# Definitions of an Algebraic Variety

I often find the following definition of an algebraic variety:

Let $A$ be a commutative ring with unity, $I\subset A$ an ideal, then the algebraic variety associated to $I$ is defined as $V(I):=\{\mathfrak{p}\in \text{Spec(A)}\mid \mathfrak{p}\supset I\}$

In other textbooks, there is also this one:

For an algebraically closed field $K$, $A:=K[x_1,...,x_n]$ and an ideal $I\subset A$, the algebraic variety associated to $I$ is the set $V(I)=\{(a_1,...,a_n)\in K^n\mid p(a_1,..,a_n)=0\,\,\forall p\in I\}$.

Since every maximal ideal of $K[x_1,...,x_n]$ is of the form $(x_1-a_1,...,x_n-a_n)$, the last definition is, morally speaking, the same as $V(I)=\{\mathfrak{m}\in \text{Maxspec}(A)\mid\mathfrak{m}\supset I\}$.

The second one seems more concrete, since maximal ideals are like points, but I don't know what is the motivation for the first one, which adds up more elements to the variety.

What is the need for two different definitions? Why not just stick with one of them? What are the advantages/disadvantages of each?

• Here's one reason. We want a ring homomorphism $\varphi: A \to B$ to induce a map $\varphi^*: \operatorname{Spec}(B) \to \operatorname{Spec}(A)$. The natural choice is $\mathfrak{p} \mapsto \varphi^{-1}(\mathfrak{p})$ and since the inverse image of a prime ideal is prime, this works. However, it would not if we insisted on maximal ideals: the inverse image of a maximal ideal need not be maximal. – André 3000 Apr 27 '17 at 19:16

This is an excellent question and can be explained using the "functor of points" point-of-view. If you know the following statements:

1. The opposite category of affine schemes is equivalent to the category of commutative rings, or commutative $\mathbb{Z}$-algebras: $\textbf{Aff}^{op} \simeq \textbf{CAlg}_\mathbb{Z}$
2. We can use the Yoneda embedding and the equivalence of categories to show that we can define $\text{Spec}(R)$ as $\text{Hom}_{\textbf{CAlg}_\mathbb{Z}}(R,-)$
3. The Hilbert Nulstellensatz, which states that points in $\mathbb{C}^n$ are in bijection with the maximal ideals. You can write this succinctly as $\text{Hom}_{\textbf{CAlg}_\mathbb{Z}}(\mathbb{C}[x_1,\ldots,x_n], \mathbb{C}) \cong \mathbb{C}^n$

your question falls out easily. For example, let $$I=(f_1(x_1,\ldots,x_n),\ldots,f_k(x_1,\ldots,x_n)) \subset \mathbb{Z}[x_1,\ldots,x_n] = R$$ be a reduced ideal (meaning the quotient ring has no nilpotents). Then $$\text{Hom}_{\textbf{CAlg}_\mathbb{Z}}(R/I,A) \cong \{p\in A^n: f_1(p)=\cdots=f_k(p) =0 \}$$ In fact, a variety can be defined as a reduced scheme of finite type over a field (although many people add an integrality assumption when you base change to an algebraically closed field).

Notice that in the second definition we assumed we were working over an algebraically closed field. If you are working over a field that is not algebraically closed then maximal ideals don't all look like $(x-a_1,x-a_2,\dots,x-a_n)$.

Also notice that in the first definition, we don't even need to be in a polynomial ring.

I guess what I am getting at is that you need to first definition to work in a more general setting.

• +1. However, the geometric intuition is easier with the second definition (which chronologically comes before the second one). – Daniel Robert-Nicoud Apr 27 '17 at 19:14