1
$\begingroup$

This is a dumb question, but it is tricky for me because my understanding of the concept of vacuous truth is still very poor. Also, note that this question is based on problem 5.3.17 in Garrity, Belshoff, Boos, et al, Algebraic Geometry: A Problem Solving Approach.

algebraic set in $\mathbb{P}^n(k)$: a set of the form $\{p \in \mathbb{P}^n(k): f(p) = 0, \forall f \in S \}$ where $S$ is a set of homogeneous polynomials in $k[x_0, \dots, x_n]$.
ideal generated by an algebraic set in $\mathbb{P}^n(k)$:
given an algebraic set in $\mathbb{P}^n(k)$, $X$, the ideal generated by $X$, denoted $I(X)$, is the ideal generated by the set: $\{f \in k[x_0, \dots,x_n]: f\text{ homogeneous}, f(p)=0\,\, \forall p \in X \}$.

Question: What is $I(\emptyset)$, the ideal generated by the empty set?

I can't decide whether or not it is all of $k[x_0, \dots, x_n]$, or just the irrelevant ideal, $\langle x_0, \dots, x_n \rangle$.

Attempt: Attempting to apply the definition directly, I get: $$I(\emptyset):= \langle\{ f \in k[x_0, \dots, x_n]: f \text{ is homogeneous, }f(p)=0\,\forall p \in \emptyset \}\rangle $$ It's the last part, the $\forall p \in \emptyset$, which trips me up, because it requires me to use vacuous truth, since no point $p \in \mathbb{P}^n(k)$ is in the empty set.

So is the generating set all $f \in k[x_0, \dots, x_n]$, regardless of whether they are homogeneous or ever equal $0$, because of vacuous truth? Is it just all homogeneous polynomials? In both of these cases, $1$ is a member of the generating set, so we get that $I(\emptyset)=k[x_0,\dots,x_n]$.

But if the generating set is just homogeneous polynomials which equal $0$ at at least one point, i.e. zero and all non-constant homogeneous polynomials (since $k$ is algebraically closed), then we conclude that $I(\emptyset) = \langle x_0, \dots, x_n \rangle$.

Here's another argument for $I(\emptyset) = \langle x_0, \dots, x_n \rangle$. The zero locus of $\langle x_0, \dots, x_n \rangle$, denoted $V(\langle x_0, \dots, x_n \rangle)$ is $\emptyset$. By the (homogeneous) Nullstellensatz, we have that $$I(\emptyset) = I(V(\langle x_0, \dots, x_n \rangle)) = \operatorname{Rad}(\langle x_0, \dots, x_n\rangle)$$ Since $\langle x_0, \dots, x_n \rangle$ is maximal, it is also prime, and therefore also radical, i.e. equal to its own radical, so we conclude that in fact $I(\emptyset) = \langle x_0, \dots, x_n \rangle$.

I am more convinced of the second argument, because I actually understand it fully. But is it actually correct? And either way, how can one modify the argument from the definition using vacuous truth to get the same result?

$\endgroup$
  • 1
    $\begingroup$ The constant function $1$ vanishes on $\emptyset$ and $1 \notin \langle x_0, \dots, x_n \rangle $ so I would say that your first argument is correct. I think also that if $ Z(I) = X \Rightarrow \sqrt I = I(X)$ probably only works if $X \neq \emptyset$, as your first argument shows. $\endgroup$ – user171326 Jun 21 '17 at 10:40
  • 1
    $\begingroup$ By looking at Hartshorne's AG book, it seems that $I(\emptyset) = k[x_0,\ldots,x_n]$. Also, there is a 1-1 correspondence between algebraic sets in $\mathbb{P}^n$ (over an algebraically closed set $k$) and homogeneous radical ideals not equal to the irrelevant maximal ideal of $k[x_0,\ldots,x_n]$. If my understanding is correct, this is because for both the whole ring, and for the irrelevant ideal, the corresponding algebraic set is $\emptyset$. So you rule out the irrelevant ideal in order to get a 1-1 correspondence given by $Z$ and $I$. $\endgroup$ – Malkoun Jun 21 '17 at 11:08
  • $\begingroup$ @Malkoun So it seems to me that you're saying that, by the hypotheses of the theorem, the Nullstellensatz doesn't apply? Is that correct? Because we need the algebraic set in question to be non-empty to satisfy the hypotheses? Where I get really confused is because otherwise the irrelevant ideal is maximal, so seemingly it should apply. EDIT: yes it seems to be the case that the hypotheses for the "projective Nullstellensatz" are different -- I wish my book had mentioned this math.stackexchange.com/questions/119434/… what I was trying to prove was (i) $\endgroup$ – Chill2Macht Jun 21 '17 at 14:17
1
$\begingroup$

One thing that can be a bit confusing here is that the $\forall p$ is written after the $f(p)$ which makes it (IMO) harder to parse.

Let us write it out a little more $f(p)=0\; \forall p\in \emptyset$ is really just $\forall p\in \emptyset \; f(p)=0$ which is the same as $\forall p \;(p\in \emptyset)\implies (f(p)=0)$ but this is true for any $f$ since the left hand side of the implication is false for all $p$ (no $p$ is in the empty set).

Now let's look at the definition of the ideal $I(\emptyset)=\{f \in k[x_0, \dots,x_n]: f\text{ homogeneous}, f(p)=0\,\, \forall p \in \emptyset \}$ that's just $\{f \in k[x_0, \dots,x_n]: f\text{ homogeneous}, \forall p \;(p\in \emptyset)\implies (f(p)=0)\}$ which is really $\{f \in k[x_0, \dots,x_n]: (f\text{ homogeneous} \wedge \forall p \;(p\in \emptyset)\implies (f(p)=0))\}$ but that's $\{f \in k[x_0, \dots,x_n]: (f\text{ homogeneous} \wedge \text{true})\}$ which is then $\{f \in k[x_0, \dots,x_n]: f\text{ homogeneous}\}$.

$\endgroup$
  • $\begingroup$ This is a really good answer explaining everything both clearly and in detail -- I appreciate it. Do you also know a reference for the Projective Nullstellensatz as stated in this question?math.stackexchange.com/questions/119434/… It states that the algebraic set has to be empty for the theorem to apply, which resolves my confusion about that, assuming that the hypothesis is true. $\endgroup$ – Chill2Macht Jun 21 '17 at 14:23
  • 1
    $\begingroup$ @Chill2Macht, you can check out the first Chapter of Hartshorne's Algebraic Geometry book, but it does not have many details about that. I mean the statements are there, though some are in the exercises, but you have to think and fill in the details. I have to admit that Hartshorne's style is very precise, and reminds me a lot of the French style in Mathematics. So he would definitely be careful about the empty set, as I feel all good Mathematicians are. $\endgroup$ – Malkoun Jun 21 '17 at 21:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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