Why is the zero locus of a polynomial ideal a variety?

Question

I'm sorry if this is a stupid question, but some books define a variety of an ideal as it's zero locus, what means that it's irreducible. I'm just new to algebra and wasn't able to prove it... Can someone help me, please?

Answer 1

I assume your definition of a variety is the following:

Definition. Let $k$ be a field, $d$ be an integer greater than $1$ and $V$ be a subset of $k^d$. $V$ is a variety if and only if there exists $f_1,\cdots,f_n$ in $k[X_1,\cdots,X_d]$ such that: $$V=\{x\in k^d\textrm{ s.t. }\forall i\in\{1,\cdots, n\},f_i(x)=0.\}.$$

Let $I$ be an ideal of $K[X_1,\cdots,X_d]$ and let $Z(I):=\{x\in k^d\textrm{ s.t. }\forall f\in I,f(x)=0\}$, you want to show that $Z(I)$ is a variety. Your problem is that $I$ is not a finite set. Can you show that:

• $I$ is generated by a finite number of polynomials, let say $f_1,\cdots,f_n$.

• $Z(I)=\{x\in k^d\textrm{ s.t. }\forall i\in\{1,\cdots, n\},f_i(x)=0\}.$