Every projective algebraic set can be written as the zero set of finitely many homogeneous polynomials of the same degree.

Question

Definition: Let $I \subset k[x_0,\ldots,x_n]$ be a homogeneous ideal (or a set of homogeneous polynomials). The set $Z(I) := \{(a_0 : \cdots : a_n)\in P^n ; f(a_0,\ldots,a_n) = 0 \ \ \forall f \in I\}$

Subsets of $P^n$ that are of the form $Z(I)$ are called algebraic sets. If $X \subseteq P_n$ is any subset, we call $I(X)$ the ideal generated by $\{f ∈ k[x_0,\ldots,x_n]: f \ \text{ homogeneous},\ f(a_0,\ldots,a_n) = 0 \forall (a_0 : \cdots : a_n) \in X\}$; by definition this is a homogeneous ideal.

A remark that is sometimes useful is that every projective algebraic set can be written as the zero set of finitely many homogeneous polynomials of the same degree.

While studying projective varieties, I read the above remark in the box. I dont understand this remark, please how to show this remark clearly. I want to learn its proof. Thank you for helping.

Answer 1

If you have a homogeneous polynomial $f \in k[x_0,\ldots,x_n]$ then the zero set of $f$ in projective space is the same as the common zero set of $x_0f, \ldots x_n f$. Iterating this, we can write the zero set of $f$ as the zero set of finitely many homogeneous polynomials of degree $d$ for arbitrary $d \geq \textrm{deg}(f)$.

A general closed set is of the common zero set of homogeneous polynomials $g_1, \ldots, g_r$ and we can apply this to every $g_i$.

Answer 2

Another way of doing this is to take powers: $Z(f) = Z(f^n)$, so if you're willing to go up to the gcd of the degrees of a finite generating set then this can be done.

It's important to note that we're not saying that each $I(X)$ is generated in a single degree.