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Stream: deprecated: mathematics

Topic: algebraic sets and affine schemes

John Baez (Apr 18 2023 at 02:24):

Jean-Baptiste Vienney said:

Do you have an idea if there is a characterization of the affine schemes which are the set of zeroes of a polynomial?

This is a slightly strange question because an affine scheme is not a set. Also, people are much more interested in collections of polynomials, not a single polynomial. But maybe you are trying to ask a hard question.

Say we have a field $k$ and a finite collection of polynomials $P_1, \dots, P_m \in k[x_1, \dots, x_n]$. We can do two different things.

First, we can look at the set $X \subseteq k^n$ where all these polynomials vanish. This sort of set is called an algebraic set. Without loss of generality we can take this collection to be finite, say $P_1, \dots, P_m$.

Second, we can form the commutative ring $k[x_1, \dots, x_n]/J$ where $J$ is the ideal generated by $P_1, \dots, P_m$. Since affine schemes correspond to commutative rings, this is an affine scheme.

So, there are two questions we could ask here:

1) which subsets of $k^n$ are algebraic sets?

2) which affine schemes are of the form $k[x_1, \dots, x_n]/J$ where $J$ is the ideal generated by $P_1, \dots, P_m$?

Question 2) has an easy answer: they are just the affine schemes whose corresponding commutative rings are finitely presented $k$-algebras. And interestingly, these are the same as finitely generated $k$-algebras.

John Baez (Apr 18 2023 at 02:27):

From what I said you might guess that algebraic sets are "the same" as finitely generated $k$-algebras. But no: nonisomorphic finitely generated $k$-algebras can give isomorphic algebraic sets. It's good (and not terribly hard) to think of an example.

Jean-Baptiste Vienney (Apr 18 2023 at 03:08):

John Baez said:

Jean-Baptiste Vienney said:

Do you have an idea if there is a characterization of the affine schemes which are the set of zeroes of a polynomial?

This is a slightly strange question because an affine scheme is not a set.

Hmm ok, I thought algebraic sets are affine schemes but is not the same kind of object.

Jean-Baptiste Vienney (Apr 18 2023 at 03:12):

John Baez said:

Jean-Baptiste Vienney said:

Do you have an idea if there is a characterization of the affine schemes which are the set of zeroes of a polynomial?

Also, people are much more interested in collections of polynomials, not a single polynomial.

I would be interested to see an example of an affine set which is not the set of zeros of a single polynomial.

Jean-Baptiste Vienney (Apr 18 2023 at 03:26):

John Baez said:

Jean-Baptiste Vienney said:

Do you have an idea if there is a characterization of the affine schemes which are the set of zeroes of a polynomial?

Question 2) has an easy answer: they are just the affine schemes whose corresponding commutative rings are finitely presented $k$-algebras.

I don't know what is a finitely presented algebra...

Jean-Baptiste Vienney (Apr 18 2023 at 03:29):

John Baez said:

From what I said you might guess that algebraic sets are "the same" as finitely generated $k$-algebras. But no: nonisomorphic finitely generated $k$-algebras can give isomorphic algebraic sets. It's good (and not terribly hard) to think of an example.

I understand that, $k[x]/((x^{2}+1)^{2})$ is not the same algebra than $k[x]/(x^2+1)$ even if they correspond to the same algebraic set. So, this is this idea of "schemes take account of the multiplicities".

Jean-Baptiste Vienney (Apr 18 2023 at 03:41):

John Baez said:

Jean-Baptiste Vienney said:

Do you have an idea if there is a characterization of the affine schemes which are the set of zeroes of a polynomial?

1) which subsets of $k^n$ are algebraic sets?

From what I've found on the internet (here), it seems that they are just all the finite subsets.

Reid Barton (Apr 18 2023 at 05:11):

Let's assume for now that $k$ is algebraically closed.

Reid Barton (Apr 18 2023 at 05:13):

There are certainly infinite algebraic sets. For example, if $n = 2$, then the points whose first coordinate is zero are the set of zeros of the polynomial $P(x,y) = x$, and there are infinitely many of those.

Reid Barton (Apr 18 2023 at 05:14):

And to answer another question, in order to cut out a single point in $k^n$ as an algebraic set, you need (at least) $n$ polynomial equations.

John Baez (Apr 18 2023 at 06:15):

Jean-Baptiste Vienney said:

I would be interested to see an example of an affine set which is not the set of zeros of a single polynomial.

Suppose we're in $\mathbb{C}^3$ and we form the subset

$\{(x,y,z) \in \mathbb{C}^3 : \; x = y = 0 \}$

This is the zero set of a pair of polynomials. Can you can express it as the set of zeros of a single polynomial? Or, conversely, can you prove it's impossible?

John Baez (Apr 18 2023 at 06:17):

Also: how about if I asked the same questions using $\mathbb{R}^3$ instead of $\mathbb{C}^3$?

John Baez (Apr 18 2023 at 06:22):

Jean-Baptiste Vienney said:

John Baez said:

Jean-Baptiste Vienney said:

Do you have an idea if there is a characterization of the affine schemes which are the set of zeroes of a polynomial?

Question 2) has an easy answer: they are just the affine schemes whose corresponding commutative rings are finitely presented $k$-algebras.

I don't know what is a finitely presented algebra...

For any sort of gadget with operations obeying equations we can talk about a "presentation" in terms of generators and relations. Examples include monoids, groups, rings, vector spaces, and what I'm talking about now: associative unital algebras over a field. We say a gadget is "finitely presented" if it has a presentation with finitely many generators and finitely many relations.

John Baez (Apr 18 2023 at 06:26):

However, to be really accurate I should have used the term "finitely presentable".

John Baez (Apr 18 2023 at 06:41):

Jean-Baptiste Vienney said:

John Baez said:

Jean-Baptiste Vienney said:

Do you have an idea if there is a characterization of the affine schemes which are the set of zeroes of a polynomial?

1) which subsets of $k^n$ are algebraic sets?

From what I've found on the internet (here), it seems that they are just all the finite subsets.

No, for example this subset of the plane $\mathbb{R}^2$

$y = x^2$

is an infinite algebraic set. Surely you've studied it: it's the parabola.

John Baez (Apr 18 2023 at 06:45):

It's the zeros of the polynomial function $y - x^2$.

Maybe you were thinking about polynomials in just one variable? An algebraic set is a subset of $k^n$ for a field $k$, where some collection of polynomial functions of the $n$ coordinates vanish.

Jean-Baptiste Vienney (Apr 18 2023 at 19:17):

John Baez said:

Also: how about if I asked the same questions using $\mathbb{R}^3$ instead of $\mathbb{C}^3$?

With $\mathbb{R}^3$, it's the zero set of $x^{2}+y^2$.

Jean-Baptiste Vienney (Apr 18 2023 at 19:19):

John Baez said:

Jean-Baptiste Vienney said:

I would be interested to see an example of an affine set which is not the set of zeros of a single polynomial.

Suppose we're in $\mathbb{C}^3$ and we form the subset

$\{(x,y,z) \in \mathbb{C}^3 : \; x = y = 0 \}$

This is the zero set of a pair of polynomials. Can you can express it as the set of zeros of a single polynomial? Or, conversely, can you prove it's impossible?

With $\mathbb{C}^3$, I don't know, I tried a bit but I didn't find. I think that the answer is: no, you can't express it as the set of zeros of a single polynomial, but I don't know how to prove it. Maybe you need to use the Nullstellensatz? Or maybe it is much simpler?

Jean-Baptiste Vienney (Apr 18 2023 at 19:22):

John Baez said:

It's the zeros of the polynomial function $y - x^2$.

Maybe you were thinking about polynomials in just one variable? An algebraic set is a subset of $k^n$ for a field $k$, where some collection of polynomial functions of the $n$ coordinates vanish.

My intuition has been based on polynomials on one variable for some time yes but now I see that things are much more interesting with several variables.

Jean-Baptiste Vienney (Apr 18 2023 at 19:31):

Jean-Baptiste Vienney said:

John Baez said:

Jean-Baptiste Vienney said:

I would be interested to see an example of an affine set which is not the set of zeros of a single polynomial.

Suppose we're in $\mathbb{C}^3$ and we form the subset

$\{(x,y,z) \in \mathbb{C}^3 : \; x = y = 0 \}$

This is the zero set of a pair of polynomials. Can you can express it as the set of zeros of a single polynomial? Or, conversely, can you prove it's impossible?

With $\mathbb{C}^3$, I don't know, I tried a bit but I didn't find. I think that the answer is: no, you can't express it as the set of zeros of a single polynomial, but I don't know how to prove it. Maybe you need to use the Nullstellensatz? Or maybe it is much simpler?

Hmm, maybe I can use what I know in $\mathbb{R}^3$ to prove it. Let me think a bit more before giving the answer.

Jean-Baptiste Vienney (Apr 18 2023 at 19:36):

No, I don't find, I give up

John Baez (Apr 18 2023 at 20:22):

Okay, the interesting thing about $\mathbb{R}^3$ is that the two equations

$x = 0, \qquad y = 0$

are equivalent to just one equation:

$x^2 + y^2 = 0$

In general, we can use this trick to take any finite number of polynomial equations in $n$ real variables and write them as just one equation!

John Baez (Apr 18 2023 at 20:24):

But this relies on the fact that in $\mathbb{R}$ we have

$a_1 = a_2 = \cdots = 0 \iff a_1^2 + a_2^2 + \cdots + a_n^2 = 0$

John Baez (Apr 18 2023 at 20:26):

In general, we say a field is formally real if

$a_1 = a_2 = \cdots = 0 \iff a_1^2 + a_2^2 + \cdots + a_n^2 = 0$

and examples include $\mathbb{Q}, \mathbb{R}$, the algebraic real numbers, the hyperreals, etc.

John Baez (Apr 18 2023 at 20:27):

Formally real fields have lots of nice properties and many nice equivalent definitions (click the link).

John Baez (Apr 18 2023 at 20:28):

But formally real fields are never algebraically complete, since $x^2 = -1$ obviously can't have a solution in a formally real field.

John Baez (Apr 18 2023 at 20:29):

(Here I'm using "obviously" to mean "think about it - I bet you can see why, since it follows pretty quickly from the definition.)

Jean-Baptiste Vienney (Apr 18 2023 at 20:30):

Okay, everything's good up to there

Jean-Baptiste Vienney (Apr 18 2023 at 20:30):

(and very interesting!)

John Baez (Apr 18 2023 at 20:31):

Okay. It's harder to show you can't describe the subset

$\{(x,y,z) \in \mathbb{C}^3 : \; x = y = 0 \}$

as the zero set of a single polynomial.

John Baez (Apr 18 2023 at 20:32):

Obviously $x^2 + y^2 = 0$ doesn't work, but it's harder to show nothing works.

John Baez (Apr 18 2023 at 20:39):

If I were an actual algebraic geometer I could probably give several proofs, but since I'm just a beginner I'll point you to

• Wikipedia, Dimension theory (algebra).

which outlines some of the technology needed to prove that you can't describe

$\{(x,y,z) \in \mathbb{C}^3 : \; x = y = 0 \}$

with fewer than two polynomial equations.

John Baez (Apr 18 2023 at 20:41):

The rough idea is that this line

$\{(x,y,z) \in \mathbb{C}^3 : \; x = y = 0 \}$

has (complex) dimension 1, and it's sitting a space of complex dimension 3, so it takes 3-1 = 2 equations to describe it, at least locally. But the hard part is defining "dimension" and using it to prove this sort of fact.

John Baez (Apr 18 2023 at 20:41):

I'll admit I've never been very interested in this.

John Baez (Apr 18 2023 at 20:43):

If I were forced to study it, I'm sure I'd enjoy it... I just enjoy other things more.

Jean-Baptiste Vienney (Apr 18 2023 at 20:45):

Hahaha you gave me a difficult homework. But I'm glad that I must try to understand all this stuff now if I want the answer.

John Baez (Apr 18 2023 at 20:49):

The Wikipedia article Dimension theory (algebra) starts out fun and then gradually gets harder.

If you want to study algebraic geometry, it's definitely worthwhile at least understanding these concepts connected to dimension theory: Krull dimension, noetherian ring, and localization.

Jean-Baptiste Vienney (Apr 18 2023 at 20:52):

I will try to :)

John Baez (Apr 18 2023 at 20:58):

I'm sure a lot of people here would enjoy discussing those ideas!

John Baez (Apr 18 2023 at 22:33):

Hey, @Jean-Baptiste Vienney - I've moved a bunch of our conversation about algebraic sets and affine schemes from "practice: our work - John Baez" to here.

John Baez (Apr 18 2023 at 22:34):

That way people can keep talking about it and have it not get mixed with me blabbing about my own work.

John Baez (Apr 18 2023 at 22:36):

I now want to find an efficient proof that there's no one polynomial $P \in \mathbb{C}[x,y,z]$ such that

$P(x,y,z) = 0 \iff (x = 0 \text{ and } y = 0)$

John Baez (Apr 18 2023 at 22:37):

Maybe I should ask Chat-GPT what it thinks. :upside_down:

But if I were pretending to be an algebraic geometer I might say something like this.

John Baez (Apr 18 2023 at 22:38):

It suffices to show that the ideal $I \subset \mathbb{C}[x,y,z]$ generated by $x$ and $y$ cannot be generated by a single element $P \in \mathbb{C}[x,y,z]$.

John Baez (Apr 18 2023 at 22:38):

That is - to use even more jargon! - it suffices to show that $I$ cannot be a [[principal ideal]].

John Baez (Apr 18 2023 at 22:47):

For this we should use Krull's principal ideal theorem which implies that if $I \subset \mathbb{C}[x,y,z]$ (strict inclusion) is a principal ideal, then each minimal prime ideal lying over $I$ has height at most one.

John Baez (Apr 18 2023 at 22:47):

(I'm just guessing this, like Chat-GPT. I don't really know it.)

John Baez (Apr 18 2023 at 22:48):

So, to finish the job we just need to find a minimal prime ideal lying over $I$ with height two or more.

John Baez (Apr 18 2023 at 22:50):

Like the name suggests, a "minimal prime ideal lying over $I$" is a prime ideal that contains $I$ for which there is no smaller prime ideal containing $I$.

John Baez (Apr 18 2023 at 23:22):

Actually I guess $I$ is itself a minimal prime ideal lying over $I$, since $I$ is a prime ideal and $J$ "lying over" $I$ doesn't require that $J$ be strictly bigger than $I$.

Jean-Baptiste Vienney (Apr 18 2023 at 23:25):

John Baez said:

Hey, Jean-Baptiste Vienney - I've moved a bunch of our conversation about algebraic sets and affine schemes from "practice: our work - John Baez" to here.

Good idea!

John Baez (Apr 18 2023 at 23:25):

So, to show $I$ Is not principal, it suffices to show $I$ has height at least two.

Jean-Baptiste Vienney (Apr 18 2023 at 23:25):

John Baez said:

I now want to find an efficient proof that there's no one polynomial $P \in \mathbb{C}[x,y,z]$ such that

$P(x,y,z) = 0 \iff (x = 0 \text{ and } y = 0)$

Okay, let's call this property $(*)$.

John Baez (Apr 18 2023 at 23:27):

Okay, I'll finish showing there's no polynomial with property (*) if I can show my ideal $I$ has height at least two. Let me try for a few more minutes.

Jean-Baptiste Vienney (Apr 18 2023 at 23:27):

John Baez said:

It suffices to show that the ideal $I \subset \mathbb{C}[x,y,z]$ generated by $x$ and $y$ cannot be generated by a single element $P \in \mathbb{C}[x,y,z]$.

Is $I$ constituted by all the polynomials which verify $(*)$?

John Baez (Apr 18 2023 at 23:28):

No, $I$ is what I said it was in the remark you just quoted: the ideal generated by $x$ and $y$. Please let me write for a couple more minutes and see if my proof works; then I'll be happy to discuss stuff.

John Baez (Apr 18 2023 at 23:31):

Okay, I claim $I$ has height two, because we have a chain of prime ideals

$0 \subset \langle x \rangle \subset \langle x, y \rangle = I$

that has length three. (There's an annoying but necessary offset in the numbering here.) Here $\langle x \rangle$ is the ideal generated by just $x$, while $\langle x, y \rangle = I$ is the ideal generated by $x$ and $y$.

John Baez (Apr 18 2023 at 23:32):

Okay, I'm done! I've shown that there's no polynomial $P$ obeying $(\ast)$ by using Krull's principal ideal theorem.

John Baez (Apr 18 2023 at 23:33):

This doesn't mean I understand what's going on: I just knew that this sort of problem is tackled using concepts like "Krull dimension" and "height of a prime ideal", and I bumped into Krull's principal ideal theorem, which looked like a good way to tackle this problem.

John Baez (Apr 18 2023 at 23:34):

But I'm done for now.

Jean-Baptiste Vienney (Apr 18 2023 at 23:34):

John Baez said:

It suffices to show that the ideal $I \subset \mathbb{C}[x,y,z]$ generated by $x$ and $y$ cannot be generated by a single element $P \in \mathbb{C}[x,y,z]$.

I still don't understand that.

John Baez (Apr 18 2023 at 23:36):

Yes, I didn't explain it. It will help if you ask a specific question.

John Baez (Apr 18 2023 at 23:39):

I'll say this: I'm using the contravariant adjunction (also known as a [[Galois connection]]) between

• the poset of algebraic subsets of $k^n$

and

• the poset of ideals of $k[x_1, \dots, x_n]$

which is the subject of the [[Nullstellensatz]]... where $k$ is an algebraically closed field.

Jean-Baptiste Vienney (Apr 18 2023 at 23:39):

Ah ok, thanks!

John Baez (Apr 18 2023 at 23:40):

Notice that $\mathbb{C}$ is algebraically closed while $\mathbb{R}$ was not. I believe that's why the proof I just gave doesn't work for $\mathbb{R}$, where we know there is a polynomial $P$ obeying $(\ast)$.

Reid Barton (Apr 19 2023 at 05:34):

Here's a low-tech, hands-on proof that the single point $(0,0) \in \mathbb{C}^2$ cannot be the zero set of a single polynomial $P(x,y)$.
Suppose it was. Write $P$ as a polynomial in $y$: $P(x,y) = P_0(x) + P_1(x)y + P_2(x)y^2 + \cdots + P_n(x)y^n$.

We know that for fixed $x = a \ne 0$, the polynomial $P(a,y) \in \mathbb{C}[y]$ has no roots. On the other hand, $\mathbb{C}$ is algebraically closed so the only way this can happen is if $P(a,y)$ is a nonzero constant polynomial: i.e. $P_0(a) \ne 0$, while $P_i(a) = 0$ for $i > 0$.

Now consider a fixed $i > 0$. We know that for every $a \ne 0$, we're supposed to have $P_i(a) = 0$. Since $\mathbb{C}$ is infinite, this can happen only if $P_i$ is the zero polynomial.

That means our original polynomial must have been of the form $P(x,y) = P_0(x)$. But then, it is not possible that $(0,0)$ is the only root of $P$. (The points $(0, b)$ for $b \ne 0$ would have to be roots also.)

Jean-Baptiste Vienney (Apr 19 2023 at 05:57):

Thanks, that's a very clear proof!

Jean-Baptiste Vienney (Apr 19 2023 at 05:59):

I will retain the trick of fixing one variable to obtain a one variable polynomial.

Jean-Baptiste Vienney (Apr 19 2023 at 06:01):

And then being able to use the algebraic closure.

John Baez (Apr 19 2023 at 06:20):

Thanks for giving a simpler proof, @Reid Barton!

John Baez (Apr 19 2023 at 06:23):

I was trying to see how to do it using general methods the way I imagine a professional would do it, but of course a real professional would know lots of ways to prove it.

John Baez (Apr 19 2023 at 06:25):

Is your idea related to the Rabinowitsch trick? It looks vaguely similar.

John Baez (Apr 19 2023 at 06:25):

(Your idea is a lot easier for me to understand than the Rabinowitsch trick.)

Jean-Baptiste Vienney (Apr 19 2023 at 06:47):

John Baez said:

I was trying to see how to do it using general methods the way I imagine a professional would do it, but of course a real professional would know lots of ways to prove it.

That was clearly useful, if you know the machinery and like theory, you can find your proof more simple too.

Reid Barton (Apr 19 2023 at 08:02):

Hmm, I'm not sure. In the background, I was thinking of constructible sets and quantifier elimination. But that is almost invisible, because it's so easy to say when a complex polynomial does (or doesn't) have a root.

Jean-Baptiste Vienney (Apr 19 2023 at 17:31):

Can you explain quickly how it is related to constructible sets and quantifier elimination? :)

John Baez (Apr 19 2023 at 18:26):

At least you're not saying "Quick question: how it is related to constructible sets and quantifier elimination?"

John Baez (Apr 19 2023 at 18:26):

I tend to say "Quick question $\implies$ long answer."

Reid Barton (Apr 22 2023 at 10:22):

If we look at the projection of the zero set of a general polynomial $P(x,y)$ to the $x$-axis [still working over an alg. closed field like the complex numbers!] , it either consists of finitely many points, or is the complement of finitely many points. The second case is the generic one.
Was that quick enough? :upside_down:

Jean-Baptiste Vienney (Apr 22 2023 at 18:04):

Thank you, it was maybe even a little bit too quick :) . I said "quickly" because I didn't want to obligate you to explain what is a constructible set and quantifier elimination but in fact I know almost nothing about these subjects and I always find it easier when someone gives me first good explanations or motivation before going into books :grinning_face_with_smiling_eyes:

Jean-Baptiste Vienney (Apr 22 2023 at 18:07):

Well, I have just an intuitive idea: I know that quantifier elimination amounts to prove that a proposition in some theory in first order classical logic is equivalent to one without quantifiers (example: a real univariate polynomial of degree 2 admits a real root if and only if $\Delta \ge 0$) and I guess that constructible sets are sets that you build by iterating practical operations (maybe taking roots of polynomials?) and not using abstract operations like in set theory, for example the general comprehension axiom.

Jean-Baptiste Vienney (Apr 22 2023 at 18:08):

But I don't know how to relate this to your quick answer.