $Y$ be the curve given parametrically by $x=t^3, y = t^4, z = t^5$. Show that $I(Y)$ is a prime ideal of height $2$ in $k[x,y,z]$ which cannot be generated by two elements.

Obviously $(x^4-y^3,x^5-z^3) \subseteq I(Y)$, but these ideals can't be equal because the one on the right is generated by two elements. So, what is the prime $I(Y)$? And how can i show its height is precisely 2?

José
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Well, implicitization is given by taking the ideal $I=\langle x-t^3,y-t^4,z-t^5\rangle$ w.t.r. the lex ordering $t>x>y>z$ and eliminating $t$ by considering the first elimination ideal $I_1 = I\cap k[x,y,z]$, which is given by the elimination theorem via its Gröbner basis $G_1 = G\cap k[x,y,z]$ where $G$ is a Gröbner basis of $I$. I've performed the calculation in Singular giving $G$ as follows:

$y^5-z^4,xz-y^2,xy^3-z^3, x^2y-z^2, x^3-yz, tz-x^2,ty-z, tx-y, t^3-x$.

user26857
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Wuestenfux
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    So does it mean that to generate the ideal we need all the polynomials on that list that don't have a $t$? – José Aug 18 '19 at 18:19
  • @José Right. But notice that some generators are superfluous. – user26857 Aug 19 '19 at 16:04
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    It's a result by Herzog that an ideal like this in $k[x,y,z]$ is generated by either 2 or 3 elements. So the exercise implies it must have 3 generators – user347489 Aug 20 '19 at 00:44