NOTES ON STATISTICS, PROBABILITY and MATHEMATICS

Galois group of a finite field extension \(\text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) \cong \mathbb{Z}/n\mathbb{Z}\) and 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 your 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 \((c_0, c_1, c_2)\) you need to write down any element.

You started with \(P(x) = x^3 + x + 1\), which should have 3 roots. You 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.

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 your 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 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 \(\alpha\).

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