Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.
share|improve this question
2  
What is a "purely numerical example"? –  Hagen von Eitzen Jun 8 '13 at 12:26
    
@HagenvonEitzen Examples involving numbers only and not polynomials involving indeterminates. –  Masroor Jun 8 '13 at 12:36
1  
Consider the following. When you do calculations involving $\sqrt2$ you, at least more often than not, don't use its "numerical value". Instead you use the property that it satisfies the equation $(\sqrt2)^2=2$. Similarly you don't use a "numerical value" of the imaginary unit $i$. You simply use the knowledge that it satisfies the equation $i^2=-1$. Similarly, when you calculate in $\mathbb{F}_4=\{0,1,\beta,\beta+1\}$ you base your calculation to the bit that $\beta^2=1+\beta$. In the other finite fields you do something similar. –  Jyrki Lahtonen Jun 9 '13 at 18:48

3 Answers 3

up vote 6 down vote accepted

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 $p$.)

share|improve this answer

There are only few candidates (up to isomorphism) for prime fields: $\mathbb Q$ and $\mathbb Z/p\mathbb Z$ with $p$ prime. If we assume that $F$ is finite, we can clearly identify it as the set $$\{0,1, 1+1,\ldots, \underbrace{1+1+\ldots+1}_{p-1}\}\subseteq F$$ where $p$ is determined by the fact that $\underbrace{1+1+\ldots +1}_p=0$. If $E$ is a subfield of $F$, then $F$ is also a vector space over $E$. Especially, if $E$ is the prime field we conclude that $|F|=p^n$, the elements of $F$ can be written as $n$-tuples of elements of $\mathbb Z/p\mathbb Z$, addition is componentwise as expected, but multiplication is not so easy to describe. For example, we can view $GF(8)$ as the set $(\mathbb Z/2\mathbb Z)^3$ with componentwise addition complicated $$(a,b,c)+(d,e,f)=(a+d,b+e,c+f) $$ and the somewhat and arbitrary multiplication $$(a,b,c)\cdot(d,e,f) = (ad+bf+ce,ae+bd+bf+ce+cf,af+be+cd+cf).$$ You are free to work with this as definition and spend an afternoon verifying the field axioms. But life is really easier if you identify $a,b,c)$ with $a+bX+cX^2$ and consider polynomials in $\mathbb Z/2\mathbb Z[X]$ modulo $x^3+x+1$ and get the field properties for free.

Again, if $E$ is a subfield, then we conclude $|E|=p^m$ for some $m$ as well, and as $F$ is a vector space over $E$, we conclude $p^n$ is a power of $p^m$, i.e. $m$ divides $n$. As $3$ is prime, there are no subfields of $F=GF(2^3)$ other than the prime field and $F$ itself.

One can show that the multiplicative group of $F$ is cyclic (if one works with polynomials, $X$ is an obvious generator) and each subfield $E$ corresponds to a subgroup of this cyclic group, i.e. is generated by some $X^d$ with $d|n$. In the above example $GF(8)$, the multiplicative group has order $7$, hence no nontrivial subgroups, hence we see once again that there are no subfields apart from $F$ and the prime field.

share|improve this answer

If you want interesting examples of finite fields and subfields, characteristic 2 offers Grundy numbers, otherwise known as Nimbers, whose arithmetic is explored in JH Conway "On Numbers and Games".

They arise as the values of impartial games, and so as genuine "numbers" rather than constructions by polynomial, and come with arithmetic operations which arise from game-theoretic constructions.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.