I’ll preface this post with the fact that I’m assuming a working knowledge of abstract algebra at the undergraduate level, realistically more about definitions than theorems.

 I’m taking a class this semester on Galois theory and I’ve been occasionally prepping throughout the summer to get back up to speed in various abstract algebra topics. The textbook, Galois Theory (2nd. edition) by Joseph Rotman, is pretty dense compared to what I’m used to; it (reasonably) assumes a familiarity with groups and jumps right into reviewing commutative rings, starting with the definition,

A commutative ring with 1 is a set $R$ equipped with two binary operations, addition: $(r, r’) \mapsto r + r’$ and multiplication: $(r, r’) \mapsto rr’$, such that…

…and the general axioms relating to commutative rings with identity follow, along with the definitions of integral domains and fields building from those. In pretty much all of the algebra literature I’ve read, these algebraic structures are defined as underlying sets equipped with operations that satisfy various properties. I think most people familiar with these concepts would agree with that generalization. I did too, but a few weeks ago, I came across an interesting concept that expanded my view on these topics a little bit. We’ll take a winding tour through the weirdness of set theory and end up on an interesting take on what a field really can be, so let’s get started.

Really infinite numbers

Start out with a boring empty set $\varnothing$ and call it 0. Now, take $\varnothing \cup \lbrace \varnothing \rbrace$ and call this 1. This successor function, recursively defined as $S(n) \mapsto n \cup \lbrace n \rbrace$, when “seeded” with the empty set, produces a set-theoretic definition of $\mathbb{N}$. But we are actually constructing the incredibly powerful ordinal numbers (specifically via the von Neumann construction), which are a generalization of the concept of “position” or “offset” to infinite cases.

 A hard requirement on all ordinal numbers is that they are transitive sets: for two given sets $x, y$ where $y \in x$, $x$ being transitive indicates that $\forall z \in y [z \in x]$. Taking a look above at our natural number construction, we have $0 = \varnothing, 1 = \lbrace \varnothing \rbrace, 2 = \lbrace \varnothing, \lbrace \varnothing \rbrace \rbrace$, so on and so forth, we can see each set contains all previous sets in the sequence as a subset, so $(0 \in 1) \wedge (0 \in 2) \wedge (1 \in 2) \ldots$ As a matter of fact, this chain can go up infinitely, and this construction illustrates the only other requirement for an ordinal: a strict well-order $\lt$ must be established on an ordinal by set inclusion. A strict order requires that elements cannot be comparable to themselves, which set inclusion satisfies as sets cannot be members of themselves, while a well-order requires a non-empty set of ordinals (which itself could be an ordinal) to have a least element and implies this set be ordered in a unique and consistent way.

 So far, the ordinals we’ve seen are examples of successor ordinals (except for the zero ordinal, which is distinct), or ordinals that can be created by iteratively applying the successor operation to other ordinals. There is one more kind of ordinal, called a limit ordinal. A limit ordinal has no largest element; for a limit ordinal $\gamma$ and an ordinal $\alpha$ where $\alpha < \gamma$, $\exists \delta [\alpha < \delta < \gamma]$. This is false for successor ordinals; given $\alpha = S(\beta)$, the largest element in $\alpha$ is clearly $\beta$. [TODO: sketch the proof of $\bigcup \Gamma = \Gamma \iff \Gamma$ is a limit ordinal]

 So let’s construct our first limit ordinal. We’ll define a finite ordinal as an ordinal $\gamma$ such that there is an $n \in \mathbb{N}$ (specifically our von Neumann construction of $\mathbb{N}$) where there is a bijection between $\gamma$ itself and $n$ (remember, ordinals are sets), and an infinite ordinal if this is not possible. Now, let us define a new ordinal $\omega = \mathbb{N}$. To prove this set is an ordinal, note that the repeated application of the successor operation well-orders this set by set inclusion and produces transitive sets in each step. Thus, we have a transitive set that is well-ordered by set inclusion and is therefore an ordinal. If $\omega$ were not a limit ordinal, there would be some $n \in \mathbb{N}$ such that $\omega = S(n)$. But then it would be simultaneously true that $S(n) \in \omega$ by definition of $\mathbb{N}$, yet sets cannot be members of themselves, so we arrive at a contradiction. Therefore, we have found our first limit ordinal, which is also the very first infinite ordinal. Now we’re ready to unlock the power of ordinal numbers.

Really infinite arithmetic

 You might be asking now how any of this ties into algebraic structures. We’re inching our way towards that goal, and we’re about to make a big step. Firstly, we need to address the initial weirdness of a successor operation on limit (or generally infinite) ordinals. If you are familiar with cardinal numbers, which characterize the size of sets, you would know that it also contains a “number” $\aleph_0$ that describes the “size” (or more accurately, the cardinality) of the natural numbers, as well as all of the natural numbers. But the next smallest cardinal number is $\aleph_1$, the cardinality of the smallest uncountable set (which may or may not be the cardinalty of the real numbers). The successor of $\omega$ is in fact just $\omega + 1$, and its successor is $\omega + 2$, so on and so forth.

 If you are aware of the Peano axioms, we can use the successor operation we previously gave to define arithmetic on the ordinal numbers themselves. For addition, we can start with the recursive definition of addition from the Peano axioms, given as $\alpha + S(\beta) = S(\alpha + \beta)$.

 Clearly these ordinals can get really big, and there’s a lot of them. So what does the set of all ordinals look like? Well, if every single ordinal was truly in this set, it would firstly be transitive. The definition of an ordinal also can be used to prove that sets of ordinals can be well-ordered, so this is a statically well-ordered transitive set, which is itself an ordinal. So the set of all ordinals is an ordinal itself, so it must contain itself. But we already know it can’t, and arrive at a contradiction. This is the Burali-Forti paradox, and stipulates that the collection of all ordinals is a proper class, an object that contains other objects (similar to a set) but does not necessarily inherit the guarantees that set theory gives to sets.

[TODO: rest of paragraph]. With ordinal arithmetic not allowing us to form fields, rings, nor even groups, and the collection of ordinal numbers being an unfamiliar type in set theory land, we’ll need to take a detour to get to our destination.

Really infinite games

 Picture this: you’re at recess in elementary school, and one very mean (but smart) kid challenges you to a game. The rules are simple: there are twenty popsicle sticks laid out on the ground, split into two piles of ten. On each turn, you can either take between one and three sticks from the left pile, or however many as you want from the right pile (as long as you take at least one stick on your turn, and you can only take from one pile per turn). He does the same after you, and whoever takes the last stick loses. If you lose, you give up your fruit snacks. He’ll even let you go first. You agree and start playing, being very careful to not put yourself in a position to hold the last stick. Despite this, he’s playing his moves suspiciously fast in response to yours, and you lose without even realizing it. You demand a rematch, and you’re crushed even quicker than the first time.

 So how did you lose so fast? As it turns out, you were challenged to a game of Nim and never stood a chance. His strategy was terrifyingly robust for a 10-year-old: after your turn, he would perform a kind of “correction” on the piles that would make them equivalent $\text{mod } 4$. At the start, both piles are equivalent to $2 \mod 4$, and unless you were aware of this strategy, your first move would unbalance this. Say your first move was taking a single stick from the pile that had a maximum of three sticks removed per turn. Your opponent could remove three sticks from that pile, in which both piles would again be equivalent to $2 \mod 4$, or he could remove a single (or 5, or 9!) stick from the unrestricted pile, making both piles equivalent to $1 \mod 4$.

Really infinite algebra

[how the nimbers form a ring, field, and algabraically closed field will be derived here]