Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $K$ be an algebraically closed field. Then let $A = K[x_1, ... , x_n]$.

If $Y \subseteq \mathbb A^n$, then the ideal of $Y$ is to defined to be $I(Y) = \{f \in A | f(P) = 0 \ \forall P \in Y \}$. What is $I(\emptyset)$?

For $T \subseteq A$, define $Z(Y) = \{P \in \mathbb A^n | f(P) = 0 \ \forall f \in A\}$. Am I correct in thinking that $Z(A) = \emptyset$?


share|cite|improve this question
up vote 2 down vote accepted

You are correct that $Z(A)=0$. Correspondingly, $I(\emptyset)=A$, as all elements of $A$ vanish on every point in $\emptyset$ (trivially, as there are none). In general, for any ideal $J\subseteq A$ we have $I(Z(J))=\sqrt{J}$, where $\sqrt{J}$ denotes the radical of $J$.

share|cite|improve this answer
Note that this last statement is one formulation of the weak Nullstellensatz. – Alex Becker Jan 20 '12 at 21:01
Thank you. I'm aware of the statement; I was checking the details of an example that showed that statement doesn't necessarily hold if $K$ wasn't algebraically closed. – Matt Jan 20 '12 at 21:05

By definition, $$I(\varnothing) = \{f\in A\mid f(P)=0\forall P\in\varnothing\} = A,$$ since every element of $A$ satisfies the condition by vacuity.

Since $P\in Z(A)$ requires $1=\mathbf{1}(P)=0$ (where $\mathbf{1}$ is the constant polynomial $1$), and this is impossible, this proves $Z(A)=\varnothing$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.