help what is an integer

• first passes
  • ghosts of ants past
  • clever constraints
• a second attempt
  • an experiment
  • interlude: generators and modules
  • the secret third thing
• coda: hitting the books

okay, I suppose I know what the integers are, but I didn’t know what an algebraic integer was until I was sitting in an algebraic number theory class for the first time. Imagine it’s the first day of class, and we see a brand-new definition that’s going to be the jumping-off point for the next fifteen weeks:

Definition (I). Let \(K\) be a finite-dimensional (field) extension of \(\QQ\). An \(\alpha \in K\) is an algebraic integer if it is the root of a monic non-zero polynomial \(f \in \ZZ[x]\).

The definition itself is simple enough… but why call it an integer, and why should we think of these things as integers?

first passes

Before we continue, we should at least check to see if this is a meaningful definition viewed from the lens of “being an integer.” And at first glance, I think it’s clear not all elements of a general \(K/\QQ\) are algebraic integers. For instance, the rational number \(\frac 12 \in \QQ \subseteq K\) is the root of \(2x - 1 \in \ZZ[x]\), but you can’t find any monic polynomial in \(\ZZ[x]\) that has \(\frac 12\) has a root. On the other hand, the integer \(17 \in \QQ\) is the root of a monic polynomial \(x - 17 \in \ZZ[x]\), and as expected, this definition says it should be an integer. Furthermore, things that seem pretty integer-y in other fields are roots of monic polynomials in \(\ZZ[x]\). For instance, if \(K = \QQ(i)\), \(i \in K\) seems like it should be integral, and it is – it’s the root of the monic polynomial \(x^2 + 1 \in \ZZ[x]\). In general, noting that \(\ZZ[i] \subseteq \QQ(i)\) is called the ring of Gaussian integers, anything of the form \(a + bi\) for \(a, b \in \ZZ\) should also be an integer under this definition. Can we find a corresponding monic polynomial that has \(a+bi\) as a root? (Maybe pause to think about it before continuing, if you can’t think of how to construct this right away.)

It’s a little contrived, but yes! Consider \(x^2 - 2ax + a^2 + b^2 \in \ZZ[x]\), which has roots \(a \pm bi\) from the quadratic formula. So this feels like a pretty reasonable distinction to make for elements of a number field (finite-dimensional extension of \(\QQ\)) \(K\), since it seems to delineate elements of \(K\) that feel reasonably integral from other elements that aren’t. So this definition feels pretty sensible to me and we can move on.

Mechanically, as the course progresses, I get used to working with this definition, but for some reason, the intuition behind why this is the right definition to describe integrality eludes me. Intuitively, I feel like the problem with this definition is that it doesn’t really mesh with any intuitive understanding of how integers work or any property that has to do with \(\ZZ\). The only real idea I have is that the integrality is surely reflected in the fact that it’s a root of a polynomial in \(\ZZ[x]\), but the monic condition doesn’t really give me any real intuition to latch onto.

One slightly modified definition that intuitively seems like it should be clearly equivalent to this first one defines algebraic integers in terms of their minimal polynomials instead:

Definition (II). Let \(K\) be a finite-dimensional field extension of \(\QQ\). An \(\alpha \in K\) is an algebraic integer if the minimal polynomial \(p_\alpha\) of \(\alpha\) over \(\QQ\) is in \(\ZZ[x]\).

I think it’s pretty clear that these two definitions are similar in flavor, and showing that Def. (II) implies Def. (I) is basically immediate. Going the other way requires a little bit of thought, though. First, pick any monic \(f \in \ZZ[x]\) that has \(\alpha\) as a root, and factor it into irreducibles over \(\ZZ[x]\). Each of these factors must be monic, and we can see that \(\alpha\) has to be a root of one of these factors since \(\ZZ[x]\) is an integral domain. So then suppose \(g \in \ZZ[x]\) is irreducible and has \(\alpha\) as a root. But then we know that by Gauss’ Lemma, \(g\) being monic and irreducible in \(\ZZ[x]\) implies that it’s irreducible in \(\QQ[x]\), so \(g\) is actually just the minimal polynomial \(p_\alpha\), and in fact \(g = p_\alpha \in \ZZ[x]\), as desired. It certainly makes intuitive sense that these definitions are equivalent since all we’re really doing is forcing the polynomial that \(\alpha\) is a root of to also be irreducible, and as it turns out, requiring irreducibility of the polynomial doesn’t change the definition at all.

The course continues, and more equivalent definitions and ways of characterizing algebraic integers are introduced, but none of them sink in for me intuitively at that point either. We also talk about a lot of other cool number theory stuff – I glimpse into the world of Dedekind domains, Minkowski’s Theorem comes back when I least expect it, my mind is blown by how nicely things Galois mesh with the theory, etc. And that’s it! I end up finding algebraic number theory very cool and interesting even though I get a little lost, and I make half-baked plans to revisit it at some point in a proper course later on (and actually doing exercises, blah blah blah). But then plans change…

ghosts of ants past

time for a timeskip!

It’s early August 2024, and I unexpectedly find myself at Hampshire College, playing junior staff for the annual Summer Studies in Mathematics¹ and teaching algebraic number theory to 17 very talented high school students. So far, we’ve been playing around in \(\ZZ[i]\) and \(\ZZ[\sqrt{-5}]\), investigating the unique prime factorization of elements in \(\ZZ[i]\) and its absence in \(\ZZ[\sqrt{-5}]\)², since \(6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})\). However, it seems to be recoverable via unique factorization of ideals into prime ideals³, since our students were able to come up with \(\idl 2 = \idl{2, 1+\sqrt{-5}}^2\) and \(\idl 3 = \idl{3, 1 + \sqrt{-5}} \idl{3, 1 - \sqrt{-5}}\), and similarly for \(\idl{1 + \sqrt{-5}}\) and \(\idl{1 - \sqrt{-5}}\). So, the pieces are now in place to start building theory towards figuring out how and when one can salvage unique prime factorization through ideals, and learning about the ring of integers \(\mathcal O_K\) and Dedekind domains.

Pedagogically, learning about the ring of integers \(\mathcal O_K\) as The Example Of A Dedekind Domain feels like the right way to go. I think it’s a good idea to introduce talking about \(\mathcal O_K\) by thinking about what the “integral” elements are in various number fields⁴, and by pointing out that the initial instinct of saying “the ‘integers’ inside \(K = \QQ(\alpha)\) is just \(\ZZ[\alpha]\)” is a bad idea. Recall⁵ that we can characterize any finite extension of the rationals by \(K = \QQ(\alpha)\) for some algebraic number \(\alpha\) (a root of some not-necessarily-monic polynomial in \(\ZZ[x]\)) using the Primitive Element Theorem. But, following this up with saying that the “correct analogue of the integers is \(\ZZ[\alpha]\)” isn’t even a well-defined construction, since the Primitive Element Theorem does not at all give a unique choice of \(\alpha\) for a given number field. As an example, consider \(\QQ(\sqrt{-3}) = \QQ(e^{\frac{2\pi i}3})\), but then this heuristic says that the “integers” inside this field should be both \(\ZZ[\sqrt{-3}]\) and \(\ZZ[e^{\frac{2\pi i}3}]\), which are distinct, non-isomorphic rings. Oops!

To continue with this line of thinking, I need to either motivate the idea of an algebraic integer to bring in \(\mathcal O_K\) as a canonical choice of the “integral elements in a number field”, or figure out how to get at \(\mathcal O_K\) without this idea. This second option is a non-starter – finding \(\mathcal O_K\) in general is a pretty non-trivial task for a general number field. The first option is also a problem, because I don’t have a very clear mental model of an algebraic integer because I didn’t do a good enough job learning algebraic number theory my first time through. But now I have no idea how to motivate the right fix to the problem of unique factorization, and I also somehow need to get my students to invent the right idea very shortly. uh oh.

After a bit of Googling in a sleep-deprived haze, I stumbled across a really clever way to introduce the definition of an algebraic integer via a few carefully chosen heuristics, which I thought was both super cool and accessible enough to be put on a problem set.

clever constraints

spoilers ahead for Hampshire 2024 students if you didn’t do this problem!⁶

the following is inspired by Bill Dubuque’s answer to this math stackexchange question, and is similar to Swinnerton-Dyer’s introduction to the ring of algebraic integers.

The idea goes as follows: we want to define a set \(\mathcal I\) of algebraic integers, right? We’re going to enforce a few conditions on \(\mathcal I\), and then show that these conditions basically force us to conclude that every element of \(\mathcal I\) satisfies one of the definitions of an algebraic integer that we’ve seen thus far. Here they are:

1. The algebraic integers \(\mathcal I\) should form a ring. After all, \(\ZZ\) is a ring, so it should be natural to assume that we should also get a whole ring of integral elements inside any finite-dimensional extension of \(\QQ\).
2. The only algebraic integers in \(\QQ\) are the regular integers \(\ZZ\); in other words, \(\mathcal I \cap \QQ = \ZZ\). This means that the only integral elements inside \(\QQ\) are the things we already know are integral, so this seems like a good sanity-check to have as an assumption.
3. If \(\alpha \in \mathcal I\), then all of its conjugates should also be in \(\mathcal I\). In other words, if \(\alpha\) and \(\alpha'\) are roots of the same minimal polynomial \(p_\alpha\) over \(\QQ\), and \(\alpha \in \mathcal I\), then \(\alpha' \in \mathcal I\). This is less of an obvious constraint, but it forces the resulting set (ring) of integral elements to be “algebraic” in the following sense. Suppose \(\mathcal I\) contained only some but not all of the conjugates of an \(\alpha \in K\). That would imply that \(\mathcal I\) would be able to distinguish between the various conjugates of a root of a minimal polynomial over \(\QQ\). In particular, the definition of \(\mathcal I\) wouldn’t be canonical anymore, since the choice of \(\mathcal I\) would have to use information about the roots themselves to decide which ones should/shouldn’t be elements of \(\mathcal I\). And in the world of algebra, making non-canonical choices is generally frowned upon⁷, and so not enforcing this constraint or enforcing some negation of it would be pretty weird as a definition of a set of algebraic integers. Maybe you can disagree with enforcing this last condition if you like – but we will need it very shortly!

Here’s how you arrive at Definition (II) from this. Suppose \(\alpha = \alpha_0 \in K \cap \mathcal I\) has some monic minimal polynomial \(p_\alpha = x^n + b_{n-1} x^{n-1} + \dots + b_0\) over \(\QQ\). All of the conjugates of \(\alpha\), \(\alpha_1, \dots, \alpha_{n-1}\) in \(K\) are also in \(\mathcal I\) by Condition 3. Over \(K\), \(p_\alpha\) factors as \((x - \alpha) (x - \alpha_1) \dots (x - \alpha_{n-1})\), but re-expanding, we see that

\[p_\alpha = x^n - \left( \sum_{k=0}^{n-1} \alpha_k \right) x^{n-1} + \dots + (-1)^n \prod_{k=0}^{n-1} \alpha_k,\]

where the coefficients \(b_i\) of \(p_\alpha\) are all in \(\mathcal I\), since by Vieta’s formulas they’re all sums of products of some \(\alpha_k\)s, and \(\mathcal I\) is a ring by Condition 1. However, each coefficient \(b_i\) is also in \(\QQ\) since \(p_\alpha \in \QQ[x]\), so \(b_i \in \mathcal I \cap \QQ\). But then by Condition 2, \(\mathcal I \cap \QQ = \ZZ\), so \(b_i\) is an integer! In particular, the minimal polynomial of \(\alpha\) is in \(\ZZ[x]\). This means \(\alpha\) exactly satisfies Definition (II).

However, going the other way and verifying that Conditions 1, 2, and 3 hold for the collection of algebraic integers satisfying one of our previous definitions gets a little sticky. Showing that Definition (II) satisfies Condition 3 is relatively quick. Condition 2 is also pretty clear – to see this, take an element \(\frac m n \in \QQ\) where \(\gcd(m, n) = 1\). \(\frac m n\) has a linear minimal polynomial over \(\QQ\), being \(x - \frac m n\), but it’s only possible to clear denominators and end up with a monic minimal polynomial in \(\ZZ[x]\) iff \(n = \pm 1\). However, verifying Condition 1 is harder – it’s not at all clear that \(\mathcal I\) as defined by Definitions (I) or (II) gives you a ring. In fact, when I saw Definition (II) for the first time, I was absolutely convinced that the addition and multiplication operations on the resulting ring of algebraic integers had to be induced from some strange other construction. There is in fact no shenanigans here – the ring structure literally is the good ol’ addition and multiplication of the underlying field, which is a little mind-boggling. Arguing this directly from Definitions (I) or (II) requires a good amount of work though, and personally, I don’t think the argument coming directly from these definitions is particularly enlightening. At least, it’s no more enlightening than showing that these definitions are equivalent to a secret third definition of an algebraic integer.

… wait, there’s a third definition?

a second attempt

the program ends with a generally successful speedrun to establish the properties of a Dedekind domain for \(\mathcal O_K = \mathcal I \cap K\), ending with examples of some terrible rings⁸, learning about prime and maximal ideals, and finally un-breaking math with the help of ideal factorization (in the right places)! Rooms are cleaned and packed, comments are written, and everyone goes home. No more algebraic number theory to be done – except for the fact that one of my students lodged the following idea in my brain, which was rattling around in there for several weeks after the program ended:

Idea: Monic polynomials are nice in \(\ZZ[x]\) because they’re the only things you can divide by in \(\ZZ[x]\) to get remainders of strictly lower degree.

This idea permanently altered my brain chemistry and is the reason why the rest of the article you’re reading now exists. I felt like this had something to do with the integrality condition and said something about why algebraic integers are the right collection of numbers to study – remember from Definition (I) that roots of monic polynomials are precisely algebraic integers! After a lot of thinking, it turns out there is kind of a connection, and it’s related to another promising intuitive way to get at an algebraic integer. But first:

an experiment

i conducted an experiment on math/physics undergraduates over Discord teehee

Inspired by my recent experiences at Hampshire, I wanted to see if one could intuitively develop Definitions (I) or (II) for an algebraic integer just by being posed the question “what makes something integral inside a number field⁴?”. Thankfully, someone actually interacted with my nonsense, and we ended up landing on the following question to ponder:

Question: Inside \(\QQ\), what is the difference between \(\frac 12\) (clearly not an integer) and \(2\) (clearly an integer)?

Answer. Having a non-trivial denominator is what separates fractions like \(\frac 12\) from integers like \(2\).

This might seem silly, but in \(\QQ\), this is a defining difference between an integer and a rational number – rational numbers that are not integers must have denominators that are not \(\pm 1\), which we’ll call “non-trivial”. But then we’ve just replaced our question by another one:

Question: How should one test in general whether or not some \(\alpha \in K\) “doesn’t have a non-trivial denominator”?

As a question, this maybe feels a little bit too open-ended for a class and gives bad “guess what the idea in my head is” vibes pedagogically. However, in this conversation, we arrived at this question after thinking about examples of several rings of integers, equipped with their corresponding norms⁹. An analysis-minded individual might then be led to think about sequences of elements in these rings.

A good candidate of a sequence to consider for any \(\alpha \in K\) is the sequence \(a_n = \alpha^n\) for \(n \in \ZZ^+\). As a silly example, let’s suppose we restrict to taking \(\alpha \in \QQ\) and \(\alpha\) is not an integer. Then in the resulting sequence, every element has a non-trivial denominator. In particular, the denominators we get are strictly increasing, and are all powers of some integer. For instance, in the case of \(\alpha = \frac 12\), when we take the sequence \(a_n = \frac 1 {2^n}\) for all \(n \in \ZZ^+\), the denominators of the elements of our sequence are \(2^n\), powers of \(2\). In particular, \(a_n\) is never an integer, which can be shown inductively from the fact that \(a_1 = \alpha = \frac 12\) has a non-trivial denominator. However, if \(\alpha\) has no non-trivial denominator (i.e. it’s a regular integer), \(a_n \in \ZZ\) for all \(n\). So the sequence of powers of \(\alpha\), \(a_n\), treats elements in \(\QQ\) with and without non-trivial denominators differently depending on whether the elements of \(a_n\) are integers or not. In this sense, \(a_n\) becomes an interesting sequence to study in this context.

Let’s consider our simplest field extension \(K = \QQ(i)\), and \(\alpha \in K\). As a first easy example, even in the case where \(\alpha = i\) (which we want to be integral), \(a_n\) is no longer all integers – but \(a_2 = i^2 = -1\) is, which is something! But it’s too much to ask that \(\alpha\) be a perfect \(m\)th root of an integer, since in general, we want the “Gaussian integer” \(\alpha = x + yi\) for arbitrary \(x,y \in \ZZ\) to also be an integral element, which may not be a perfect root of an integer. What happens to \(a_n\) in this general case?

We can start by analyzing a few values of this sequence. First, we can compute \(a_2\). Note that \(a_2 = \alpha^2 = (x+yi)^2 = x^2 - y^2 + 2xyi\) is no longer a pure integer, but it still looks pretty integral since it has real part \(x^2 - y^2\) and imaginary part \(2xy\), both of which are integers. We can kind of “correct” the imaginary part of \(a_2\) to get an actual element of \(\ZZ\) by subtracting off \(2x\) times \(a_1\) from \(a_2\). Computing,

\[a_2 - 2x a_1 = x^2 - y^2 + 2xyi - 2x(x + yi) = x^2 - y^2 - 2x^2 = -x^2 - y^2\]

which definitely is a normal integer! Let’s call \(a_2 - 2x a_1 = C \in \ZZ\). We can now use this identity recursively to show that \(a_n\) can be added to an integer multiple of \(a_1\) to form an integer for all \(n \geq 2\). For instance,

\[a_3 = \alpha a_2 = \alpha(2x a_1 + C) = 2x a_2 + C a_1.\]

But now we can write \(a_2\) in terms of \(a_1\) and integers, so now

\[a_3 = 2x (2x a_1 + C) + C a_1 = (4x^2 + C) a_1 + 2x C.\]

and therefore \(a_3 - (4x^2 + C)a_1 = 2x C\), an integer. One could similarly imagine doing this for \(a_4, a_5, \dots\). Notice that once we find a way to add an integer multiple of \(a_1\) to \(a_2 = \alpha^2\) to get an element of \(\ZZ\), we can do this for any power of \(\alpha^n\) for \(n \geq 2\) by recursion. Precisely, for \(n \geq 2\), there exists some integer multiple of \(a_1\) such that \(a_n + M_n \alpha_1 \in \ZZ\) for some \(M_n \in \ZZ\). This works in the case of \(\alpha = i\) in a silly way since \(a_2 = i^2 = -1 \in \ZZ\), but we actually need to add non-trivial multiples of \(a_1 = i\) to say, \(a_3\), to get an element of \(\ZZ\).

Note that something similar never holds for \(\alpha \in \QQ\) not an integer. Revisiting the case of \(\alpha = \frac 12\), note that \(a_n + \sum_{k=1}^{n-1} M_k a_k \not \in \ZZ\) for any choice of integers \(M_k \in \ZZ\). Suppose for the sake of contradiction that \(\frac 1{2^n} + \sum_{k=1}^{n-1} \frac{M_k}{2^k} = C \in \ZZ\), but then \(1 + \sum_{k=1}^{n=1} 2^{n-k} M_k = 2^n C\). In mod \(2\) (which is exactly the denominator of \(\alpha\)), we see that this equation never holds. The same argument also holds for other rational numbers, like \(\alpha = \frac 32\). Here, the same argument goes through, with the critical step being that the numerator \(3\) is relatively prime to the denominator \(2\).

So maybe this is a little contrived, but a possible answer to our question is:

Answer. A number \(\alpha \in K\) “doesn’t have a nontrivial denominator” iff there exists some power of \(\alpha\), \(\alpha^n\) for \(n \in \ZZ^+\), for which \(\alpha^n + \sum_{k=1}^{n-1} c_k \alpha^k \in \ZZ\) for some \(c_k\) integers. It then follows that for all \(m \geq n\), \(\alpha^m + \sum_{k=1}^{n-1} d_k \alpha^k \in \ZZ\) for some suitably-chosen \(d_k \in \ZZ\).

Let’s view this in the following way from the point of view of rings. For any \(\alpha\), the sequence \(a_n = \alpha^n\) additively generates the elements of the ring \(\ZZ[\alpha]\), by definition. When \(\alpha\) feels integral, like in the case when \(\alpha = x + yi \in K = \QQ(i)\) for \(x,y\) integers, since we showed that \(\alpha^2 + 2x \alpha \in \ZZ\) and therefore for all \(n \geq 2\), \(\alpha^n + M_n \alpha = C_n \in \ZZ\) for some \(M_n \in \ZZ\), we see that all of the elements of \(\ZZ[\alpha]\) can be written in terms of integer linear combinations of \(\alpha\) and \(1\). On the other hand, when \(\alpha\) doesn’t feel integral, like when \(\alpha = \frac 12\) or \(\frac 32\), the rings \(\ZZ[\frac 12]\) and \(\ZZ[\frac 32]\) feel “big” in a certain sense. (Note: \(\ZZ[\frac 32]\) is actually the exact same ring as \(\ZZ[\frac 12]\) since \(\frac 32 = 1 + \frac 12\), and so any expression with \(\frac 32\)s in it can be rewritten in terms of \(\frac 12\)s and other integers, and vice versa.) Since we can never write \(\frac 1{2^n}\) as an integer linear combination of smaller powers of \(\frac 12\), we might have to use an arbitrarily large power of \(\frac 12\) to write down an arbitrary element in \(\ZZ[\frac 12]\) as an integer linear combination of other ones. This is reflected by the idea that we get some redundancy from relations like \(i^2 = - 1\) in \(\ZZ[i]\) or \((x+yi)^n = -M_n (x+yi) + C_n\) in \(\ZZ[x+yi]\), but no such relation holds in \(\ZZ[\frac 12]\). This is the exact property we want that somehow makes \(i\) and \(x+yi\) integers but \(\frac 12\) and \(\frac 32\) not integers, and will give us our third definition!

interlude: generators and modules

but first, we have to talk about generators. If you already know the language of modules over a ring, definitely go ahead and skip to the next section. Also, if you’re familiar with generators of a group, you should mostly be fine. The exposition that follows is the minimum amount of content needed to understand what follows, and is especially targeted towards Hampshire students, so some statements might not be phrased in their full generality, oops.

expand me for some exposition about generators and modules

One might see generators in the context of groups first:

Def. A group \(G\) is generated by a set \(S\) (denoted \(G = \idl S\)) if every element \(g \in G\) is a product of elements \(s_1 s_2 \dots s_n\) where \(s_i \in S\) for all \(1 \leq i \leq n\). \(G\) is finitely-generated if there exists a finite set \(S\) that generates \(G\).

Now, we are talking about rings here, but rings are abelian groups under addition. As such, it makes sense to talk about rings as being “finitely-generated” as an abelian group, and actually this would be enough for our purposes. You can stop reading now if you want. However, people often state this property differently and tend not to reference groups at all, instead opting to say this in the context of a more abstract setting. In particular, people tend to talk about this property by using more linear algebra-style intuition. Recall:

Def. A field \(K / \QQ\) is spanned by a set \(S \subseteq K\) if every element in \(K\) is a rational linear combination of elements in \(S\), i.e. for all \(k \in K\), \(k = \sum_{s \in S} q_s \cdot s\) where all of the \(q_s \in \QQ\). \(K\) is finite-dimensional if \(S\) can be chosen to be a finite set.

From here, people generalize to not just allow \(K\) to be any field extension of \(\QQ\), but allow \(K\) to be an arbitrary abelian group under addition that also has a multiplication by elements in \(\QQ\). This yields what people call a vector space over \(\QQ\). You can even generalize one step further and allow the multiplication on \(K\) to be by elements in any ring \(R\), and you get:

Def. A module over a (commutative) ring \(R\) (with unit \(\Bbb 1\)) is an abelian group \((M, +)\) with a scalar multiplication by elements of \(R\) that is associative, has an identity \(\Bbb 1\), and distributes over addition. \(M\) is also called an \(R\)-module.

Def. An \(R\)-module \(M\) is generated by a set \(S \subseteq M\) if every element in \(M\) can be written as an \(R\)-linear combination of elements in \(S\), i.e. for all \(m \in M\), \(m = \sum_{s \in S} r_s \cdot s\) where all of the \(r_s \in R\). \(M\) is finitely-generated if \(S\) can be chosen to be finite.

We don’t really need think about \(R\)-modules for a general ring \(R\) – what we really care about is whether elements of \(\ZZ[\alpha]\) can be expressed as \(\ZZ\)-linear combinations of a finite set of elements in \(\ZZ[\alpha]\), so we really only need to view \(\ZZ[\alpha]\) as a \(\ZZ\)-module. But it turns out \(\ZZ\)-modules are just a fancy word for abelian groups! Any abelian group \((G, +)\) can be thought of as a \(\ZZ\)-module in the following way – given an arbitrary element \(g \in G\) and \(n \in \ZZ^+\), define \(n \cdot g = \sum_{i=1}^n g = g + g + \dots + g\), in the same way that multiplying a number by 17 means adding up 17 copies of that number. By extension, we should define \(0 \cdot g = e\), where \(e\) is the identity in \(G\), since we’re adding zero things together. Extrapolating further, \((-n) \cdot g = \sum_{i=1}^n g^{-1}\), since \(g^{-1}\) can be thought of as “\(-g\)” in an abelian group if the (additive) identity is “\(0\)”, and \((-n) \cdot g\) is kinda like \(n \cdot (-g)\) (which should really be \(n \cdot g^{-1}\)). This gives you a scalar multiplication by \(\ZZ\) on \(G\) that automatically turns it into a \(\ZZ\)-module. (And definitionally, any \(\ZZ\)-module is an abelian group, so these notions are equivalent.)

Moreover, it’s not too hard to check for yourself that the concept of being finitely-generated in terms of \(\ZZ\)-modules is the exact same thing as being finitely-generated as an abelian group! So, we don’t really need to bring in the abstract language of modules when we can get away with thinking about abelian groups, and if you’re more comfortable with that, you can kind of ignore the above.

As an aside, though, modules are the language that is most commonly used when describing linear combinations with coefficients in an arbitrary ring, and they describe a much more general class of algebraic structures. Just to briefly justify why people like modules, modules are to the vector spaces of linear algebra as rings are to fields. Recall that rings can be pretty wild and wacky and have a lot of general theory, whereas fields are more constrained in the kinds of behaviors they can exhibit. In the same way, modules encode a really broad class of algebraic structures, whereas vector spaces are relatively constrained in their structure. Modules therefore are more flexible in the kinds of objects that they can represent. Personally, I was initially skeptical about why one should ever think about modules when I first saw them, but as I’ve gotten more used to thinking about them, the more I think they have a good amount of complexity to be fun and interesting to play with while also being useful.

…

Enough digression for now – let’s get to

the secret third thing

for a given \(\alpha \in K\) with \(K\) finite-dimensional over \(\QQ\), to think about the integrality of \(\alpha\), we’re thinking about how “big” \(\ZZ[\alpha]\) is as a \(\ZZ\)-module/abelian group. In our previous examples with \(\alpha = \frac 12\) versus \(\alpha = i\), note that \(\ZZ[i]\) is finitely-generated with generating set \(\set{1, i}\). However, note that \(\ZZ[\frac 12]\) is weird because it’s not finitely generated! A generating set for \(\ZZ[\frac 12]\) is \(\set{\frac 1 {2^n} : n \in \NN}\), but can there be any finite generating set? Turns out, no – if we did, we would have an upper bound on the largest possible denominator of a fraction we could achieve with \(\ZZ\)-linear combinations, and we know that \(\ZZ[\frac 12]\) has fractions with arbitrarily large denominators.

This leads us to a new definition for integrality from this point of view:

Definition (III). Let \(K\) be a finite-dimensional field extension of \(\QQ\). An \(\alpha \in K\) is an algebraic integer if \(\ZZ[\alpha]\) is a finitely-generated \(\ZZ\)-module (abelian group, under addition).

This is pretty different from Definitions (I) and (II)! I like this definition intuitively because the integrality is suggested by the fact that \(\alpha\) now has to “play nicely” in some sense with \(\ZZ\). On the other hand, this definition feels like a more abstract condition, whereas Definitions (I) and (II) feel more concrete since they deal with polynomials. Speaking of which, does Definition (III) really describe the same collection of numbers as Definitions (I) and (II)?

Suppose you were to take Definition (I) or Definition (II) as your definition of an algebraic integer. Recall⁵ that by the First Isomorphism Theorem that \(\ZZ[\alpha] \cong \ZZ[x] / \idl{p_\alpha}\), where \(p_\alpha\) is the minimal polynomial of \(\alpha\) over \(\ZZ\) with degree \(n\). Assuming Definition (II) of an algebraic integer, \(p_\alpha\) has to be monic! Now we can use the Idea from above – when dividing any polynoimal \(f \in \ZZ[x]\) by \(p_\alpha\), we can write \(f = q p_\alpha + r\) for \(q, r \in \QQ[x]\) where \(\deg r < n\) because \(p_\alpha\) is monic! Actually, because \(p_\alpha\) is monic, we can also force \(q, r \in \ZZ[x]\) if we think a little harder. Now, if we consider the ring homomorphism from \(\ZZ[x] \to \ZZ[\alpha]\) that evaluates any polynomial \(f\) at \(\alpha\), we see that \(f\) gets sent to \(f(\alpha) = q(\alpha) p_\alpha(\alpha) + r(\alpha) = q(\alpha) \cdot 0 + r(\alpha) = r(\alpha)\), and since this homomorphism is surjective, we see that \(\ZZ[\alpha]\) consists of polynomials in \(\alpha\) that have degree less than \(n\)! Therefore, the set \(\set{1, \alpha, \dots, \alpha^{n - 1}}\) is a generating set for \(\ZZ[\alpha]\) and so \(\ZZ[\alpha]\) is finitely generated, as required by Definition (III). A similar vibes-based way to see this is that \(p_\alpha\) gives an expression for \(\alpha^n\) as an integer linear combination of all of the lower powers of \(\alpha\), so using this one can express all powers of \(\alpha\) in terms of integer linear combinations of \(\set{1, \alpha, \dots, \alpha^{n-1}}\). So Definition (I)/Definition (II) certainly imply Definition (III)!

Let’s try this on our non-example for \(\alpha = \frac 12\) – here, \(p_\alpha = 2x - 1\). We actually saw this on a problem set at Hampshire – when you consider \(\ZZ[x] / \idl{2x - 1}\), you actually can’t use the quotiented ideal to cancel out powers of \(x\) with odd coefficients, so \(\set{1, x, x^2, \dots}\) are all nonzero elements in this ring and the higher powers of \(x\) can’t be written in terms of the lower ones with integer coefficients! This is exactly like how \(\set{1, \frac 12, \frac 14, \dots}\) are all nonzero elements in \(\ZZ[\frac 12]\), and you can’t write \(\frac 1{2^n}\) as an integral linear combination of \(\set{1, \frac 12, \dots, \frac 1{2^{n-1}}}\), which is kind of what we expect under the isomorphism \(\ZZ[x] / \idl{2x - 1} \cong \ZZ[\frac 12]\) given by substituting \(x\) for \(\frac 12\). In general, any element with a non-monic minimal polynomial over \(\ZZ\) won’t admit a general way to write a power of \(\alpha\) as an integer combination of its lower powers, which is a good heuristic for seeing that \(\ZZ[\alpha]\) is infinitely generated.

But how do we actually show Definition (III) implies Definition (I)/(II)? Ah… this is a little trickier and requires a little bit of matrix magic. Let’s consider as an example \(\alpha = 1 + 4i\), and so \(\ZZ[\alpha] = \ZZ[1+4i] = \ZZ[4i]\), since the \(1\) already is an integer. Note that this ring has generating set \(\set{1, 4i}\). Now here comes a bit of a trick: we’re going to multiply \(\alpha\) by each of the generators of \(\ZZ[4i]\), and write the result in terms of integer multiples of the generators (which must be possible, since they were generators, after all):

\[\alpha \cdot 1 = 1 + 4i = 1 \cdot 1 + 1 \cdot 4i \qquad \alpha \cdot 4i = -16 + 4i = -16 \cdot 1 + 1 \cdot 4i\]

Let’s rewrite these equations like so:

\[(\alpha - 1) \cdot 1 + (-1) \cdot 4i = 0 \qquad 16 \cdot 1 + (\alpha - 1) \cdot 4i = 0\]

I’m writing these equations in a weird way, because I want to rewrite these equations in matrix form in the following way:

\[\bmat{\alpha - 1 & -1 \\ 16 & \alpha - 1} \bmat{1 \\ 4i} = \bmat{0 \\ 0}\]

Now, I claim that the matrix \(A = \tbmat{\alpha - 1 & -1 \\ 16 & \alpha - 1}\) is not invertible. If it had some inverse \(B\), then I could multiply by \(B\) on both sides to get \(\bmat{1 \\ 4i} = B \bmat{0 \\ 0} = \bmat{0 \\ 0}\), which is a contradiction. Therefore, \(A\) is not invertible, so its determinant must be zero. This implies that \((\alpha - 1)^2 + 16 = 0\), which magically yields a polynomial with integer coefficients that has \(\alpha\) has a root. This polynomial expands to be \(\alpha^2 - 2\alpha + 17\), which is also monic! In the specific case of \(\alpha = 1 + 4i\), then, we can do this kind of linear algebra magic to cook up a monic polynomial that has \(\alpha\) as a root.

This argument indicates how one can show that Definition (III) implies Definition (I). I’m going to hide it in case you don’t want to see the general argument:

expand me if you do want to see how to do this in general

First, we know that by Definition (III) that \(\ZZ[\alpha]\) is finitely generated, perhaps with generating set \(\set{b_1, \dots, b_k}\) for some \(k \in \NN\). In particular, for every \(1 \leq i \leq k\), \(\alpha b_i\) is some integer combination \(\sum_{j=1}^k a_{ij} b_j\) for some \(a_{ij} \in \ZZ\). If we form the coefficients \(a_{ij}\) into a matrix \(A\) such that the \((i, j)\)th entry of \(A\) is \(a_{ij}\), and letting \(\mathbf b = \tbmat{b_1 \\ \vdots \\ b_k}\), this system of equations can be thought of as the matrix equation \(\alpha I_k \cdot \mathbf b = A \mathbf b\). Doing some rearranging, we can say that \((\alpha I_k - A) \cdot \mathbf b = 0\), so by the same logic as above, \(\alpha I_k - A\) is not invertible and has determinant zero. Note that \(f(x) = \det(x I_k - A)\) is a monic polynomial in \(x\) with root \(\alpha\), so \(f\) is the desired polynomial we want for Definition (I). This is a little messy, but it works!

… but now we’ve shown these definitions are equivalent! Nice!

The neat thing about Definition (III) is that it really quickly allows you to see that algebraic integers are a ring. Let’s knock this out. Suppose \(\alpha\) and \(\beta\) are algebraic integers. Then, by Definition (III) we know that \(\ZZ[\alpha]\) and \(\ZZ[\beta]\) are finitely generated, perhaps with generators \(\set{a_1, \dots, a_k}\) and \(\set{b_1, \dots, b_l}\), respectively. Now, we claim that \(\ZZ[\alpha, \beta]\) is generated by \(\set{a_i b_j : 1 \leq i \leq k, 1 \leq j \leq l}\), where \(\ZZ[\alpha, \beta]\) is obtained by adding all the powers of \(\alpha\) and \(\beta\) and their pairwise products to \(\ZZ\). Note that for any non-negative integers \(e, f\), \(\alpha^e = \sum_{i=1}^k c_i a_i\) and \(\beta^f = \sum_{j=1}^l d_j b_j\) for some integer coefficients \(c_i, d_j\). Then, \(\alpha^e \beta^f = \sum_{i=1}^k \sum_{j=1}^l c_i d_j a_i b_j\), so all products of powers of \(\alpha\) and \(\beta\) are generated by integer linear combinations of this finite generating set of \(kl\) elements. Moreover, \(\ZZ[\alpha + \beta]\) and \(\ZZ[\alpha \beta]\) are both subrings of \(\ZZ[\alpha, \beta]\), so these rings must also be finitely generated \(\ZZ\)-modules (abelian groups). But then \(\alpha + \beta\) and \(\alpha \beta\) are algebraic integers by Definition (III), so the algebraic integers must form a ring!

I hope the relative slickness of this argument at least shows that Definition (III) has a lot of utility as another definition for an algebraic integer, and that the preceding exposition makes it feel more intuitive while also meshing well with the idea of being a root of a monic polynomial that pervades Definitions (I) and (II). Not to say that the first two definitions aren’t intuitive, but to believe them intuitively I feel like requires a few convictions about what an integer should be that are reasonable but might not be what you first come up with. Both definitions have their uses and different definitions are nicer for exhibiting different facts about the ring of algebraic integers \(\mathcal I\) (or \(\mathcal O_K\)). However, the fact that they’re equivalent through some manipulations is really neat and unexpected, in my opinion.

and that’s it! I think I get what an integer is now!

coda: hitting the books

in researching and thinking about this topic, I wanted to see how people came up with this idea in the first place. I went down a bit of a rabbit hole looking for the places where people first came up with this idea, and it turns out that a lot of this stuff came from some guys in the 19th century named Ernst Kummer and Richard Dedekind.

Kummer came first in time, studying higher reciprocity laws. One might remember from an elementary number theory class (or program) that Carl Gauss had tons of arguments for quadratic reciprocity, which tells one how to relate the existence of solutions to \(x^2 \equiv p \mod q\) to \(x^2 \equiv q \mod p\) for primes \(p\) and \(q\). The natural generalization is to then think about whether the equations \(x^m \equiv p \mod q\) and \(x^m \equiv q \mod p\) have related sets of solutions for any \(m > 2\). Several people (Gauss, Jacobi, Eisenstein) had tackled the case of \(m = 3\) and \(m = 4\) before Kummer by examining the rings \(\ZZ[i]\) and \(\ZZ[e^{\frac{2\pi i}3}]\), and Kummer was looking to build on this work and find higher reciprocity laws for any \(m\). Kummer was kind of successful in this project, but really needed to use \(\ZZ[\omega]\) for \(\omega = e^{\frac{2\pi i}m}\) to do this and assuming that \(\ZZ[\omega]\) had unique factorization. Unfortunately, as one might suspect, this isn’t true for arbitrary \(m\) – and I’ll probably revisit this in another post because we’ll probably talk about this in my abstract algebra class next quarter LOL. In any case, Kummer came up with the idea of “ideal numbers” to handle the loss of unique factorization, which is the direct historical predecessor to the modern idea of an ideal, now formalized by Dedekind.

In formalizing and extending Kummer’s work, Dedekind is generally credited with developing the modern concept of a ring, a theory of ideals (and prime ideals), and trying to extend Kummer’s ideas to the right kinds of rings. Dedekind thought the right setting for Kummer’s ideas was the “domain of integers of algebraic number fields,” which is what we would today call the ring of integers \(\mathcal O_K\) of a finite-dimensional extension \(K / \QQ\). But how did Dedekind come up with the right definition of \(\mathcal O_K\)? The following passage (along with much of the above) comes from Israel Kleiner’s book A History of Abstract Algebra:

But Dedekind did not motivate his definition of the domain of integers of an algebraic number field, as historian of mathematics Edwards laments: “Insofar as this is the crucial idea of the theory, the genesis of the theory appears, therefore, to be lost.”

… dang. thanks Dedekind.

I guess it’s good that we developed our own intuition for why integers are integers!

---