NOTES ON STATISTICS, PROBABILITY and MATHEMATICS

Galois group of a finite field extension \(\text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) \cong \mathbb{Z}/n\mathbb{Z}\) versus

the absolute Galois group of all finite extensions \(G_{\mathbb{F}_p} = \text{Gal}(\bar{\mathbb{F}}_p/\mathbb{F}_p) \cong \widehat{\mathbb{Z}}\):

The Galois group of a single finite extension \(\text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) \cong \mathbb{Z}/n\mathbb{Z}:\)

The Galois group of a single finite extension \(\text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) \cong \mathbb{Z}/n\mathbb{Z}\) is the finite/local view. We are restricting our vision to a single “sub-universe” generated by one polynomial (like a degree-3 polynomial giving \(\mathbb{F}_{p^3}\) in the example below). In that local sub-universe, every symmetry is just an integer power of Frobenius modulo \(n\).

Let’s work with a concrete base field: \(\bf\mathbb F_p=\mathbb{F}_2 = \{0, 1\}\) (so \(p = 2\)).

Consider the polynomial:

\[P(x) = x^3 + x + 1\]

Is \(P(x)\) irreducible over \(\mathbb{F}_2\)?

Plug in \(x = 0 \implies 0^3 + 0 + 1 = 1 \neq 0\).

Plug in \(x = 1 \implies 1^3 + 1 + 1 = 1 \neq 0\).

Neither \(0\) nor \(1\) is a root. Since it has degree \(3\) and no roots in \(\mathbb{F}_2\), \(P(x)\) cannot be factored. It is irreducible.

Since \(P(x)\) has no roots in \(\mathbb{F}_2,\) we invent a formal symbol \(\alpha\) and declare:

\[\alpha^3 + \alpha + 1 = 0 \implies \alpha^3 = \alpha + 1\]

We adjoin only this single root \(\alpha\) to \(\mathbb{F}_2\). We write the new set as \(\mathbb{F}_2(\alpha)\). What does an element in \(\mathbb{F}_2(\alpha)\) look like? Any power of \(\alpha\) higher than \(2\) can be simplified using \(\alpha^3 = \alpha + 1\). So every element in \(\mathbb{F}_2(\alpha)\) can be written uniquely as a quadratic expression in \(\alpha\): (🪁)

\[c_0 + c_1\alpha + c_2\alpha^2 \quad \text{where } c_0, c_1, c_2 \in \{0, 1\}\] How many choices of elements are there?

\(c_0\) has 2 choices (\(0\) or \(1\))

\(c_1\) has 2 choices (\(0\) or \(1\))

\(c_2\) has 2 choices (\(0\) or \(1\))

Total elements \(= 2 \times 2 \times 2 = 2^3 = 8.\) This is why the size is \(p^3 = 2^3 = 8\). The exponent \(3\) comes from the degree of the polynomial, which determines the number of coefficients \((c_0, c_1, c_2)\) we need to write down any element.

Note that we really don’t deal with any numerical value of . remains a formal symbol (an indeterminate satisfying \(\alpha^3 + \alpha + 1 = 0\)). What we really manipulate is the coefficient for \(\alpha\) and any other power of \(\alpha\) that we may need to include to build a field (closure under addition and multiplication): the coefficients of the polynomial expressions \(c_0 + c_1\alpha + c_2\alpha^2\). Each coefficient is an element of the base field \(\mathbb Z_p=\{0,1,2,\dots,p-1\}.\) The elements of the extension are determined entirely by the vector of coefficients \((c_0, c_1, \dots, c_{n-1})\) over the base field.

We started with \(P(x) = x^3 + x + 1\), which should have \(3\) roots. Yet we only adjoined one root, \(\alpha\). Where are the other two? Here is the magic of finite fields: they were automatically created inside \(\mathbb{F}_2(\alpha)\)!

Because we are in characteristic \(2,\) applying the Frobenius map (squaring \(x \mapsto x^2\)) to our single root \(\alpha\) produces the other roots:

First root: \(\alpha\)

Second root: \(\alpha^2\)

Third root: \((\alpha^2)^2 = \alpha^4\)

Let’s test if \(\alpha^2\) is really a root of \(x^3 + x + 1\):

\[( \alpha^2 )^3 + (\alpha^2) + 1 = \alpha^6 + \alpha^2 + 1\]

Using \(\alpha^3 = \alpha + 1\), square both sides: \(\alpha^6 = (\alpha + 1)^2 = \alpha^2 + 1\). Substitute \(\alpha^6\) back in:

\[(\alpha^2 + 1) + \alpha^2 + 1 = 2\alpha^2 + 2 = 0 \pmod 2\]

It equals 0! So \(\alpha^2\) is a root! If you square it once more, \(\alpha^4 = \alpha(\alpha^3) = \alpha(\alpha + 1) = \alpha^2 + \alpha\), which is the 3rd root. If you square it a 3rd time, \(\alpha^8 = \alpha\), bringing you back to the 1st root.

Whether we start with a base finite field \(\mathbb{F}_q\) (where \(q = p^k\) is a prime power), or just with \(\mathbb{F}_p\), for every degree \(n \ge 1\), there is exactly one extension field of \(\mathbb{F}_q\) of degree \(n\) up to isomorphism, denoted \(\mathbb{F}_{q^n}\). Just as over \(\mathbb{F}_p\), we can construct \(\mathbb{F}_{q^n}\) by picking any irreducible polynomial of degree \(n\) over \(\mathbb{F}_q\). It does not matter which polynomial we choose: all irreducible polynomials of degree \(n\) over \(\mathbb{F}_q\) factor completely inside the same master splitting field defined by:

\[x^{q^n} - x = 0\]

because the roots of any such irreducible polynomial naturally generate a finite field extension of degree \(n,\) which is uniquely isomorphic to \(\mathbb{F}_{q^{n}},\) the exact set of roots forming the splitting field of \(x^{q^n} - x.\)

Any field with \(q^n\) elements is precisely the root set (splitting field) of the single polynomial:

\[f(x) = x^{q^n} - x\]

Because the multiplicative group \(\mathbb{F}_{p^n}^\times\) is finite and cyclic (order \(p^n - 1\)), every non-zero element \(x\) satisfies \(x^{p^n-1} = 1\), making every element in the field a root of \(x^{p^n} - x\). No matter which irreducible polynomial \(P(x) \in \mathbb{F}_p[x]\) of degree \(n\) we pick to adjoin a root \(\alpha\), the generated field \(\mathbb{F}_p(\alpha)\) must have \(p^n\) elements. Since splitting fields are unique up to isomorphism, every degree-\(n\) extension collapses into the exact same field \(\mathbb{F}_{p^n}\).

If we start at \(K = \mathbb{F}_{p^k}\), the relevant Frobenius automorphism is the \(q\)-Frobenius map:

\[\sigma_q : \bar{\mathbb{F}}_p \to \bar{\mathbb{F}}_p, \quad x \mapsto x^q \quad (\text{where } q = p^k)\]

The map \(\sigma_q(x) = x^q\) fixes an element \(x\) if and only if:

\[x^q = x \quad \iff \quad x^q - x = 0\]

The set of all solutions to \(x^q - x = 0\) in \(\bar{\mathbb{F}}_p\) is precisely your base field \(\mathbb{F}_q = \mathbb{F}_{p^k}\). So \(\sigma_q\) leaves every element inside \(\mathbb{F}_{p^k}\) completely untouched:

\[\sigma_q(\alpha) = \alpha \quad \text{for all } \alpha \in \mathbb{F}_{p^k}\]

Every element \(\beta \in \bar{\mathbb{F}}_p\) that lies outside \(\mathbb{F}_{p^k}\) is moved by \(\sigma_q\):

\[\sigma_q(\beta) = \beta^q \neq \beta \quad \text{for all } \beta \notin \mathbb{F}_{p^k}\]

To fix an element in a larger subfield \(\mathbb{F}_{q^n} = \mathbb{F}_{p^{k n}}\) (where \(n > 1\)), we have to apply \(\sigma_q\) a total of \(n\) times:\[(\sigma_q)^n(\beta) = \beta^{q^n} = \beta\]

The absolute Galois group of all finite extensions \(G_{\mathbb{F}_p} = \text{Gal}(\bar{\mathbb{F}}_p/\mathbb{F}_p) \cong \widehat{\mathbb{Z}}\)

Now we are ready to make the final leap directly from our root \(\alpha\) to the absolute Galois group \(G_{\mathbb{F}_2}\).

Here is the step-by-step path from a single root \(\alpha\) to the infinite, global picture.

Step 1: The root \(\alpha\) and the “rotary dial” on \(\mathbb{F}_{2^3}\)

As we saw, adjoining a single root \(\alpha\) of a degree-3 irreducible polynomial gives the field \(\mathbb{F}_{2^3}\) of size \(2^3 = 8\). The \(3\) roots of that polynomial are:

\[\alpha \xrightarrow{\text{Frob}_2} \alpha^2 \xrightarrow{\text{Frob}_2} \alpha^4 \xrightarrow{\text{Frob}_2} \alpha^8 = \alpha\]

The Frobenius map \(\text{Frob}_2(x) = x^2\) acts like a rotary dial on these roots.

Turn the dial 0 clicks (\(\text{Frob}_2^0 = \text{id}\)): \(\alpha \mapsto \alpha\)

Turn the dial 1 click (\(\text{Frob}_2^1\)): \(\alpha \mapsto \alpha^2\)

Turn the dial 2 clicks (\(\text{Frob}_2^2\)): \(\alpha \mapsto \alpha^4\)

Turn the dial 3 clicks (\(\text{Frob}_2^3\)): \(\alpha \mapsto \alpha^8 = \alpha\) (Back to start!)

So for the extension \(\mathbb{F}_{2^3}\), the Galois group \(\text{Gal}(\mathbb{F}_{2^3}/\mathbb{F}_2)\) is just the 3 settings on this rotary dial:

\[\{0, 1, 2\} \cong \mathbb{Z}/3\mathbb{Z}\]

To build the full algebraic closure \(\bar{\mathbb{F}}_2\) (which contains every possible root of every polynomial over \(\mathbb{F}_2\)), we adjoin roots of polynomials of all degrees \((n=1,2,...):\)

A degree-2 root \(\alpha_2\) gives \(\mathbb{F}_{2^2}\) \(\implies\) dial with 2 settings (\(\mathbb{Z}/2\mathbb{Z}\))

Our degree-3 root \(\alpha_3\) gives \(\mathbb{F}_{2^3}\) \(\implies\) dial with 3 settings (\(\mathbb{Z}/3\mathbb{Z}\))

A degree-4 root \(\alpha_4\) gives \(\mathbb{F}_{2^4}\) \(\implies\) dial with 4 settings (\(\mathbb{Z}/4\mathbb{Z}\))

A degree-6 root \(\alpha_6\) gives \(\mathbb{F}_{2^6}\) \(\implies\) dial with 6 settings (\(\mathbb{Z}/6\mathbb{Z}\))

Here is where the connection happens: Some fields sit inside larger fields. Because \(3\) divides \(6\), our field \(\mathbb{F}_{2^3}\) is a subfield of \(\mathbb{F}_{2^6}\). That means our root \(\alpha_3\) (from \(\mathbb{F}_{2^3}\)) also lives inside \(\mathbb{F}_{2^6}\)!

If an automorphism \(\sigma\) turns the dial on the big field \(\mathbb{F}_{2^6}\) by \(k_6\) clicks (\(x \mapsto x^{2^{k_6}}\)), what does it do to our root \(\alpha_3\)? Since \(\alpha_3\) comes from a degree-3 polynomial, turning its dial 3 times returns it to where it started (\(\alpha_3^{2^3} = \alpha_3\)). Therefore, turning the big dial \(k_6\) times acts on \(\alpha_3\) as:

\[k_3 \equiv k_6 \pmod 3\]

Any symmetry of the entire field \(\bar{\mathbb{F}}_2\) is a sequence of dial settings \((k_1, k_2, k_3, k_4, \dots)\) across all degree fields, subject to the rule that whenever \(m\) divides \(n\), \(k_n \pmod m = k_m\).

An element of the absolute Galois group \(G_{\mathbb{F}_2} = \text{Gal}(\bar{\mathbb{F}}_2 / \mathbb{F}_2)\) is an automorphism that acts on every root simultaneously.

Pick a single integer like \(k = 5\).

It turns every dial by 5 clicks:

On our root \(\alpha_3 \in \mathbb{F}_{2^3}\): turns dial \(5 \equiv 2 \pmod 3\) times (\(\alpha_3 \mapsto \alpha_3^4\)).

On a root in \(\mathbb{F}_{2^2}\): turns dial \(5 \equiv 1 \pmod 2\) times.

This simple “turn every dial \(k\) times” generates the countable subgroup \(\mathbb{Z} \subset G_{\mathbb{F}_2}\).

Is every compatible sequence of dial choices \((k_1, k_2, k_3, \dots)\) achievable by turning every dial a fixed integer \(k\) times?

No. By the Chinese Remainder Theorem, you can choose a sequence of dial settings—say \(1 \pmod 2\), \(2 \pmod 3\), \(3 \pmod 5\), \(4 \pmod 7\), and so on, where the required number of clicks grows infinitely. No single finite integer \(k\) can equal all those congruences simultaneously, yet when given any specific root (like our \(\alpha_3\)), this choice hands you a clear, unambiguous answer (\(k_3 = 2\), so \(\alpha_3 \mapsto \alpha_3^4\)).

We go from a single root \(\alpha\) to the absolute Galois group by recognizing that \(\alpha\) lives in \(\mathbb{F}_{2^3}\) where Frobenius acts as a 3-setting dial (\(\mathbb{Z}/3\mathbb{Z}\)); taking all finite extensions together gives an infinite collection of compatible dials (\(\varprojlim \mathbb{Z}/n\mathbb{Z}\)); and the absolute Galois group \(G_{\mathbb{F}_2} = \widehat{\mathbb{Z}}\) is the complete set of all valid, compatible ways to set those infinite dials simultaneously.

The absolute Galois group of a finite field \(\mathbb{F}_p\) is the inverse limit over all finite extension degrees \(n\):

\[G_{\mathbb{F}_p} = \text{Gal}(\bar{\mathbb{F}}_p / \mathbb{F}_p) \cong \varprojlim \mathbb{Z}/n\mathbb{Z} = \widehat{\mathbb{Z}}\]

At this point the all extensions degrees makes it sound like there is one extension for every degree, but aren’t there infinite polynomials for each degree?

For any finite field \(\mathbb{F}_{p}\) and any positive integer \(n,\) there is only one extension field of degree \(n\) up to isomorphism, not an infinite number (😱). All irreducible polynomials of degree \(n\) over \(\mathbb{F}_{p}\) yield the exact same splitting field, which is why the inverse limit accurately reflects a single extension for each degree \(n.\)

Just as with Galois group of finite field extensions, to build the algebraic closure \(\bar{\mathbb{F}}_p\) — and thus understand its absolute Galois group \(G_{\mathbb{F}_p} \cong \widehat{\mathbb{Z}}\) — we need one extension of every degree \(n\), which we can construct by picking one irreducible polynomial of each degree \(n\). It does not matter which irreducible polynomial of degree \(n\) we pick, because all irreducible polynomials of degree \(n\) over \(\mathbb{F}_p\) yield the exact same splitting field \(\mathbb{F}_{p^n}\) up to isomorphism.

If we choose \(P(x)\) of degree \(n\) and our friend chooses \(Q(x)\) of degree \(n\) (both irreducible over \(\mathbb{F}_p\)), both \(P(x)\) and \(Q(x)\) divide the master polynomial

\[x^{p^n} - x\]

Adjoining a root of \(P(x)\) generates the unique field \(\mathbb{F}_{p^n}\). Adjoining a root of \(Q(x)\) also generates \(\mathbb{F}_{p^n}\). The automorphism group \(\text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)\) is always the unique cyclic group \(\mathbb{Z}/n\mathbb{Z}\) generated by Frobenius (\(x \mapsto x^p\)), regardless of how you named the elements or which polynomial generated the basis.

While the extensions are unique?

The reason all irreducible polynomials of degree \(n\) generate the same field is that they all share roots within a single, larger “master” polynomial:

\[x^{p^n} - x\]

Every irreducible polynomial of degree \(n\) over \(\mathbb{F}_{p}\) is a clean, algebraic factor of this master polynomial. Therefore, any root of your chosen polynomial is automatically a root of \(x^{p^n} - x,\) trapping it inside the exact same field.

Concrete Example: \(\mathbb{F}_{2}\) and Degree \(n = 3:\)

Let’s look at the base field \(\mathbb{F}_{2}\) (integers modulo \(2)\) and extensions of degree \(3.\)

There are exactly two irreducible polynomials of degree \(3\) over \(\mathbb{F}_{2}:\)

\(P_1(x) = x^3 + x + 1\)

\(P_2(x) = x^3 + x^2 + 1\)

If we build fields by adjoining a root of each, we will see they are just different perspectives of the exact same 8-element field, \(\mathbb{F}_{8}.\)

The master polynomial for degree \(3\) is \(x^{2^3} - x = x^8 - x.\) Over \(\mathbb{F}_{2},\) this polynomial factors completely into all irreducible polynomials whose degrees divide \(3\) (degrees \(1\) and \(3):\)

\[x^{8}-x=x(x-1)(x^{3}+x+1)(x^{3}+x^{2}+1)\]

Notice that both \(P_1(x)\) and \(P_2(x)\) are sitting right there inside it.

Adjoining a root of \(P_1(x)\)

Let \(\alpha\) be a root of \(P_1(x),\) so \(\alpha^3 + \alpha + 1 = 0,\) or \(\alpha^3 = \alpha + 1.\) The field generated is

\[\mathbb{F}_2(\alpha) = \{0, 1, \alpha, \alpha^2, \alpha+1, \alpha^2+\alpha, \alpha^2+\alpha+1, \alpha^2+1\}\]

Finding the root of \(P_2(x)\) inside \(\mathbb{F}_2(\alpha):\)

We do not need to create a new universe to find a root for \(P_2(x).\) It already exists inside the field we just made! Let’s test an element from our field, like \(\beta = \alpha + 1,\) and plug it into \(P_2(x):\)

\[\begin{aligned}P_{2}(\alpha +1)&=(\alpha +1)^{3}+(\alpha +1)^{2}+1\\ &=(\alpha ^{3}+3\alpha ^{2}+3\alpha +1)+(\alpha ^{2}+2\alpha +1)+1\end{aligned}\]

Since our coefficients are in \(\mathbb{F}_{2},\) all even numbers become \(0\) and odd numbers become \(1\) \(3 \equiv 1\) and \(2 \equiv 0:\)

\[\begin{aligned}P_{2}(\alpha +1)&=(\alpha ^{3}+\alpha ^{2}+\alpha +1)+(\alpha ^{2}+1)+1\\ &=\alpha ^{3}+\alpha +1\end{aligned}\]

Because \(\alpha\) is a root of \(P_1(x),\) we know \(\alpha^3 + \alpha + 1 = 0.\) Therefore:

\[P_{2}(\alpha +1)=0\]

The element \(\beta = \alpha + 1\) is a perfectly valid root of \(P_2(x).\) If you start with a root of \(P_1(x),\) you can construct a root of \(P_2(x)\) by adding \(1.\) If you start with a root of \(P_2(x),\) you can construct a root of \(P_1(x)\) by adding \(1.\) The choice of polynomial changes your coordinate system (which element you choose to label as your primary generator), but it does not change the algebraic structure of the field itself.

How do finite fields compare to \(\mathbb{Q}\):

Level \(1:\) Finite Extensions (Adjoining Roots):

At this level, we look at extending a base field by adjoining roots of polynomials to build a single, finite-dimensional extension field.

1. Finite Fields \((\mathbb{F}_{p^n} / \mathbb{F}_p)\):

  • Number of degree-\(n\) extensions: Exactly 1. Adjoining roots of any degree-\(n\) irreducible polynomial over \(\mathbb{F}_p\) lands you in the exact same field \(\mathbb{F}_{p^n}\).

  • Galois Group structure: Always a cyclic group \(\mathbb{Z}/n\mathbb{Z}\).

  • Generator: Generated by a single map—the Frobenius \(x \mapsto x^p\).

  • Example: For \(n=3\), \(\text{Gal}(\mathbb{F}_{p^3}/\mathbb{F}_p) \cong \mathbb{Z}/3\mathbb{Z} = \{0, 1, 2\}\).

2. Number Field Extensions over \(\mathbb{Q}\)

  • Number of degree-\(n\) extensions: Infinitely many. Adjoining roots of different degree-\(n\) polynomials over \(\mathbb{Q}\) produces completely different, non-isomorphic fields (e.g., \(\mathbb{Q}(\sqrt{2})\) vs. \(\mathbb{Q}(\sqrt{3})\) vs. \(\mathbb{Q}(\sqrt{-5})\) are all degree-\(2,\) but entirely distinct).

  • Galois Group structure: Can be almost any finite group of order \(n\) or a multiple of \(n\) (cyclic, symmetric groups like \(S_n\), dihedral groups, etc.).

  • Generator: Generally requires multiple complex permutations of roots; not generated by a simple power map.

Level \(2:\) Absolute Galois Groups (the whole closure)

At this level, we take the inverse limit over all possible extensions at once to study the symmetry group of the entire algebraic closure (\(\bar{\mathbb{F}}_p\) or \(\bar{\mathbb{Q}}\)).

1. Absolute Galois Group of a Finite Field (\(G_{\mathbb{F}_p} = \text{Gal}(\bar{\mathbb{F}}_p/\mathbb{F}_p)\))

  • What it combines: The inverse limit over all the cyclic finite groups \(\mathbb{Z}/n\mathbb{Z}\) from Level 1.Resulting Group: \(G_{\mathbb{F}_p} \cong \widehat{\mathbb{Z}}\) (the profinite completion of \(\mathbb{Z}\)).

  • Key Properties:

    • Abelian: \(a \cdot b = b \cdot a\) (because all the building blocks \(\mathbb{Z}/n\mathbb{Z}\) are abelian).
    • Topologically Cyclic: Still generated by a single map—the global Frobenius \(x \mapsto x^p\).
    • Element Representation: The infinite vectors \((a_1, a_2, a_3, \dots) \in \widehat{\mathbb{Z}}\)
2. Absolute Galois Group of Rational Numbers (\(G_\mathbb{Q} = \text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\))

  • What it combines: The inverse limit over the infinitely many, highly varied finite Galois groups over \(\mathbb{Q}\) from Level 1.

  • Resulting Group: An uncountably infinite, non-abelian, extremely complex profinite group.

  • Key Properties:

    • Highly Non-Abelian: Composition order matters (\(a \cdot b \neq b \cdot a\)).
    • Infinitely Generated: Cannot be generated by a single Frobenius map or any finite set of elements.

🪁: The term \(\alpha^2\) must be included because \(\alpha^2\) cannot be simplified into anything simpler, and without it, the set of elements would fail to be closed under multiplication (it wouldn’t be a field).

When we declared \(\alpha\) to be a root of \(x^3 + x + 1 = 0\), we established one single reduction rule (working mod \(2):\)

\[\alpha^3 = \alpha + 1\]

This rule tells us how to replace \(\alpha^3\) (and any higher power like \(\alpha^4, \alpha^5, \dots\)) with lower-degree terms. It does not tell us how to reduce \(\alpha^2\). \(\alpha^2\) is a power strictly less than \(\alpha^3\). Because \(\alpha^2\) cannot be rewritten as a combination of \(1\) and \(\alpha\), \(\alpha^2\) is an independent term, just like \(1\) and \(\alpha\).

Suppose we tried to build a set using only \(c_0 + c_1\alpha\) (leaving out \(c_2\alpha^2\)). That set would only contain \(4\) elements:

\[\{0, 1, \alpha, 1+\alpha\}\]

Now, test if this set is closed under multiplication (a fundamental requirement for a field): Take the element \(\alpha\) and multiply it by itself:

\[\alpha \times \alpha = \alpha^2\]

Since \(\alpha^2\) is not \(0\), not \(1\), not \(\alpha\), and not \(1+\alpha,\) the product \(\alpha^2\) falls outside our \(4\)-element set! To make multiplication work, we are forced to add \(\alpha^2\) to our building blocks.

If we multiply \(\alpha \cdot \alpha^2\):

\[\alpha^3 = \alpha + 1\]

We don’t need a \(c_3\alpha^3\) term because \(\alpha^3\) instantly collapses back into terms we already have (\(1\) and \(\alpha\)). What about \(\alpha^4\)?

\[\alpha^4 = \alpha \cdot \alpha^3 = \alpha(\alpha + 1) = \alpha^2 + \alpha\]

Again, it collapses into a combination of \(\alpha^2\) and \(\alpha\).

😱 While the field extension \(\mathbb{F}_{p^n}\) of degree \(n\) is unique up to isomorphism, the automorphisms of that extension are not unique — and the elements of \(\widehat{\mathbb{Z}}\) represent these Galois automorphisms, not the fields themselves. To see why there are infinitely many vectors in \(\widehat{\mathbb{Z}}\), let’s trace how the geometric/algebraic structure connects to the sequence vector representation:

\[\text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) = \{\text{id}, \sigma, \sigma^2, \dots, \sigma^{n-1}\} \cong \mathbb{Z}/n\mathbb{Z}\]

The \(n\)-th entry of a vector in \(\widehat{\mathbb{Z}}\) corresponds to choosing which specific automorphism \(\sigma^a \in \text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)\) acts on the degree-\(n\) subextension.mSince there are \(n\) distinct choices for \(a \pmod n\) at level \(n\), there are \(n\) distinct valid choices for the \(n\)-th component of the vector.

To see how different vector entries give distinct permutations of elements, let’s look at a concrete finite field — say \(\mathbb{F}_8 = \mathbb{F}_{2^3}\) (so \(p=2, n=3\)).

Here, the local Galois group is cyclic of order 3:

\[\text{Gal}(\mathbb{F}_8/\mathbb{F}_2) = \{\text{id}, \sigma, \sigma^2\} \cong \mathbb{Z}/3\mathbb{Z}\]

There are 3 possible choices for the \(n=3\) entry of a vector in \(\widehat{\mathbb{Z}}\):

\[a_3 \in \{0, 1, 2\}\]

Let’s pick an explicit element \(\beta \in \mathbb{F}_8\) satisfying \(\beta^3 + \beta + 1 = 0\) (this polynomial comes from constructing the field \(\mathbb{F}_8\) explicitly. To build \(\mathbb{F}_8 = \mathbb{F}_{2^3}\), we need to adjoin a root of an irreducible polynomial of degree 3 over \(\mathbb{F}_2\)).

The full field consists of 8 elements: \(\{0, 1, \beta, \beta^2, \beta + 1, \beta^2 + 1, \beta^2 + \beta, \beta^2 + \beta + 1\}.\)

Here is how each choice of \(a_3\) permutes the non-base elements of \(\mathbb{F}_8\):

Case 1:

\(a_3 = 0 \pmod 3\) (Identity Map \(\sigma^0\))Formula: \(g(x) = x^{2^0} = x\)

Action on \(\beta\): \(g(\beta) = \beta\)

Full Permutation:

\[\begin{aligned} \beta &\mapsto \beta \\ \beta^2 &\mapsto \beta^2 \\ \beta^2 + \beta &\mapsto \beta^2 + \beta \end{aligned}\]

Permutation Cycle Structure: \((1)(1)(1)\dots\) — Every element stays fixed.

Case 2: \(a_3 = 1 \pmod 3\) (Standard Frobenius \(\sigma^1\))

Formula: \(g(x) = x^{2^1} = x^2\)

Action on \(\beta\): \(g(\beta) = \beta^2\)

Action on \(\beta^2\): \(g(\beta^2) = (\beta^2)^2 = \beta^4 = \beta^2 + \beta\)

Action on \(\beta^2 + \beta\): \(g(\beta^2 + \beta) = (\beta^2 + \beta)^2 = \beta^4 + \beta^2 = (\beta^2 + \beta) + \beta^2 = \beta\)

Full Permutation:

\[\beta \;\longrightarrow\; \beta^2 \;\longrightarrow\; \beta^2 + \beta \;\longrightarrow\; \beta\]

Permutation Cycle Structure: \((\beta \quad \beta^2 \quad \beta^2 + \beta)\) — A 3-cycle shifting roots forward.

Case 3: \(a_3 = 2 \pmod 3\) (Square Frobenius \(\sigma^2\))

Formula: \(g(x) = x^{2^2} = x^4 = x^2 + x\)

Action on \(\beta\): \(g(\beta) = \beta^4 = \beta^2 + \beta\)

Action on \(\beta^2 + \beta\): \(g(\beta^2 + \beta) = (\beta^2 + \beta)^4 = \beta^{16} + \beta^4 = \beta^2 + (\beta^2 + \beta) = \beta\)

Action on \(\beta^2\): \(g(\beta^2) = (\beta^2)^4 = \beta^8 = \beta\)

Full Permutation:\[\beta \;\longrightarrow\; \beta^2 + \beta \;\longrightarrow\; \beta^2 \;\longrightarrow\; \beta\]

Permutation Cycle Structure: \((\beta \quad \beta^2 + \beta \quad \beta^2)\) — The reverse 3-cycle.

For a fixed degree \(n\), the extension \(\mathbb{F}_{p^n}/\mathbb{F}_p\) is indeed unique. However, its Galois group \(\text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)\) is a cyclic group of order \(n\), generated by the Frobenius automorphism \(\sigma(x) = x^p\):

\[\text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) = \{\text{id}, \sigma, \sigma^2, \dots, \sigma^{n-1}\} \cong \mathbb{Z}/n\mathbb{Z}\]

The \(n\)-th entry of a vector in \(\widehat{\mathbb{Z}}\) corresponds to choosing which specific automorphism \(\sigma^a \in \text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)\) acts on the degree-\(n\) subextension. Since there are \(n\) distinct choices for \(a \pmod n\) at level \(n\), there are \(n\) distinct valid choices for the \(n\)-th component of the vector.

Recall the inverse limit definition of the profinite completion of the integers:

\[\widehat{\mathbb{Z}} = \varprojlim_{n} \mathbb{Z}/n\mathbb{Z} = \left\{ (a_1, a_2, a_3, \dots) \in \prod_{n=1}^\infty \mathbb{Z}/n\mathbb{Z} \;\middle\vert{}\; a_m \equiv a_n \pmod n \text{ whenever } n \mid m \right\}\]

An element in \(\widehat{\mathbb{Z}}\) is an infinite tuple \((a_n)_{n \ge 1}\) where \(a_n \in \mathbb{Z}/n\mathbb{Z}\), satisfying the compatibility constraint that if \(n\) divides \(m\), \(a_m \pmod n = a_n\).

The “Zero” Vector: \(a = (0, 0, 0, \dots)\) represents the identity automorphism across the entire algebraic closure \(\bar{\mathbb{F}}_p\) (it fixes every element of every field).

The “Canonical Frobenius” Vector: \(a = (1, 1, 1, \dots)\) represents the map \(x \mapsto x^p\) acting on every \(\mathbb{F}_{p^n}\). The map \(x \mapsto x^p\) fixes only the base field \(\mathbb{F}_p = \mathbb{F}_{p^1}\). It does not fix any larger subfield \(\mathbb{F}_{p^n}\) for \(n > 1\). An element \(x \in \bar{\mathbb{F}}_p\) is fixed by the Frobenius map \(\sigma(x) = x^p\) if and only if it satisfies: \(x^p = x \quad \implies \quad x^p - x = 0.\) By definition, the set of roots of \(x^p - x = 0\) in \(\bar{\mathbb{F}}_p\) is precisely the prime field \(\mathbb{F}_p\).

Infinitely Many Other Vectors: For any arbitrary \(k \in \mathbb{Z}\), \(a = (k \bmod 1, k \bmod 2, k \bmod 3, \dots)\) is a valid vector corresponding to \(\sigma^k\). Furthermore, because \(\widehat{\mathbb{Z}}\) is the uncountable topological completion of \(\mathbb{Z}\), there are uncountably many non-integer, infinite limit vectors (such as elements that act as “fractional” or \(q\)-adic powers of Frobenius).

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NOTE: These are tentative notes on different topics for personal use - expect mistakes and misunderstandings.