When Equations Have Secrets: A Woody Calculus Introduction to Galois Theory

At first, algebra looks like the art of solving equations.

We learn to solve linear equations. Then quadratic equations. Eventually, we discover the quadratic formula:

\[
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.
\]

This formula feels like a victory. No matter what quadratic equation appears, as long as

\[
ax^2+bx+c=0,
\]

we have a universal procedure.

For a long time, mathematicians searched for similar formulas for higher-degree equations. There are formulas for cubic equations. There are formulas for quartic equations. These formulas are much more complicated than the quadratic formula, but they exist.

So the natural question became:

\[
\text{Is there a formula for the general quintic equation?}
\]

That is, can we solve every equation of the form

\[
ax^5+bx^4+cx^3+dx^2+ex+f=0
\]

using addition, subtraction, multiplication, division, and radicals?

The answer is no.

But the deeper question is not merely that the answer is no. The deeper question is:

\[
\textit{Why is the answer no?}
\]

Galois theory answers this question.

Galois theory says that every polynomial has a hidden structure. The roots may look like numbers, but beneath them lies a system of symmetries. These symmetries form a group. That group tells us what kind of equation we are really dealing with.

In other words, Galois theory does not simply ask:

\[
\text{What are the roots?}
\]

It asks:

\[
\text{What are the symmetries among the roots?}
\]

That shift changed algebra forever.

This article is written for undergraduate students studying Abstract Algebra, especially students beginning to connect groups, fields, polynomial roots, splitting fields, and solvability by radicals. This is one of the places where Abstract Algebra becomes breathtaking.

The Basic Idea: Roots Have Symmetry

Consider the polynomial

\[
f(x)=x^2-2.
\]

Its roots are

\[
\sqrt{2}
\quad \text{and} \quad
-\sqrt{2}.
\]

At first, these seem like two separate numbers. But from the point of view of rational numbers, they are almost indistinguishable.

Why?

Because both roots satisfy the same equation:

\[
x^2-2=0.
\]

If all we are allowed to use are rational numbers, there is no rational reason to prefer \(\sqrt{2}\) over \(-\sqrt{2}\). The two roots are algebraic twins.

There is a symmetry:

\[
\sqrt{2}\longmapsto -\sqrt{2},
\]

and this symmetry preserves every rational algebraic relationship.

This is the beginning of Galois theory.

\[
\boxed{\text{A Galois group is a group of symmetries of the roots of a polynomial.}}
\]

But to make this precise, we need the language of fields.

Fields: The Natural Home of Algebra

A field is a set in which we can add, subtract, multiply, and divide by nonzero elements.

The most familiar examples are

\[
\mathbb{Q}, \quad \mathbb{R}, \quad \mathbb{C}.
\]

The rational numbers \(\mathbb{Q}\) form a field because we can add, subtract, multiply, and divide rational numbers without leaving \(\mathbb{Q}\), as long as we do not divide by zero.

In Galois theory, we usually begin with a base field such as \(\mathbb{Q}\).

Then we enlarge it by adding roots of polynomials.

For example, \(\sqrt{2}\notin \mathbb{Q}\), so if we want a field containing both rational numbers and \(\sqrt{2}\), we write

\[
\mathbb{Q}(\sqrt{2}).
\]

This field contains all numbers of the form

\[
a+b\sqrt{2},
\]

where

\[
a,b\in \mathbb{Q}.
\]

So

\[
\mathbb{Q}(\sqrt{2})
=
\{a+b\sqrt{2}:a,b\in \mathbb{Q}\}.
\]

This is called a field extension.

If \(F\) is a field and \(E\) is a larger field containing \(F\), then \(E/F\) is called a field extension.

So

\[
\mathbb{Q}(\sqrt{2})/\mathbb{Q}
\]

is a field extension.

Automorphisms: Symmetries of Fields

An automorphism of a field \(E\) is a bijective function

\[
\sigma:E\to E
\]

such that for all \(a,b\in E\),

\[
\sigma(a+b)=\sigma(a)+\sigma(b),
\]

and

\[
\sigma(ab)=\sigma(a)\sigma(b).
\]

In other words, an automorphism is a way of rearranging the elements of a field while preserving addition and multiplication.

In Galois theory, we usually care about automorphisms that fix the base field.

Let \(E/F\) be a field extension. An automorphism

\[
\sigma:E\to E
\]

is said to fix \(F\) if

\[
\sigma(a)=a
\]

for every \(a\in F\).

So if we are working over \(\mathbb{Q}\), then our automorphisms must leave every rational number unchanged.

The First Example: \(x^2-2\)

Let

\[
f(x)=x^2-2.
\]

The roots are

\[
\sqrt{2},\quad -\sqrt{2}.
\]

The splitting field of \(f(x)\) over \(\mathbb{Q}\) is

\[
\mathbb{Q}(\sqrt{2}).
\]

Every element of this field has the form

\[
a+b\sqrt{2},
\]

where \(a,b\in\mathbb{Q}\).

Now suppose

\[
\sigma:\mathbb{Q}(\sqrt{2})\to \mathbb{Q}(\sqrt{2})
\]

is an automorphism fixing \(\mathbb{Q}\).

Since \(\sigma\) fixes rational numbers,

\[
\sigma(2)=2.
\]

Also,

\[
(\sqrt{2})^2=2.
\]

Applying \(\sigma\) gives

\[
\sigma\left((\sqrt{2})^2\right)=\sigma(2).
\]

Since \(\sigma\) preserves multiplication,

\[
\sigma(\sqrt{2})^2=2.
\]

Therefore,

\[
\sigma(\sqrt{2})=\sqrt{2}
\quad \text{or} \quad
\sigma(\sqrt{2})=-\sqrt{2}.
\]

So there are two possible automorphisms:

\[
\sigma_1(\sqrt{2})=\sqrt{2},
\]

and

\[
\sigma_2(\sqrt{2})=-\sqrt{2}.
\]

The first is the identity automorphism. The second sends

\[
a+b\sqrt{2}
\longmapsto
a-b\sqrt{2}.
\]

Thus

\[
\operatorname{Gal}(\mathbb{Q}(\sqrt{2})/\mathbb{Q})
=
\{\sigma_1,\sigma_2\}.
\]

This group has two elements, so it is isomorphic to \(C_2\), the cyclic group of order \(2\).

\[
\boxed{
\operatorname{Gal}(\mathbb{Q}(\sqrt{2})/\mathbb{Q})\cong C_2.
}
\]

This is the simplest example of a Galois group.

Splitting Fields

To understand the Galois group of a polynomial, we need the field where the polynomial fully factors.

Let \(F\) be a field, and let

\[
f(x)\in F[x].
\]

A splitting field of \(f(x)\) over \(F\) is the smallest field extension \(E/F\) such that \(f(x)\) factors completely into linear factors over \(E\).

For example,

\[
x^2-2
\]

does not split over \(\mathbb{Q}\) because

\[
\sqrt{2}\notin \mathbb{Q}.
\]

But it does split over \(\mathbb{Q}(\sqrt{2})\):

\[
x^2-2=(x-\sqrt{2})(x+\sqrt{2}).
\]

So the splitting field is

\[
\mathbb{Q}(\sqrt{2}).
\]

The Galois Group of a Polynomial

Let \(f(x)\in F[x]\), and let \(E\) be the splitting field of \(f(x)\) over \(F\). The Galois group of \(f(x)\) over \(F\) is

\[
\operatorname{Gal}(E/F),
\]

the group of field automorphisms of \(E\) that fix every element of \(F\).

So the Galois group is not just a random group. It is the group of algebraic symmetries of the splitting field.

Every automorphism in the Galois group must send roots to roots.

Why?

Suppose

\[
f(x)\in F[x],
\]

and \(\alpha\) is a root of \(f(x)\). Then

\[
f(\alpha)=0.
\]

If \(\sigma\) fixes \(F\), then it fixes the coefficients of \(f(x)\). Applying \(\sigma\) gives

\[
\sigma(f(\alpha))=\sigma(0)=0.
\]

Since \(\sigma\) fixes the coefficients,

\[
f(\sigma(\alpha))=0.
\]

Therefore, \(\sigma(\alpha)\) is also a root of \(f(x)\).

Thus:

\[
\boxed{
\text{Elements of the Galois group permute the roots of the polynomial.}
}
\]

This is why Galois groups are often viewed as subgroups of permutation groups.

If \(f(x)\) has roots

\[
\alpha_1,\alpha_2,\ldots,\alpha_n,
\]

then each automorphism permutes these roots. So the Galois group embeds into

\[
S_n,
\]

the symmetric group on \(n\) objects.

\[
\operatorname{Gal}(f)\leq S_n.
\]

But not every permutation is allowed. Only the permutations that preserve all algebraic relationships among the roots are allowed.

That is the secret.

The Real Meaning of a Galois Group

A Galois group measures which roots can be exchanged without changing the algebraic world seen by the base field.

If we work over \(\mathbb{Q}\), then rational numbers are fixed.

The roots may move, but rational truths cannot move.

So the Galois group answers this question:

\[
\boxed{
\text{Which permutations of the roots preserve all rational algebraic relationships?}
}
\]

This is why Galois theory is so powerful.

The polynomial gives us roots.

The roots give us a field.

The field gives us automorphisms.

The automorphisms give us a group.

The group tells us the hidden structure of the equation.

A Cubic Example: \(x^3-2\)

Now consider

\[
f(x)=x^3-2.
\]

Let

\[
\alpha=\sqrt[3]{2}.
\]

One root is

\[
\alpha.
\]

But over the complex numbers, there are three roots:

\[
\alpha,\quad \omega\alpha,\quad \omega^2\alpha,
\]

where

\[
\omega=e^{2\pi i/3}
=
-\frac12+\frac{\sqrt{3}}{2}i.
\]

The number \(\omega\) is a primitive cube root of unity, so

\[
\omega^3=1,
\]

but

\[
\omega\neq 1.
\]

Also,

\[
1+\omega+\omega^2=0.
\]

The splitting field of \(x^3-2\) over \(\mathbb{Q}\) is

\[
E=\mathbb{Q}(\alpha,\omega).
\]

This field contains all three roots:

\[
\alpha,\quad \omega\alpha,\quad \omega^2\alpha.
\]

An automorphism

\[
\sigma\in \operatorname{Gal}(E/\mathbb{Q})
\]

must send roots of \(x^3-2\) to roots of \(x^3-2\).

Therefore,

\[
\sigma(\alpha)\in \{\alpha,\omega\alpha,\omega^2\alpha\}.
\]

Also, \(\omega\) satisfies

\[
\omega^2+\omega+1=0.
\]

So \(\sigma(\omega)\) must be another root of

\[
x^2+x+1.
\]

Therefore,

\[
\sigma(\omega)\in\{\omega,\omega^2\}.
\]

There are three choices for \(\sigma(\alpha)\) and two choices for \(\sigma(\omega)\), producing six automorphisms.

Thus

\[
|\operatorname{Gal}(E/\mathbb{Q})|=6.
\]

Since the Galois group permutes the three roots, it is a subgroup of \(S_3\). Since it has six elements,

\[
\operatorname{Gal}(E/\mathbb{Q})\cong S_3.
\]

So

\[
\boxed{
\operatorname{Gal}(x^3-2\text{ over }\mathbb{Q})\cong S_3.
}
\]

This is already much richer than the quadratic example.

The Discriminant Connection

For a cubic polynomial, the discriminant gives a quick way to distinguish between \(A_3\) and \(S_3\) in many cases.

Suppose

\[
f(x)\in \mathbb{Q}[x]
\]

is an irreducible cubic.

Then the Galois group of \(f(x)\) over \(\mathbb{Q}\) is a transitive subgroup of \(S_3\).

The transitive subgroups of \(S_3\) are:

\[
A_3
\quad \text{and} \quad
S_3.
\]

The discriminant tells us which one occurs.

If \(f(x)\in\mathbb{Q}[x]\) is an irreducible cubic and the discriminant of \(f(x)\) is a square in \(\mathbb{Q}\), then

\[
\operatorname{Gal}(f)\cong A_3.
\]

If the discriminant is not a square in \(\mathbb{Q}\), then

\[
\operatorname{Gal}(f)\cong S_3.
\]

For

\[
f(x)=x^3-2,
\]

the discriminant is

\[
\Delta=-108.
\]

Since \(-108\) is not a square in \(\mathbb{Q}\), we get

\[
\operatorname{Gal}(x^3-2)\cong S_3.
\]

This shows something important:

\[
\boxed{
\text{The Galois group is not guessed from the degree alone.}
}
\]

It depends on the hidden algebraic relationships among the roots.

Normal and Separable Extensions

The cleanest version of Galois theory works for extensions that are both normal and separable.

Since most undergraduate introductions focus on fields of characteristic zero, especially \(\mathbb{Q}\), separability is usually automatic.

A field extension \(E/F\) is separable if every element of \(E\) is the root of a separable polynomial over \(F\).

A polynomial is separable if it has no repeated roots in its splitting field.

Over fields of characteristic zero, such as

\[
\mathbb{Q},\quad \mathbb{R},\quad \mathbb{C},
\]

every irreducible polynomial is separable.

So over \(\mathbb{Q}\), separability is not the dangerous part.

Normality is the more visible condition.

A field extension \(E/F\) is normal if every irreducible polynomial in \(F[x]\) that has one root in \(E\) splits completely over \(E\).

A splitting field is the standard example of a normal extension.

A finite field extension \(E/F\) is called a Galois extension if it is normal and separable.

So if \(E\) is the splitting field of a polynomial over \(\mathbb{Q}\), then \(E/\mathbb{Q}\) is a Galois extension, provided the polynomial has no repeated roots. In characteristic zero, this is the usual situation.

For finite Galois extensions, the size of the Galois group matches the degree of the field extension.

\[
\boxed{
|\operatorname{Gal}(E/F)|=[E:F].
}
\]

This fact is one reason Galois theory is so powerful. The algebraic degree of the field extension is reflected exactly in the number of symmetries of the extension.

The Fundamental Theorem of Galois Theory

This is the heart of the subject.

Galois theory builds a bridge between fields and groups.

On one side, we have intermediate fields:

\[
F\subseteq K\subseteq E.
\]

On the other side, we have subgroups of the Galois group:

\[
H\leq \operatorname{Gal}(E/F).
\]

The magic is that these two worlds correspond to each other.

Fundamental Theorem of Galois Theory. Let \(E/F\) be a finite Galois extension, and let

\[
G=\operatorname{Gal}(E/F).
\]

Then there is an inclusion-reversing correspondence between intermediate fields

\[
F\subseteq K\subseteq E
\]

and subgroups

\[
H\leq G.
\]

The correspondence is given by

\[
K \longmapsto \operatorname{Gal}(E/K),
\]

and

\[
H \longmapsto E^H,
\]

where

\[
E^H=\{x\in E:\sigma(x)=x\text{ for every }\sigma\in H\}.
\]

The field \(E^H\) is called the fixed field of \(H\).

It consists of all elements of \(E\) that are fixed by every automorphism in \(H\).

The correspondence is inclusion-reversing because larger fields correspond to smaller groups.

If

\[
F\subseteq K_1\subseteq K_2\subseteq E,
\]

then

\[
\operatorname{Gal}(E/K_2)\leq \operatorname{Gal}(E/K_1).
\]

Why?

Because fixing more elements gives an automorphism less freedom.

A bigger field means more elements must stay fixed.

More restrictions mean fewer automorphisms.

Thus:

\[
\boxed{
\text{Bigger fields correspond to smaller groups.}
}
\]

The Main Galois Correspondence Picture

The entire theorem can be remembered visually:

\[
\begin{array}{ccc}
\textbf{Fields} & & \textbf{Groups} \\[6pt]
E & \longleftrightarrow & \{e\} \\[6pt]
\cup & & \cap \\[6pt]
K & \longleftrightarrow & H \\[6pt]
\cup & & \cap \\[6pt]
F & \longleftrightarrow & G
\end{array}
\]

Notice the reversal.

The largest field \(E\) corresponds to the smallest subgroup \(\{e\}\).

The smallest field \(F\) corresponds to the largest subgroup \(G\).

Normal Subgroups and Galois Subextensions

There is one more important part of the Fundamental Theorem of Galois Theory.

Let \(E/F\) be a finite Galois extension, and let

\[
G=\operatorname{Gal}(E/F).
\]

If \(K\) is an intermediate field,

\[
F\subseteq K\subseteq E,
\]

then \(K/F\) is Galois if and only if

\[
\operatorname{Gal}(E/K)
\]

is a normal subgroup of \(G\).

In that case,

\[
\operatorname{Gal}(K/F)\cong G/\operatorname{Gal}(E/K).
\]

This is extremely important.

The first part of the Fundamental Theorem tells us that intermediate fields correspond to subgroups.

This second part tells us which intermediate fields are themselves Galois over the base field.

\[
\boxed{
K/F \text{ is Galois}
\quad \Longleftrightarrow \quad
\operatorname{Gal}(E/K)\triangleleft \operatorname{Gal}(E/F).
}
\]

So normal subgroups correspond to Galois intermediate extensions.

This is why normality matters so much.

A subgroup may exist, but if it is not normal, then the corresponding intermediate field is not Galois over the base field.

Fixed Fields: What Does a Subgroup See?

Let \(E/F\) be a Galois extension with Galois group \(G\).

If \(H\) is a subgroup of \(G\), then

\[
E^H
\]

is the set of elements that \(H\) cannot move.

This is a powerful idea.

A subgroup represents a limited set of symmetries.

The fixed field represents the quantities that remain unchanged under those symmetries.

So:

\[
\boxed{
\text{Groups describe motion. Fixed fields describe invariance.}
}
\]

This is one of the most beautiful ideas in Abstract Algebra.

Example: The Fixed Fields of \(\mathbb{Q}(\sqrt{2})\)

Let

\[
E=\mathbb{Q}(\sqrt{2}),
\]

and

\[
F=\mathbb{Q}.
\]

We found

\[
G=\operatorname{Gal}(E/F)\cong C_2.
\]

So

\[
G=\{1,\sigma\},
\]

where

\[
\sigma(\sqrt{2})=-\sqrt{2}.
\]

The subgroups of \(G\) are:

\[
\{1\}
\quad \text{and} \quad
G.
\]

The fixed field of \(\{1\}\) is all of \(E\):

\[
E^{\{1\}}=E=\mathbb{Q}(\sqrt{2}).
\]

The fixed field of \(G\) is:

\[
E^G=\mathbb{Q}.
\]

Indeed, if

\[
a+b\sqrt{2}
\]

is fixed by \(\sigma\), then

\[
a+b\sqrt{2}=a-b\sqrt{2}.
\]

Thus

\[
2b\sqrt{2}=0,
\]

so

\[
b=0.
\]

Therefore, the fixed elements are exactly the rational numbers.

A Small Field/Subgroup Lattice

\[
\begin{array}{c@{\qquad\qquad}c}
\textbf{Fields} & \textbf{Subgroups} \\[8pt]
\begin{array}{c}
\mathbb{Q}(\sqrt{2}) \\[10pt]
| \\[10pt]
\mathbb{Q}
\end{array}
&
\begin{array}{c}
\{e\} \\[10pt]
| \\[10pt]
C_2
\end{array}
\end{array}
\]

The field lattice has only two fields, and the subgroup lattice has only two subgroups.

The extension

\[
\mathbb{Q}(\sqrt{2})/\mathbb{Q}
\]

is the simplest possible nontrivial Galois extension.

Even in this simple example, we see the whole philosophy:

\[
\boxed{
\text{The base field is what remains fixed under all symmetries.}
}
\]

Example: The Galois Group of \(x^4-2\)

Now consider

\[
f(x)=x^4-2.
\]

Let

\[
\alpha=\sqrt[4]{2}.
\]

The roots are

\[
\alpha,\quad -\alpha,\quad i\alpha,\quad -i\alpha.
\]

The splitting field is

\[
E=\mathbb{Q}(\alpha,i).
\]

An automorphism fixing \(\mathbb{Q}\) must send \(\alpha\) to another root of \(x^4-2\), so

\[
\alpha\mapsto \alpha,\,-\alpha,\,i\alpha,\,-i\alpha.
\]

It must also send \(i\) to a root of

\[
x^2+1.
\]

So

\[
i\mapsto i
\quad \text{or} \quad
i\mapsto -i.
\]

The Galois group has order \(8\) and is isomorphic to the dihedral group of order \(8\):

\[
\operatorname{Gal}(x^4-2\text{ over }\mathbb{Q})\cong D_4.
\]

This is the symmetry group of a square.

That is not an accident.

The roots

\[
\alpha,\quad i\alpha,\quad -\alpha,\quad -i\alpha
\]

sit like the vertices of a square in the complex plane.

The Roots of \(x^4-2\) in the Complex Plane

\[
\begin{array}{ccccc}
& & i\alpha & & \\[8pt]
& \diagup & & \diagdown & \\[-2pt]
-\alpha & & & & \alpha \\[-2pt]
& \diagdown & & \diagup & \\[8pt]
& & -i\alpha & &
\end{array}
\]

The roots of \(x^4-2\) sit at the vertices of a square in the complex plane.

The Galois group acts by algebraic symmetries of this square.

This is why the Galois group is isomorphic to \(D_4\).

So once again:

\[
\boxed{
\text{Algebraic roots secretly organize themselves geometrically.}
}
\]

A More Detailed Lattice Example: \(x^3-2\)

Now return to

\[
f(x)=x^3-2.
\]

Let

\[
\alpha=\sqrt[3]{2}
\]

and

\[
\omega=e^{2\pi i/3}.
\]

The splitting field is

\[
E=\mathbb{Q}(\alpha,\omega).
\]

We showed that

\[
\operatorname{Gal}(E/\mathbb{Q})\cong S_3.
\]

The group \(S_3\) has six elements. Its subgroup structure is simple but rich enough to show the real power of Galois theory.

Since \(E/\mathbb{Q}\) is a finite Galois extension,

\[
[E:\mathbb{Q}]=|\operatorname{Gal}(E/\mathbb{Q})|=6.
\]

The subgroups of \(S_3\) are:

\[
\{e\},
\]

three subgroups of order \(2\),

\[
\langle (12)\rangle,\quad \langle (13)\rangle,\quad \langle (23)\rangle,
\]

one subgroup of order \(3\),

\[
A_3=\langle (123)\rangle,
\]

and the full group

\[
S_3.
\]

By the Fundamental Theorem of Galois Theory, these correspond to intermediate fields between

\[
\mathbb{Q}
\quad \text{and} \quad
E.
\]

Subgroups of order \(2\) correspond to intermediate fields of degree

\[
[\text{fixed field}:\mathbb{Q}]
=
\frac{|S_3|}{2}
=
3.
\]

Subgroups of order \(3\) correspond to intermediate fields of degree

\[
[\text{fixed field}:\mathbb{Q}]
=
\frac{|S_3|}{3}
=
2.
\]

Thus the three order-\(2\) subgroups correspond to the three cubic fields:

\[
\mathbb{Q}(\alpha),
\quad
\mathbb{Q}(\omega\alpha),
\quad
\mathbb{Q}(\omega^2\alpha).
\]

The order-\(3\) subgroup \(A_3\) corresponds to the quadratic field:

\[
\mathbb{Q}(\omega).
\]

The Field Lattice for \(x^3-2\)

\[
\begin{array}{c}
E=\mathbb{Q}(\alpha,\omega) \\[12pt]
\begin{array}{cccc}
\mathbb{Q}(\alpha) &
\mathbb{Q}(\omega\alpha) &
\mathbb{Q}(\omega^2\alpha) &
\mathbb{Q}(\omega)
\end{array} \\[12pt]
\mathbb{Q}
\end{array}
\]

All four intermediate fields sit at the same level. None of them contains another one.

This diagram shows the intermediate fields between \(\mathbb{Q}\) and the splitting field

\[
E=\mathbb{Q}(\alpha,\omega).
\]

The three cubic fields come from choosing one of the three roots:

\[
\alpha,\quad \omega\alpha,\quad \omega^2\alpha.
\]

The quadratic field

\[
\mathbb{Q}(\omega)
\]

comes from adjoining the primitive cube root of unity.

All four intermediate fields are incomparable. None of them contains another one. They all sit strictly between

\[
\mathbb{Q}
\quad \text{and} \quad
E.
\]

The Corresponding Subgroup Lattice

Because the Galois correspondence reverses inclusion, the subgroup lattice is upside down relative to the field lattice.

\[
\begin{array}{c}
\{e\} \\[12pt]
\begin{array}{cccc}
\langle(12)\rangle &
\langle(13)\rangle &
\langle(23)\rangle &
A_3=\langle(123)\rangle
\end{array} \\[12pt]
S_3
\end{array}
\]

The subgroup lattice is inclusion-reversing relative to the field lattice.

The top of the field lattice is the splitting field \(E\).

The top of the subgroup lattice is the trivial group \(\{e\}\).

The bottom of the field lattice is the base field \(\mathbb{Q}\).

The bottom of the subgroup lattice is the full group \(S_3\).

This is the inclusion-reversing nature of Galois theory.

\[
\boxed{
\text{Fields go up while groups go down.}
}
\]

Normality in the \(x^3-2\) Example

Now we can see the normal subgroup idea in action.

The subgroup

\[
A_3\triangleleft S_3
\]

is normal. It corresponds to the intermediate field

\[
\mathbb{Q}(\omega).
\]

Therefore,

\[
\mathbb{Q}(\omega)/\mathbb{Q}
\]

is a Galois extension.

In fact,

\[
\operatorname{Gal}(\mathbb{Q}(\omega)/\mathbb{Q})
\cong
S_3/A_3
\cong C_2.
\]

On the other hand, the subgroups of order \(2\),

\[
\langle(12)\rangle,\quad
\langle(13)\rangle,\quad
\langle(23)\rangle,
\]

are not normal in \(S_3\).

These correspond to the cubic fields

\[
\mathbb{Q}(\alpha),
\quad
\mathbb{Q}(\omega\alpha),
\quad
\mathbb{Q}(\omega^2\alpha).
\]

Therefore, these cubic extensions are not Galois over \(\mathbb{Q}\).

For example,

\[
\mathbb{Q}(\alpha)/\mathbb{Q}
\]

is not Galois because it contains one root of

\[
x^3-2,
\]

but not all three roots.

It contains

\[
\alpha,
\]

but it does not contain

\[
\omega\alpha
\quad \text{or} \quad
\omega^2\alpha.
\]

So the polynomial

\[
x^3-2
\]

does not split over \(\mathbb{Q}(\alpha)\).

This is exactly what normality detects.

\[
\boxed{
\text{Normal subgroups correspond to intermediate fields that are Galois over the base field.}
}
\]

Solvability by Radicals

Now we return to the original question.

What does it mean to solve an equation by radicals?

Roughly, it means that the roots can be expressed using:

\[
+,\quad -,\quad \times,\quad \div,\quad \sqrt{\phantom{x}},\quad \sqrt[3]{\phantom{x}},\quad \sqrt[4]{\phantom{x}},\quad \ldots
\]

For example, the quadratic formula solves every quadratic equation by radicals:

\[
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.
\]

The cubic formula solves every cubic equation by radicals.

The quartic formula solves every quartic equation by radicals.

But there is no general radical formula for degree \(5\) polynomials.

This result is known as the Abel–Ruffini theorem.

Galois theory explains why.

The reason is group-theoretic.

A group \(G\) is called solvable if there exists a chain of subgroups

\[
\{e\}=G_0\triangleleft G_1\triangleleft \cdots \triangleleft G_n=G
\]

such that each quotient

\[
G_{k+1}/G_k
\]

is abelian.

The word “solvable” is not a coincidence. It is directly connected to solving equations.

Over a field of characteristic zero, a polynomial is solvable by radicals if and only if its Galois group is a solvable group.

This theorem is one of the great moments in mathematics.

It says:

\[
\boxed{
\text{An equation is solvable by radicals exactly when its symmetry group is solvable.}
}
\]

This is stunning.

A question about formulas becomes a question about groups.

Why Quadratics, Cubics, and Quartics Work

The general polynomial of degree \(n\) has Galois group

\[
S_n
\]

in the generic case.

Here, “generic” means that we are talking about the general polynomial with no special algebraic relationships among its coefficients or roots. Individual polynomials can have smaller Galois groups.

For example,

\[
x^3-3x+1
\]

has Galois group

\[
C_3,
\]

not \(S_3\).

So the statement is not that every degree-\(n\) polynomial has Galois group \(S_n\). The point is that the fully general degree-\(n\) equation has maximal symmetry, and that maximal symmetry is represented by \(S_n\).

For small \(n\), the symmetric groups are solvable:

\[
S_1,\quad S_2,\quad S_3,\quad S_4
\]

are solvable groups.

This matches the classical facts:

\[
\text{degree }2:\text{ quadratic formula exists},
\]

\[
\text{degree }3:\text{ cubic formula exists},
\]

\[
\text{degree }4:\text{ quartic formula exists}.
\]

But for degree \(5\), something changes.

The group

\[
S_5
\]

is not solvable.

The reason is connected to the alternating group

\[
A_5.
\]

The group \(A_5\) is simple and nonabelian. This means it has no nontrivial proper normal subgroups, and it is not abelian.

This prevents \(S_5\) from having the kind of normal subgroup chain required for solvability.

Therefore,

\[
S_5
\]

is not solvable.

So the general quintic cannot be solved by radicals.

\[
\boxed{
\text{The obstruction to the quintic formula is the nonsolvability of }S_5.
}
\]

This does not mean no quintic equation can be solved by radicals.

Some quintics can be solved by radicals.

For example,

\[
x^5-2=0
\]

has the obvious real root

\[
\sqrt[5]{2}.
\]

The statement is more subtle:

\[
\boxed{
\text{There is no single radical formula that solves every quintic equation.}
}
\]

The general quintic has too much symmetry.

Its Galois group is usually too complicated.

The Deep Pattern

Here is the structure of the whole story:

\[
\text{Polynomial}
\longrightarrow
\text{Roots}
\longrightarrow
\text{Splitting Field}
\longrightarrow
\text{Automorphisms}
\longrightarrow
\text{Galois Group}
\longrightarrow
\text{Solvability}.
\]

The Galois Theory Flowchart

\[
\begin{array}{ccccccccc}
\text{Polynomial}
& \longrightarrow &
\text{Roots}
& \longrightarrow &
\text{Splitting Field}
& \longrightarrow &
\text{Automorphisms}
& \longrightarrow &
\text{Galois Group}
\\[8pt]
& & & & & & & & \downarrow \\[8pt]
& & & & & & & &
\text{Solvability}
\end{array}
\]

That is Galois theory in one picture.

The polynomial gives us roots.

The roots generate a field.

The field has symmetries.

The symmetries form a group.

The group tells us whether the equation can be solved by radicals.

This is why Galois theory is not just another chapter of Abstract Algebra.

It is the moment when algebra becomes structural.

A Concrete Computation: \(x^2-5\)

Let

\[
f(x)=x^2-5.
\]

The roots are

\[
\sqrt{5}
\quad \text{and} \quad
-\sqrt{5}.
\]

The splitting field is

\[
E=\mathbb{Q}(\sqrt{5}).
\]

The Galois group has two automorphisms:

\[
\sigma_1(\sqrt{5})=\sqrt{5},
\]

and

\[
\sigma_2(\sqrt{5})=-\sqrt{5}.
\]

Thus

\[
\operatorname{Gal}(E/\mathbb{Q})\cong C_2.
\]

The fixed field of the entire Galois group is

\[
E^{\operatorname{Gal}(E/\mathbb{Q})}=\mathbb{Q}.
\]

So the only elements that remain unchanged under all symmetries are rational numbers.

Again, the base field is the field of invariants.

A Concrete Computation: \(x^2+1\)

Let

\[
f(x)=x^2+1.
\]

The roots are

\[
i
\quad \text{and} \quad
-i.
\]

The splitting field over \(\mathbb{Q}\) is

\[
\mathbb{Q}(i).
\]

The nontrivial automorphism is complex conjugation:

\[
i\mapsto -i.
\]

So

\[
\operatorname{Gal}(\mathbb{Q}(i)/\mathbb{Q})\cong C_2.
\]

This example reveals something important.

Galois automorphisms generalize complex conjugation.

Complex conjugation is the symmetry that swaps

\[
i
\quad \text{and} \quad
-i
\]

while fixing every rational number.

Galois theory extends this idea to more complicated algebraic worlds.

A Concrete Computation: \(x^3-3x+1\)

Consider

\[
f(x)=x^3-3x+1.
\]

This polynomial is irreducible over \(\mathbb{Q}\) by the rational root test.

Indeed, the only possible rational roots are

\[
\pm 1.
\]

But

\[
f(1)=1-3+1=-1\neq 0,
\]

and

\[
f(-1)=-1+3+1=3\neq 0.
\]

So \(f(x)\) is irreducible over \(\mathbb{Q}\).

For a cubic polynomial

\[
ax^3+bx^2+cx+d,
\]

the discriminant is

\[
\Delta=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd.
\]

For

\[
f(x)=x^3-3x+1,
\]

we have

\[
a=1,\quad b=0,\quad c=-3,\quad d=1.
\]

Thus

\[
\Delta
=
0^2(-3)^2
–
4(1)(-3)^3
–
4(0)^3(1)
–
27(1)^2(1)^2
+
18(1)(0)(-3)(1).
\]

So

\[
\Delta
=
0
–
4(-27)
–
0
–
27
+
0.
\]

Therefore,

\[
\Delta=108-27=81.
\]

Since

\[
81=9^2,
\]

the discriminant is a square in \(\mathbb{Q}\).

Because \(f(x)\) is irreducible cubic and its discriminant is a square, its Galois group is

\[
A_3.
\]

Since

\[
A_3\cong C_3,
\]

we get

\[
\boxed{
\operatorname{Gal}(x^3-3x+1\text{ over }\mathbb{Q})\cong C_3.
}
\]

This example contrasts beautifully with

\[
x^3-2,
\]

whose Galois group is

\[
S_3.
\]

Both are irreducible cubics.

But one has Galois group \(C_3\), and the other has Galois group \(S_3\).

The degree is the same.

The hidden symmetry is different.

Why This Matters for Abstract Algebra Students

Galois theory brings together many ideas that may seem separate at first:

• fields,
• field extensions,
• roots of polynomials,
• irreducibility,
• automorphisms,
• groups,
• subgroups,
• normal subgroups,
• quotient groups,
• solvable groups,
• symmetric groups,
• alternating groups.

At first, these topics may feel disconnected.

Galois theory reveals that they are part of one story.

A polynomial is not merely an expression.

It is a doorway into a field.

That field has symmetries.

Those symmetries form a group.

The structure of that group tells us what the polynomial can and cannot do.

This is why Galois theory is one of the crowning achievements of Abstract Algebra.

The Woody Calculus Perspective: Structure Before Computation

In most math classes, students are trained to compute.

Compute the derivative.

Compute the integral.

Compute the determinant.

Compute the eigenvalues.

Compute the roots.

But advanced mathematics requires a deeper skill:

\[
\boxed{
\text{See the structure before you compute.}
}
\]

Galois theory is one of the greatest examples of this principle.

Instead of forcing a formula onto every polynomial, Galois theory asks:

\[
\text{What structure controls this equation?}
\]

Instead of blindly solving, it studies the symmetries.

Instead of focusing only on the roots, it studies the relationships among the roots.

This is the same transition students must make in Abstract Algebra.

Early math often asks:

\[
\text{Can you get the answer?}
\]

Abstract Algebra asks:

\[
\text{What structure makes the answer possible?}
\]

That is a higher level of thinking.

A Summary of the Main Ideas

1. A field is a number system where addition, subtraction, multiplication, and division work.
2. A field extension adds new algebraic elements to a base field.
3. The splitting field of a polynomial is the smallest field containing all of its roots.
4. A field automorphism is a symmetry preserving addition and multiplication.
5. A Galois group is the group of automorphisms of the splitting field that fix the base field.
6. Elements of the Galois group permute the roots of the polynomial.
7. Not every permutation is allowed. Only permutations preserving algebraic relationships are allowed.
8. For finite Galois extensions, the size of the Galois group equals the degree of the extension.
9. The Fundamental Theorem of Galois Theory connects intermediate fields with subgroups of the Galois group.
10. The correspondence between fields and subgroups is inclusion-reversing.
11. Normal subgroups correspond to intermediate fields that are Galois over the base field.
12. A polynomial is solvable by radicals precisely when its Galois group is solvable.
13. The Abel–Ruffini theorem says there is no general radical formula for the quintic.
14. The general quintic has Galois group \(S_5\), which is not solvable.
15. Therefore, there is no general radical formula for all quintic equations.

Final Reflection: Equations Have Secrets

A polynomial may look simple:

\[
x^5-4x+2.
\]

But hidden inside that expression is a universe of structure.

There are roots.

There are fields.

There are automorphisms.

There are groups.

There are subgroups.

There are fixed fields.

There are symmetries that reveal what can be solved and what cannot.

This is the beauty of Galois theory.

It shows us that equations have secrets.

Some equations open easily.

Some equations resist.

Some equations cannot be solved by the methods we hoped would always work.

But even that resistance is not chaos.

It has structure.

It has symmetry.

It has a reason.

And that is the deeper lesson.

Mathematics is not only about finding answers.

It is about discovering why certain answers are possible in the first place.

\[
\boxed{
\text{Galois theory is the study of the hidden symmetries that decide the fate of equations.}
}
\]

Need Help With Abstract Algebra?

If you are taking Abstract Algebra, Galois Theory, Linear Algebra, Real Analysis, Differential Equations, Calculus 2, Calculus 3, or another advanced mathematics course, Woody Calculus was built to help serious students train with structure.

Inside the Woody Calculus Mastery Lab, students get access to detailed step-by-step instruction, exam-style solutions, structured repetition, and the kind of pattern recognition that makes advanced mathematics much more manageable.

You can also learn more at BrianWoody.com, read the 5-star Google reviews, or visit the Woody Calculus Private Professor community on Skool.