[Math] Understanding examples of subfield and prime subfield of a finite field

abstract-algebrafield-theoryfinite-fields

I have already taken a look at this answer. Somehow it did not answer my question.

As I can find, in various literatures,

Now, if we consider the operations of the field to be $+ \bmod n$ and $\times \bmod n$. We find that $\mathbb{Z}_2$ and $\mathbb{Z}_5$ are both fields under these operations.

But in order to get a good feeling of subfields, we try to consider $\mathbb{Z}_{3^2} = \mathbb{Z}_9$. We find that this not a field under the afore stated operations.
Not all the non-zero elements, notably 3 and 9 ($\gcd(3,9) \not=1$ and $\gcd(6,9) \not=1$), do not have multiplicative inverses.

Indeed, as Wikipedia states,

Even though all fields of size $p$ are isomorphic to $\mathbb{Z}/p\mathbb{Z}$, for $n \ge 2$ the
ring $\mathbb{Z}/p^n\mathbb{Z}$ (the ring of integers modulo $p^n$) is not a field. The
element $p$ $(\bmod\ p^n)$ is nonzero and has no multiplicative inverse.

Looking for examples, we find one here for $GF(2^3)$. This is based on polynomials.

Now, coming to my original point on (understanding) subfield or prime subfield
of finite fields, please tell me,

  1. Whether it is totally impossible to construct purely numerical examples of
    fields of size $p^n$.
  2. Given a (non-numerical) field of size $p^n$, (one can be found in page 90 (16) of this document), what is the best way to identify the subfield(s) and prime subfield? I appreciate an answer which nurtures my intuition, not a theoretical one which puts me deep in difficult mathematical terms.

Best Answer

Let me address your first question. First, I want to argue that there is no precise meaning of "involving numbers only". For example, given a finite field $F$ of size $4$ constructed in the usual manner (quotient of a polynomial ring over $\mathbb{Z}/2\mathbb{Z}$), I can choose a set of numbers, say $$S=\{37,\tfrac{5}{19},\pi,e\}$$ and, choosing a bijection of $S$ with $F$, use transport of structure to give $S$ the structure of a field. The field structure does not depend in any way on what the underlying set is "made of".

However, along the lines of what I think you are ultimately after, you can obtain finite fields of any possible order using larger rings of integers. For example, $\mathbb{Z}[i]/(3)$ is a finite field of size $9$, and $\mathbb{Z}[i]$ consists of very reasonable numbers, $$\mathbb{Z}[i]=\{a+bi\mid a,b\in\mathbb{Z}\}.$$

Now let me addressr your second question. Let's use $\mathbb{F}_p$ to mean $\mathbb{Z}/p\mathbb{Z}$, a finite field of order $p$ - it is a very common notation that is slightly less cumbersome, but doesn't mean anything different, they are exact synonyms.

A finite field of order $p^n$ is often constructed by taking the polynomial ring $\mathbb{F}_p[x]$, choosing an irreducible polynomial $f\in \mathbb{F}_p[x]$ of degree $n$, and then making the field $$F=\mathbb{F}_p[x]/(f).$$ Now, the division algorithm for polynomials tells you that each equivalence class in this quotient can be uniquely identified by a representative of degree $<n$. In other words, $$\begin{align*} F&=\{a_0+a_1x+\cdots +a_{n-1}x^{n-1}+(f)\mid a_0,a_1,\ldots,a_{n-1}\in\mathbb{F}_p\}\\\\ &=\left\{\,\overline{a_0+a_1x+\cdots +a_{n-1}x^{n-1}}\,\;\middle\vert\;a_0,a_1,\ldots,a_{n-1}\in\mathbb{F}_p\right\}\\\\\\ &=\{a_0+a_1\overline{x}+\cdots +a_{n-1}\overline{x}^{n-1}\mid a_0,a_1,\ldots,a_{n-1}\in\mathbb{F}_p\} \end{align*}$$ Letting the symbol $\alpha$ be a stand-in for $\overline{x}$, you can think of $F$ as being $\mathbb{F}_p$ with a new element "$\alpha$" added in, where $\alpha$ is a root of $f$, and you can write $F=\mathbb{F}_p[\alpha]$.

Now, the prime subfield of $F$ is just the "constant" polynomials, i.e. the ones with no $\alpha$'s in them: $$\text{the prime subfield of }F=\{a_0+0\alpha+\cdots+0\alpha^{n-1}\mid a_0\in\mathbb{F}_p\}$$ and for each divisor $d\mid n$, the unique subfield of $F$ of order $p^d$ is the collection of polynomials in $\alpha$ whose terms are those of exponents that are multiples of $n/d$: $$\text{the subfield of }F\text{ of order }p^d=\{a_0+a_1\alpha^{n/d}+\cdots+a_{d-1}\alpha^{(d-1)n/d}\mid a_0,a_1,\ldots,a_{d-1}\in\mathbb{F}_p\}$$ (clearly, the above set has cardinality $p^d$, because it takes $d$ elements of $\mathbb{F}_p$ to specify a given element of the above set, namely, each of the coefficients of the powers of $\alpha$. To see that it is a field, remember that $(a+b)^p=a^p+b^p$ in a field of characteristic $p$.)

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