Theorem. (Ax-Grothendieck)An injective polynomial map is surjective.

Stated.

The Ax-Grothendieck theorem feels sensible to everyone who took Complex Analysis. However, they tend to find the proof, and the role of model theory, a bit mysterious. Here, I attempt to explain Ax’s proof to those who speak the language of Abstract Algebra, but not the language of Model Theory. I’ll try to cordon off the logical ingredient into a single, well-defined step. We’ll see that anyone could prove the analogue of this theorem over algebraically closed fields of characteristic . Transferring the result to characteristic zero may feel hopeless: one does not have homomorphisms between fields of different characteristic. Fortunately, Model Theory steps in to bridge this gap.

First, let’s recall the following facts:

Theorem. (Fundamental, of Algebra)Every non-constant polynomial over has a root in .

Stated.

We call fields that satisfy (their analogue of) the Fundamental Theorem of Algebra *algebraically closed fields*. Interestingly, the Fundamental Theorem of Algebra implies the case of the Ax-Grothendieck theorem.

Exercise.If a polynomial function is injective then it is surjective.

Solution.

Since is injective, it is not constant. Take an arbitrary . Proving that has a preimage w.r.t. suffices to prove that is surjective. Let’s use the Fundamental Theorem of Algebra to construct this preimage: the polynomial has a root $a \in \mathbb{C}$, giving , or equivalently .

Solved.

Another important fact about the field of complex numbers:

Theorem. (Categoricity)If the algebraically closed field has characteristic zero and is bijective with , then it is isomorphic to .

Proof.

Exercise. Hint: how would you prove that two vector spaces with the same dimension are isomorphic? Similary, choose transcendence bases for and over .

Qed.

Using a clever trick, we can obtain the analogue of the Ax-Grothendieck theorem for algebraically closed fields of positive characteristic.

Theorem.Let denote the algebraic closure of . An injective polynomial map is surjective.

Proof.

Take any such polynomial map and any element . We need to show that has a preimage w.r.t. . But that’s easy!

Recall that the algebraic closure can be constructed by iteratively adjoining roots of polynomials to : thus, it can be written as the union of a chain of finite fields . One of these finite fields in the chain, say , must contain both and the coefficients of .

So our polynomial maps this finite field into itself. But injective functions that map finite sets to themselves are surjective on that set (consider this an exercise or see here)! In particular, has a preimage with respect to $f$. However, was arbitrary so the function must be surjective.

Qed.

Unfortunately, this says nothing about the case of the complex numbers. Or does it? One one hand, we feel that transfer of facts between different characteristics should be possible. On the other hand, we know that the theories of finite and infinite characteristic do not interact via homomorphisms. Logic sidesteps this problem by making the two theories interact at the level of proofs.

Here, we allude to the logical ingredients: first, compactness (related, but not identical to the topological notion). In our case, compactness asserts that if you have a countable set of laws describing the behavior of some hypothetical operations, and you can exhibit examples of operations that satisfy any finite subset of the laws, then you can also exhibit an example of operations that satisfy every law in the countable set. If you find this unintuitive, you should think of it as a logic analogue of the Erdõs-de Bruijn theorem on graph coloring. Second, the Löwenheim-Skolem theorem, which shows that you can exhibit such operations on a set bijective with .

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Theorem. (Ax-Grothendieck)An injective polynomial map is surjective.

Proof.

We will construct an “ideal” algebraically closed field of characteristic zero that satisfies the Ax-Grothendieck theorem and is bijective with . By the Categoricity Theorem above, will turn out to be isomorphic to , i.e. too satisfies the Ax-Grothendieck theorem.

Consider a structure with two operations, additions and multiplication, that satisfy the laws of a field (finitely many), the laws asserting that the characteristic is zero (, countably many), the laws asserting algebraic closure (, one for each degree, countably many), and the laws saying that every polynomial map over the field that is injective is surjective (one for each degree, countably many).Any finite set of these laws contains only finitely many (say $p$) instances of the laws asserting that the characteristic is zero, so the addition and multiplication operations of satisfy all laws in the finite subset. By the compactness and Löwenheim-Skolem theorems, there is a set bijective with that has addition and multiplication operations satisfying all of the laws! This means that is an algebraically closed field of characteristic zero that satisfies the Ax-Grothendieck theorem. By the Categoricity Theorem, it is isomorphic to .

Qed.

1. Basic questions

1.1. What is the purpose of this FAQ?

1.2. What’s the intuition behind Smooth Infinitesimal Analysis?

1.3. Making infinitesimals nilsquare seems arbitrary. Is there a justification for this?

**2. Logic Matters**

2.1. I find Intuitionistic Logic… unintuitive. What kind of reasoning am I allowed to use?

2.2. I’ve heard in Algebra that fields don’t have nilpotent elements. Yet my SIA book calls the real line a field. Isn’t that a contradiction?**

2.3. Can you show me a non-zero nilsquare infinitesimal?

** 3. Microcancellation **

3.1. How do I prove microcancellation?

3.2. I believe I’ve found a contradiction using microcancellation. What do I do?

**4. Infinitesimal Calculus**

4.1. I don’t know how to prove the chain rule. Can you help?

4.2. How do I prove that the exponential function is its own derivative?

4.3. What is the deal with the square root function?

4.4. Infinitesimals don’t have reciprocals, so how does the quotient rule work?

**5. Counterexamples to Classical Theorems**

5.1. Can I prove the Extreme Value Theorem in SIA?

5.2. Can I prove the Intermediate Value Theorem in SIA?

5.3. Can I prove the non-infinitesimal form of Taylor’s theorem in SIA?

**6. Technical Stuff**

6.1. Where can I find all the axioms of Smooth Infinitesimal Analysis?

6.2. Is Smooth Infinitesimal Analysis constructive?

Smooth Infinitesimal Analysis (abbreviated SIA) is a variant of Non-classical Real Analysis which uses nilpotent infinitesimal quantities to deal with concepts such as continuity and differentiability. The subject had its origins in Alexander Grothendieck’s work on Algebraic Geometry, as interpreted by F. William Lawvere, Anders Kock, John L. Bell and others.

The orginal formulation of Smooth Infinitesimal Analysis, and the wider Synthetic Differential Geomerty, was topos-theoretic. Later on, SIA was given a readily accessible axiomatic presentation. Since Kock’s textbook, “Synthetic Differential Geometry” became freely available, there’s been much interest in Smooth Infinitesimal Analysis. Unfortunately, SIA has a fair amount of initial hurdles: the necessary use of Intuitionistic Logic and the failure of many classical theorems of Analysis; worst of all, the real payoff (definitions that are easier to state, theorems that are easier to prove) comes much later, when one gets to the treatment of Differential Geometry.

This FAQ is not a self-contained introduction to SIA: its purpose is to get independent learners (readers of Kock’s book) past the aforementioned initial hurdles by answering the questions that often come up. The questions in this FAQ were collected in part from fellow students, and in part from Reddit, Stack Exchange and various online mailing lists. If you have a question that you’d like answered, feel free to write a comment.

“Under the microscope”, the graph of a differentiable function is *approximately* a straight line. The idea is to extend the real numbers with infinitesimal values. Viewed at infinitesimal scales, the graph of every smooth function becomes an *exact* (i.e. not merely approximate) straight line.

In Smooth Infinitesimal Analysis, the set of real numbers is replaced with the *real line* . The real line contains a set of infinitesimal numbers . The fact that every function is linear is formalized by the Kock-Lawvere axiom:

**Axiom.** (Kock-Lawvere). Every function has a unique slope such that for all nilsquare infinitesimals , .

**Stated.**

This unique slope is called the *derivative at zero* of the function . Generally, we can define the *derivative at point * as the derivative at zero of the translated function .

Starting from here, infinitesimals allow us to do differential and integral calculus without invoking limits. Things get even better once we get to differential geometry: notions such as tangent vectors and jets can be given single-sentence definitions using infinitesimals.

The end goal is not nilsquare infinitesimals: the end goal is to define infinitesimals that are so small that every smooth function looks linear on their scale. It just happens that as long as infinitesimals

- Obey the Kock-Lawvere axiom
- Come in positive-negative pairs, that is and are both infinitesimal

they are *necessarily* nilsquare. Why? Well, the function is even, that is, for all . Let be an infinitesimal that obeys the too conditions above. By the Kock-Lawvere axiom, we have both and . Thus, by algebra .

It follows that .

First of all, watch Andrej Bauer’s Five Stages of Accepting Constructive Mathematics, where he clears up some common misconceptions about intuitionistic logic.

In my experience, there’s one more thing that trips people up: switching back and forth between affirmative and negative sentences. The good news is that in most cases, you *can* use De Morgan’s laws in Intuitionistic Logic. The bad news is that you’ll have to remember to avoid the three exceptions listed below (needless to say, valid replacements work “the other way round” in the hypotheses).

Replacing … | with … | is … |
---|---|---|

the book is thick | it is not the case that the book is not thick | valid. |

it is not the case that the book is not thick | the book is thick | INVALID. |

the book is either not thick or not blue | the book is not thick and the book is not blue | valid. |

the book is not both thick and blue | the book is either not thick or not blue | INVALID. |

there is a book that is not thick | not all books are thick | valid. |

not all books are thick | there is a book that is not thick | INVALID. |

Both are right, from a certain point of view. In Classical Logic, fields don’t have nilpotent elements.

**Thm.** Let . Then .

**Proof.**

Induction on . The base case with is trivial.

In the inductive case, we know that . There are two possibilites. Either as well, or . In the former case, by inductive hypothesis. In the latter case, has a multiplicative inverse , and therefore .

**Qed.**

Fortunately, the proof above relies on the statement that “either or “. When we work inside Intuitionistic Logic, this is a non-admissible instance of the Law of Excluded Middle. So the proof does not go through in the context of Smooth Infinitesimal Analysis, and the contradiction is avoided.

No. Imagine that you produce an infinitesimal that is non-zero, i.e. . Since it is non-zero, it has to have a multiplicative inverse, . That gives us the equation , a contradiction.

Infinitesimals are *not non-zero*. This is not the same as being zero, since we’re working in intuitionistic logic. However, this does mean that they are so close to zero that we can’t tell them apart from zero. The only way to produce an infinitesimal is to pick an arbitrary one: infinitesimals only occur in proofs, introduced by the sentence “take any “.

When proving non-trivial equalities in SIA, the trick is usually to pick a suitable function, calculate its derivative in two ways, and argue that the results must be equal due to the uniqueness clause of the Kock-Lawvere axiom.

**Thm.** (Microcancellation). Let be real numbers (i.e. elements of the real line) such that for *every* nilsquare infinitesimal . Then .

**Proof.**

Consider the function . On one hand, the derivative of is , since . On the other hand, since , we also have , so . Therefore, .

**Qed.**

Since SIA is known to be consistent, chances are you have misused microcancellation. A common “paraproof” goes as follows:

Take an arbitrary infinitesimal . We get the equality , and by microcancellation , contradicting the fact that zero is not the only infinitesimal.

What’s the mistake here? Stated formally, the microcancellation principle says the following: . The scope of the quantifier () is the parenthesized expression only. That is, if you’re using microcancellation correctly, cannot occur in either or .

Hint: every multiple of a nilsquare infinitesimal is also a nilsquare infinitesimal.

**Thm.** (Chain rule). The derivative of the function is .

**Proof.**

We have by the Kock-Lawvere axiom. Let . Notice that , so we can apply the Kock-Lawvere axiom to . Observe that . On the other hand, by the Kock-Lawvere axiom, we have .

This gives us the equality and therefore, by microcancellation .

**Qed.**

For any exponential function , we have by the Kock-Lawvere axiom, and clearly depends on b only. Thus, we can take to be the defining property of the number .

**Thm.** The derivative of is itself.

**Proof.**

We have by the algebraic properties of exponentials. Since the defining property of the number is , we get . On the other hand, by the Kock-Lawvere axiom, .

This gives us the equality and therefore, by microcancellation .

**Qed.**

Now, what is the domain of the square root function? Negative numbers don’t have square roots, so the square root function has domain . It is not true that (since that would imply ). However, since is an arbitrary nilsquare infinitesimal, you can’t prove that . Therefore, is outside the domain of the square root function, and is undefined.

This is a common source of confusion. Many people think that you cannot divide by , since the value may be an infinitesimal.

However, the quotient rule works only when the function in the denominator is non-zero (this is the same in ordinary real analysis). Fortunately, if , then you are free to divide by (why? see Question 2.3).

To talk about the Extreme Value Theorem, we have to discuss the definition of the partial order on . We define to mean the same thing as . In this case, the Extreme Value Theorem is not merely unprovable, but explicitly *false* in SIA. Informally: we can’t tell if an infinitesimal is positive or negative, so we can’t find the maximum of a linear function with infinitesimal slope. Formally:

**Thm.** (neg-EVT) Not every function attains its maximum.

**Proof.**

Assume that every such function attains a maximum. Choose an arbitrary infinitesimal . Then in particular the function attains its maximum at some . This means that for any , , and thus . As a special case, we have , so if , then . Otherwise, , and thus . In both cases, since infinitesimals cannot be provably greater than zero, we get .

**Qea.**

However, in the literature it is more common to let denote the preorder given by the negation of . In this case, the Extreme Value Theorem is still not provable. Can we actually prove neg-EVT in this case? I don’t know, but please contact me with any information.

No. The Intermediate Value Theorem is *false* in Smooth Infinitesimal Analysis. There is an explicit counterexample (given in Bell’s book), but the proof is quite involved.

Yes! Initially, most people think that this can’t be done. I certainly used to believe that, until I actually tried to prove it and realized that the simplest inductive argument goes through.

**Thm.** (Taylor’s theorem) Let , . Choose any natural number . Then we have .

**Proof.**

Induction on . The base case is , which reduces to the identity . For the inductive case, notice that integration by parts on the error term produces the next term of the Taylor series and a higher order error term.

**Qed.**

The following three axioms are sufficient for a course in multivariable calculus or undergraduate differential geometry.

*Algebra*. The real line is an Archimedean ordered field (its elements are called real numbers). Every positive real number has a square root, there is no polynomial of odd degree that has no roots, and every element of gives rise to a corresponding exponential function with the usual properties.*Kock-Lawvere*. Every function is linear.*Integration*. Every function has a unique antiderivative such that . If the function is positive, so is the antiderivative .

For all twenty axioms, see the book Models of Smooth Infinitesimal Analysis (the additional axioms concern the general form of the Kock-Lawvere axiom, the properties of smooth natural numbers, non-standard analysis, and topology).

No. Like constructive mathematics, Smooth Infinitesimal Analysis is built on *intuitionistic logic*. However, SIA itself is not constructive. For example, the Kock-Lawvere axiom asserts that every smooth function has a slope, without giving us any construction (~ method of finding) of that slope. One may hope, and I personally believe that this is the case, that Smooth Infinitesimal Analysis will ultimately inspire a new constructive treatment of differentiability. For example,it seems that SIA is somehow related to numerical stability, so a constructive interpretation could potentially lead to type systems that can check the numerical stability of simulations, etc.

* Anachronistically, one can say that we introduce the real numbers to get rid of an annoying approximation process. Functions such as have many “approximate” roots, but no exact roots, so we fill up the rationals with exact roots of all kinds of functions, yielding the real numbers . We could totally do analysis on , but it would be a pain. For example, continuous functions of a rational variable on a closed interval are sometimes not uniformly continuous, but they are always “approximately” uniformly continuous.

The above (again, totally anachronistic) motivation for SIA does something similar: we have a zooming-in approximation process that we make exact by introducing some new numbers. At the risk of sounding controversial: doing Differential Calculus without the machinery of nilpotent infinitesimals is as much an artificial difficulty as doing Real Analysis without irrational numbers.

]]>** Actually, in the constructive setting, the notion of field falls apart into two rather different notions. Weak fields have the law that “every non-zero element has a multiplicative inverse”, while strong fields also have the law that “every element is either zero or it has a multiplicative inverse”.

Problem Set 7: First Isomorphism Theorem

Problem Set 6: Chinese Remainder Theorem, Casting out Nines

Problem Set 5: Fibers, Conjugacy, Normal Subgroups

Problem Set 4: Klein 4-group, Inner Automorphisms

Problem Set 3: Subgroups of the Integers, Cyclic Groups, Group Homomorphisms

Problem Set 2: Lagrange’s Theorem, Laws of Composition

Problem Set 1: Matrices

Problem Set 11 coming soon

Problem Set 10: Linear Groups, Change of Basis as Conjugation

Problem Set 9: Linear Independence in Function Spaces, Subspace, Dimension

Problem Set 8: Field Homomorphisms, Wilson’s Theorem

** Proof **

Let be the identity element of the group ;

let be an element that behaves like the identity for for ;

Take any (since , we also have ).

We know that since ,

and also that , since .

Equating the two, , and a right-multiplication by gives , **Qed**.

The argument could also be: since the identity element is unique in any group, and since a subgroup consists only of the elements of the “parent” group, the identity elements of the two groups must coincide.

— M

]]>One has to be careful with that second argument! Sure, there is only one identity element in every group, but that doesn’t mean (in itself) that there’s can’t be a faux-identity, which works like the identity for elements of a subgroup , but works differently for some !

In fact, faux-identities can’t exist, and the argument above establishes that.

If we read the proof carefully, we’ll realise that it can be simplified even further! Instead of considering an arbitrary , we can just take . We get that . Now multiply by to conclude .

The argument, in plain words: a faux-identity cannot exist, because it would square to itself, and only the actual identity does that.— Z

Your hosts are: Benjamin Merlin Bumpus, Zoltan Attila Kocsis and Pernille Rytterhuus.

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