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I need to prove that the ideal $$ I = (xz -y^2, x^2t^2 -yz^3, x^2yt^2 -xz^4) \subset R = \mathbb{K}[x,y,z,t]$$ is generated by a $R$-regular sequence. How can I do it?

I don't know if this can help, but $I = J \cap (x,y)$, where $J$ is generated by the $2\times 2$ minors of $$\begin{pmatrix}x & y & z^3 \\ y & z & xt^2 \end{pmatrix}.$$

In addition, I've calculated the Hilbert series of $R/I$ and I get $$\frac{1 + 2x + 2x^2 + 2x^3 + x^4}{(1-x)^2}$$ so $\dim (R/I) = 2$.

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The last generator is superfluous, as

$$(x^2yt^2 - xz^4) - y(x^2t^2 - yz^3) + z^3(xz - y^2) = 0$$

Thus $I = (xz-y^2, x^2t^2-yz^3)$ is generated by a regular sequence (indeed, $xz-y^2, x^2t^2 - yz^3$ are both prime elements).

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  • $\begingroup$ Simpler than I've thought! Actually I've taken the problem (too) seriously, and wonder how could one prove that a height 2 ideal is generated by a regular sequence, and the answer was $\beta_2(R/I)=1$, that is, it's enough to show that $R/I$ is Gorenstein. $\endgroup$ – user26857 Aug 27 '14 at 20:08
  • $\begingroup$ Yes, I saw your comment before that was precisely to that effect (since Gorenstein = complete intersection in codimension 2). The relation wasn't obvious to me on first sight either, but it just happened to pop out after a little work $\endgroup$ – zcn Aug 27 '14 at 20:11

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