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Let $R = K[X_1, …, X_n]$ and $I$ be an ideal of R.

Definition 1: the vanishing set of $I$ is the set of prime ideals of $R$ containing $I$: $V(I) = \{P \in \operatorname{Spec}(R) : P \supseteq I\}$

Definition 2: the vanishing set of $I$ is the set of all points in $K^n$ where every polynomial in $I$ vanishes: $V(I) = \{(k_1, \ldots, k_n) \in K^n : f(k_1, \ldots, k_n) = 0 \ \forall f \in I\}$

Are these definitions both correct and equal? If so, how? It is especially confusing me that the elements of V(S) in definition 1 are ideals while in definition 2, the elements are in $K^n$.

KReiser
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lucid_mathing
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  • The first definition is needed to define a scheme, the second is needed to define a variety. Schemes are a generalization of varieties. See here. – Soumik Mukherjee Mar 07 '25 at 06:19
  • @lucid_mathing: If $K$ is algebraically closed, it follows a maximal ideal $m \subseteq A:=K[x_1,..,x_n]$ is on the form $m:=(x_1-a_1,..,x_n-a_n)$ for $(a_1,..,a_n) \in K^n$. Hence there is a 1-1 correspondence between maximal ideals in$A$ and points $a \in K^n$. Definition 1 speaks about a set of prime ideals, definition 2 speaks about maximal ideals - hence these two sets are different, – hm2020 Mar 07 '25 at 09:38

1 Answers1

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You're right that these are not literally equal - they don't live in the same place, so they can't be equal as sets. But they are both correct, and they are morally the same construction seen from different perspectives: the first construction is the answer to "what is the vanishing set of an ideal" from a more schemey viewpoint, like the third or fourth definition at What is an algebraic variety? while the second construction is the answer from a more variety-theoretic viewpoint, like the first two definitions at that link.

To see how they compare, consider the map $\varphi:K^n \to \operatorname{Spec} K[x_1,\cdots,x_n]$ by $(k_1,\cdots,k_n) \mapsto (x_1-k_1,\cdots,x_n-k_n)$. For a fixed ideal $I\subset R$, I claim that $(k_1,\cdots,x_n)\in V(I)\subset K^n$ (with $V(I)$ used in the second sense) iff $(x_1-k_1,\cdots,x_n-k_n)\in V(I)\subset\operatorname{Spec} R$ (with $V(I)$ used in the first sense). Indeed, $f(k_1,\cdots,k_n)=0$ for a polynomial $f$ iff $f\in (x_1-k_1,\cdots,x_n-k_n)$: by induction on the degree of $f$, we can show that there is a polynomial $p$ such that $$f(x_1,\cdots,x_n) = p(x_1-k_1,\cdots,x_n-k_n) + f(k_1,\cdots,k_n),$$ and the RHS is in $(x_1-k_1,\cdots,x_n-k_n)$ iff $f(k_1,\cdots,k_n) = 0$. So every element of $I$ evaluates to $0$ on $(k_1,\cdots,k_n)$ iff every element of $I$ is in $(x_1-k_1,\cdots,x_n-k_n)$, and we've proven the claim.

Under the identification of $K^n$ with a subset of $\operatorname{Spec} K[x_1,\cdots,x_n]$, we have that $V(I)\subset K^n$ is naturally a subset of $V(I)\subset\operatorname{Spec} R$. But there are many places where this inclusion after identification is proper:

  • if $K$ is not algebraically closed, there can be other points in $V(I)\subset\operatorname{Spec} R$, like if $K=\Bbb R$ and $I=(x_1^2+x_2^2)$, the prime ideal $(x_1^2+1,x_2-1)$ is in $V(I)\subset\operatorname{Spec} R$ but not in the image of $\varphi$,
  • if $V(I)$ is positive-dimensional, there are other points in $V(I)\subset\operatorname{Spec} R$, such as if $n=3$ and $I=(x_1)$, then the prime ideal $(x_1,x_2)$ is in $V(I)$ but it is not in the image of $\varphi$ because it is not maximal.

So even after identification, the second definition (usually) only gets you a subset of the first definition. But the extra stuff we pick up is quite useful - being able to do algebraic geometry in a more intrinsic way over a non-algebraically-closed field (or even a general ring) as in the first bullet, and being able to keep track of generic points as in the second bullet are both big advances that make a lot of further algebraic geometry possible.

KReiser
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