# What is Galois Theory Anyway?

/Perhaps you've heard of Évariste Galois? (Pronounced "GAL-wah.") You know, the French mathematician who died tragically in 1832 in a duel at the tender age of 20? (Supposedly over a girl! C*'est romantique, n'est-ce pas?*) Well, today we're taking a bird's-eye view of his most well-known contribution to mathematics: the appropriately named Galois theory. The goal of this post is twofold:

## one

If you are a student *about* to study Galois theory, I hope the info below will serve as a small appetizer to your main course. In the "From English to Math" section below, we'll take a brief survey of the ideas that appear in a standard graduate course so that when you start doing exercises, you at least have a bird's-eye-view of what's going on.

## two

Even if you're *not* about to study Galois theory and are just curious, this post is also for you! The info here should be accessible to anyone with at least an undergrad background in abstract algebra. I'm going to leave out a lot of technical details (but you don't that mind, do you?) as my goal here is just to convey some main ideas. I hope this may whet your appetite to study further.

In a word, Galois Theory uncovers a relationship between the structure of groups and the structure of fields. It then uses this relationship to describe how the roots of a polynomial relate to one another.

More specifically, we start with a polynomial $f(x)$. Its roots live in a field (called the

*splitting field*of $f(x)$). These roots display a symmetry which is seen by letting a certain group (called the

*Galois group*of $f(x)$) act on them. And we can gather information

*about*the group's structure

*from*the field's structure and vice versa via the Fundamental Theorem of Galois:

*can the roots of an nth degree polynomial be written down as some algebraic combination $(+,-,\times,\div,\sqrt)$ of the polynomial's coefficients?*It turns out the answer is "Yes!" when $n\leq 4$, but "No," for any $n\geq 5$.

It was precisely Galois' study of permutation groups of the roots of polynomials that led to his discovery of a necessary and sufficient condition for finding a such a formula.* The condition (which eluded mathematicians for over 300 years!) becomes elegantly clear when the problem is translated *from* the language of field theory to that of group theory. Galois theory is the dictionary which makes this possible.

## The Field Story

Suppose $F$ is a field. Then the polynomial ring $F[x]$ is a (Euclidean domain and hence a) unique factorization domain. This means any polynomial $f(x)\in F[x]$ can be factored uniquely as a product of irreducible polynomials. Now we know that any root of $f$ must be a root of one of those irreducible factors, but we may not know how and where to find those roots. It turns out that passing to field *larger* than $F$ helps produce those roots.

**Example:**Consider the irreducible polynomial $f(x)=x^2+1$ with coefficients in $\mathbb{Q}$. One of its roots is $i=\sqrt{-1}$. Even though $i\not\in\mathbb{Q}$, we can find a bigger field which contains it, namely $\mathbb{Q}(i)=\{a+bi:a,b\in\mathbb{Q}\}$. ($\mathbb{Q}(i)$ is "bigger" since $\mathbb{Q}\subset\mathbb{Q}(i)$.)

In the example above, we say $\mathbb{Q}(i)$ is an *extension field* of $\mathbb{Q}$. And in general, any field $K$ which contains a smaller field $F$ is called an **extension** of $F$, and $F$ is referred to as the **base** (or **ground**)** field**. But so far we've only talked about a *single* root of our polynomial $f(x)$. What about *all* of its roots? Can we find an extension of $F$ which contains all of the roots of a general $f(x)$? Again the answer is *yes*, and that field is called the **splitting field** of $f(x)$.
(Technically, the splitting field the *smallest* extension of $F$ which contains all the roots of $f(x)$.)

(*Aside:* We could also turn this question on its head. Suppose we have a field $F$ and an extension $K$, and we pick a random element** $\alpha\in K$. *Can we find a polynomial in $F[x]$ which has $\alpha$ as a root?* This time, the answer is not always yes! But in the cases when it is, we say $\alpha$ is an **algebraic element**. Moreover, if the answer is yes for *every* element of $K$, we say $K$ is an **algebraic field**. As we've discussed before, algebraic elements are sort of like limit points in topology/analysis.)

**adjoin**to $F$ the roots of $f(x)$ until we obtain the field $K$ (the splitting field) which contains all the roots of $f$. I'm doing a lot of hand-waving here, but we eventually obtain a tower of fields which looks something like this:

where $F_{i+1}$ is bigger than $F_i$ because it contains (at least) one more root of $f(x)$. It's the structure of *this* tower of fields which is mirrored in the structure of the Galois group associated with $f(x)$. And that group is our next topic of discussion.

## The Group Story

*VERB: It's what you do.*

*do*stuff. In particular, a group

**acts**on a set by shuffling elements around. We can gather a lot of information about the group and about the set it acts on via this action. So if you ever want to create a TV commercial about mathematical groups,

*GROUPS: It's what they do.*

How does this relate to Galois theory? Well the **Galois group** $G$ associated to an $n$th degree polynomial $f(x)\in F[x]$ has its own action. In particular, the elements of $G$ are automorphisms $\sigma:K\to K$ where $K$ is the splitting field of $f$. (Recall a **field automorphism** is just an isomorphism from a field to itself.) This group is isomorphic to (i.e. has the same structure as) a subgroup, called a **permutation group**, of the symmetric group $S_n$.

Since the *raison d'être* of a permutation group is to permute things, it's not too hard to believe that $G$ *acts on* the splitting field $K$ by permuting the roots of $f(x)$. So suppose for a moment that $f(x)\in F[x]$ can be written as a product of irreducible factors $f(x)=f_1(x)f_2(x)\cdots f_k(x)$. Then $G$ will permute the roots of $f_1$ among themselves, and the roots of $f_2$ among *themselves*, and so on. In other words, $G$ is said to **act transitively** on the irreducible factors of $f$. For instance, in the drawing below, $G$ is some subgroup of $S_n$ (for the sake my illustration, suppose*** $n\geq 9$) which contains a product of two 2-cycles, $(13)(24)$, one three cycle, $(567)$, and a transposition $(n \; n-1)$.

In general, once we know the Galois group of $f$, we can analyze its subgroup structure. Galois' insight was to notice that *if* the structure of $G$ was such that it has a chain of subgroups

(where $e$ is the identity) such that $H_i\triangleleft H_{i+1}$ is normal and $H_{i+1}/H_{i}$ is abelian, *then and only then* can we write down an explicit algebraic expression (like the quadratic formula) for the roots of $f(x)$. In this case, we say the group $G$ is **solvable**. By the Fundamental Theorem of Galois, the ability to write down this chain of subgroups corresponds to the ability to write down a *particular* tower of subfields of the splitting field $K$ of $f$. So the solvability of our polynomial $f$ amounts to knowing something about the structure of its splitting field, or equivalently the structure of its Galois group. The fact that there is no algebraic formula for the roots of polynomials of degree 5 and higher is due to the fact that the symmetric group $S_n$ for $n\geq 5$ is not solvable!

*Basic Abstract Algebra*(2 ed., ch. 15-17) by Bhattacharya, Jain, and Nagpaul. This book is very approachable at the undergraduate level. And of course,

*Abstract Algebra*(3 ed., ch. 13-14) by Dummit and Foote is a classic for both undergraduate and graduate students.

*Footnotes*

* In fact, Galois himself invented the concept of a group for this very purpose!

****Edited May 21, 2016**: This sentence originally read, "...a random element $\alpha\in F$" and did not mention the extension field $K$ (a rather boring case)! Many thanks to a reader for pointing this out.

*** In a typical course, you'll usually play with polynomials of degree two, three, four, and five since the group structures of $S_2$, $S_3$, $S_4$, and $S_5$ are more manageable than that of $S_n$ for $n\geq 6$.