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Consider the homogeneous ideal $$ I=\left(x_0 x_3-x_1 x_2, x_1^2-x_0 x_2, x_2^2-x_1 x_3\right) \subset k\left[x_0, x_1, x_2, x_3\right] $$ and let $X:=V(I) \subset \mathbb{P}^3$.

One can show that this variety (twisted cubic) is isomorphic to $P^1$ and thus have dimension 1. So I was told in a lecture that this variety cannot be generated by two elements, thus this is an example of a 1 dimension variety in a 3-dimensional ambient space that cannot be cut out by two (homogeneous) polynomials but the explanation was unclear so I would like to ask why this is the case?

Also a quick question on homogeneous ideals. So homogeneous ideals are defined to be ideals where there exists a set of homogeneous generators, but it is still possible for a set of nonhomogeneous polynomials to generate a homogeneous ideal correct?

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  • $\begingroup$ Pick two generators, you get the twisted cubic and a line. The last one is "just" to get rid of the extra line. $\endgroup$ Feb 12, 2023 at 16:20
  • $\begingroup$ I would assume the main question has been asked and answered on this site before. Let me answer the second, unrelated question in this comment. The answer is yes, e.g. $x-y^2$ and $y+y^2-x$ are two non-homogeneous polynomials that generate the homogeneous ideal $(x,y)$. $\endgroup$
    – Thorgott
    Feb 12, 2023 at 16:27

2 Answers 2

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For a homogeneous ideal $I\subset k[x_0,\dots,x_n]$, we have a decomposition $I=\oplus_l I_l,\ I_l=I\cap k[x_0,\dots,x_n]_l$, the degree-decomposition of $I$. In your case, your ideal $I$ has $I_0=I_1=(0)$ and $\dim_k I_2=3$, the last space spanned over $k$ by the three generators. This shows that $I$ cannot be generated by two elements, since two elements cannot span the 3-dimensional space $I_2$.

For the second question, the inhomogeneous elements $x_0+x_1^2,x_0-x_1^2$ generates a homogeneous ideal $(x_0,x_1^2)$ (when char $k\neq 2$). So, a homogeneous ideal can be generated by inhomogeneous elements.

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I'll use exercise I.7.2 in Hartshorne. The arithmetic genus of a complete intersection of a degree $a$ and degree $b$ hypersurface in $\mathbb{P}^3_k$ is $\frac{1}{2} ab (a+b-4) + 1$.

Since the twisted cubic is not on a plane, neither $a$ nor $b$ can be $1$. They must be at least $2$.

Therefore, the arithmetic genus is at least $1$.

But $\mathbb{P}^1_k$ has genus $0$.

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