# The homogeneous ideal $I=\left(x_0 x_3-x_1 x_2, x_1^2-x_0 x_2, x_2^2-x_1 x_3\right) \subset k\left[x_0, x_1, x_2, x_3\right]$ must have 3 generators

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?

• Pick two generators, you get the twisted cubic and a line. The last one is "just" to get rid of the extra line. Feb 12, 2023 at 16:20
• 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 $(x,y)$. Feb 12, 2023 at 16:27

## 2 Answers

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.

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 $$0$$.