Subfields and extension fields

A subfield of a field is a subset of the field that is also a field. Thus, for example, $\Qbb$ is a subfield of $\Rbb$.

A more potent concept is that of an extension field. Viewed one way, it simply turns the idea of a subfield around: an extension field $E$ of a field $F$ is a field that contains every element of $F$, so that $F$ forms a subfield of $E$. The field $F$ in this case is said to be the base field. But more importantly is the way that the extension field is created. Most commonly, extension fields are created to determine roots of polynomials that do not have roots in the base field.

In this definition, the ring (or field) in which the polynomial is irreducible makes a difference. For example, the polynomial $f(x) = x^2-2$ is irreducible over $\Qbb$, but over the real numbers we can write

$$f(x) = (x+\sqrt{2})(x-\sqrt{2}).$$

We will demonstrate the construction of the familiar field of complex numbers as an extension of the real field. The polynomial $p(x) \in \Rbb[x]$ with real coefficients

$$p(x) = x^2 + 1$$

is irreducible over the real numbers. Additionally (and not quite the same thing), it has no solution over the real numbers. That is, there is no $x\in \Rbb$ such that $p(x) = 0$. We can create a new field, an extension to $\Rbb$, essentially by adjoining a new element to the field that is specifically the root of $p(x)$. In this new field, we must carefully and consistently define the operations of addition, multiplication, and so forth.

Let $\alpha$ be an indeterminate. Let us create a field of polynomials, with multiplication modulo $\alpha^2 + 1$. We will denote this field (for the moment) as $\la \Rbb[\alpha]\ra_{\alpha^2+1}$. . We must verify that it in fact forms a field and not a ring. All elements in the field are of the form

$$a + b\alpha.$$

(Why?) Addition of elements of this form in the field is straightforward (i.e., polynomial addition)

$$(a + b\alpha) + (c+d\alpha) = (a+c) + (b+d)\alpha.$$

Multiplication of these elements modulo $\alpha^2 + 1$ can be written as

$$(a+b\alpha)(c+d\alpha) \pmod{\alpha^2+1} = (ac - bd) + (ad+bd)\alpha.$$

The multiplicative inverse of the nonzero element $a+b\alpha$ can be verified to be

$$(a+b\alpha)^{-1} = \frac{(a-b\alpha)}{a^2+b^2}.$$

Note that for the element $\alpha \in \la \Rbb[\alpha]\ra_{\alpha^2+1}$,

$$(\alpha)(\alpha) \pmod{\alpha^2+1} = -1,$$

so that $\alpha$ is a root of the polynomial equation $x^2+1=0$. This field has the same rules of arithmetic as does the complex field $\Cbb$. In fact, they are the same field. It is conventional to denote the indeterminate $\alpha$ as $i$ (the unit imaginary number) or as $j$.

The point of this is that if a polynomial exists which has no solution in a field $F$, a new field can be constructed in which a solution does exist. Related to this particular example, there are some other observations that can be made.

1. Miraculously enough, once the real field is extended to the complex field, all polynomials with coefficients either from $\Rbb$ or $\Cbb$ have solutions in the field. There is thus no need in usual computations to form extensions to larger fields. (This fact tends to make the idea of extension fields a little foreign at first, since we have a large enough field for most purposes at hand.) This fact is known as the fundamental theorem of algebra.
2. Consider as an example the polynomial $q(x) \in \Qbb[x]$ with
   
   $$q(x) = x^2-2,$$
   
   
   The polynomial $q(x)$ has no zeros in $\Qbb$, and so an extension field can be created in which $q(x)$ has a zero. Elements in this field are of the form $a + b\sqrt{2}$, where $a,b \in \Qbb$. Arithmetic in this field is done modulo the polynomial $x^2-2$; This field is an extension of $\Qbb$; it is large enough to contain roots of $q(x)$, but not large enough to contain roots of every polynomial in $\Qbb[x]$. For example, $r(x) = x^2 -3$ does not have roots in this field, so another extension is necessary.

In this discussion about extension fields, the extension obtained has been stated to be a field, and seems to obey the properties of a field for the cases examined. That the extensions are in fact fields may be rigorously established, but requires some theoretical machinery (regarding maximal ideals) which we are not ready for yet.

Galois fields

In addition from providing some interesting insight into the structure of the numbers and equations we commonly deal with, the idea of extension fields provides a means of describing all fields of finite order, or finite fields. We have already observed that $\la \Zbb_p, +,\cdot\ra$ forms a field when $p$ is prime. It turns out that all finite fields have $p^m$ elements in them, where $p$ is prime. For $m>1$, the finite fields are obtained as extension fields to $\Zbb_p$ using an irreducible polynomial in $\Zbb_p[x]$ of degree $m$. These finite fields are usually denoted by $GF(p^m)$ or $GF(q)$ where $q = p^m$, where $GF$ stands for ``Galois field,'' named after the French mathematician Everiste Galois.

Before introducing and proving some key properties of Galois fields, it is interesting to see a construction of one such field, $GF(2^3)$. As may be verified by direct substitution, the polynomial $p(x) = x^3+x+1$ is irreducible over $GF(2)$. (The polynomial is also primitive). We will form the extension field by adjoining the root of $p(x)$. Let $\alpha$ be such a root; then $p(\alpha) = \alpha^3+\alpha+1 = 0$, so $\alpha^3 = \alpha+1$. The the elements of $GF(2^3)$ are the polynomials of the form $a + \alpha b + \alpha^2 c$ for $a,b, c \in GF(2)$. Another representation is simply as a 3-tuple $(a,b,c)$. We observe that there must therefore be 8 elements in $GF(2^3)$. Addition is performed as usual (element-by-element, just as in polynomial addition). Multiplication is performed modulo the irreducible polynomial that was used to create the extension field. (Point out analogy with forming fields modulo a number). In our example, the elements are These are

$$0, 1, \alpha, 1+\alpha, \alpha^2, 1+\alpha^2, \alpha+\alpha^2, 1+\alpha+\alpha^2$$

These field elements can be expressed as triplets of the coefficients:

$$\begin{aligned} 0 &\rightarrow (0,0,0) \\ 1 & \rightarrow (0... ...,1,1) \\ 1+\alpha+\alpha^2 & \rightarrow (1,1,1) \end{aligned}$$

Addition is easily accomplished in either the polynomial form or in the equivalent triplet form. From this form, we recognize that the elements of the Galois field form a vector space over the base field $GF(2)$. Observe that for any element $\beta \in GF(2^3)$, $\beta+\beta=0$. Recalling the definition of the characteristic of a ring (which also applies to fields), we see that the characteristic of this field is $2$.

Multiplication in the field is polynomial multiplication modulo $p(\alpha)$. For example,

$$(1+\alpha^2)(\alpha+\alpha^2) = \alpha + \alpha^2 + \alpha^3 + \alpha^4 \pmod{\alpha^3+\alpha+1} = 1+\alpha$$

Another useful representation is as powers of $\alpha$. Since $\alpha^3 = \alpha+1$, we can form the following list of the nonzero elements in the field:

$$\begin{aligned} \alpha^0 &= 1 \\ \alpha^1 &= \alpha \\ \alp... ...5 &= \alpha^2+\alpha+1 \\ \alpha^6 &= \alpha^2+1 \end{aligned}$$

The next power is $\alpha^7 = \alpha^3+\alpha = 1$, so the list is complete. All of the nonzero elements of the field are generated by $\alpha$; $\alpha$ is said to be a primitive element of the field. The fact that $\alpha$ is the root of the polynomial $p(x)$ and also a primitive element is because $p(x)$ is a primitive polynomial.

In the exponential notation, multiplication of field elements is easy. For example, since $1+\alpha^2 = \alpha^6$ and $\alpha+\alpha^2 = \alpha^4$, we have

$$(1+\alpha^2)(\alpha+\alpha^2) = \alpha^6 \alpha^4 = \alpha^{10} = \alpha^{7}\alpha^{3} = \alpha^3 = \alpha+1.$$

Having presenting an examples, we now present some important ideas associated with Galois fields.



Note: the notation $a\vert b$ means: $a$ divides $b$, and $(a,b)$ is the greatest common divisor of $a$ and $b$.

lemma 6   If $\beta \in GF(q)$ and $\beta \neq 0$ then $\order(\beta)\vert(q-1)$.

lemma 7   If $\alpha \in GF(q)$ and $\beta \in GF(q)$ with $\beta = \alpha^i$ for some $i$, and if $\order(\alpha) = t$ then

$$\order(\beta) = \frac{t}{(i,t)}$$

In other words, a primitive element has the highest possible order. The question of whether there are any primitive elements in $GF(q)$, and how many, is now addressed.



It can be shown that the $\phi$ function can be written as

$$\phi(n) = n\prod_{p\vert n}(1-\frac{1}{p})$$

where the product is taken over all primes $p$ dividing $n$. For example,

$$\phi(56) = \phi(2\cdot 2 \cdot 2 \cdot 7) = 56(1-1/2)(1-1/7) = 24.$$

We observe that:

1. $\phi(p) = p-1$ if $p$ is prime.
2. $\phi(p_1 p_2) = (p_1-1)(p_2-1)$ for primes $p_1$ and $p_2$.
3. $\phi(p^m) = p^{m-1}(p-1)$ for $p$ prime.
4. $\phi(p^m q^n) = p^{m-1}q^{n-1}(p-1)(q-1)$ for distinct primes $p$ and $q$.

theorem 2   For a Galois field $GF(q)$:

1. If $t \notdiv (q-1)$, then there are no elements of order $t$ in $GF(q)$
2. if $t \vert q-1$ then there are $\phi(t)$ elements of order $t$ in $GF(q)$

From this theorem we make the following observation: there are $\phi(q-1)$ primitive elements in $GF(q)$.

Collecting our thoughts, we observe that in $GF(q)$, there are $\phi(q-1)>1$ primitive elements, and that all non-zero elements of the field can be constructed as powers of the primitive element. We will frequently denote the primitive element in the field as $\alpha$.

lemma 8   The characteristic of a Galois field is always a prime integer.

(Recall that the characteristic is the smallest positive integer such that $m(1) = 1+1+\cdots + 1 = 0$.)

On the basis of this lemma, we can observe that in a field $GF(q)$, there are $p$ elements ($p$ a prime number) $\{0,1,2,\ldots,(p-1)(1)\}$ which behave as a field (i.e., we can define addition and multiplication on them as a field). Thus $\Zbb_p$ (or something isomorphic to it, which is the same thing) is a subfield of every Galois field $GF(q)$. In fact, a stronger assertion can be made:

theorem 3   The order $q$ of every finite field $GF(q)$ must be a power of a prime.

This points the way to constructing every finite field. To construct $Gf(p^m)$, we a polynomial degree $m$ irreducible over $GF(p)$, and form the extension field for this polynomial, as we did for the example of $GF(2^3)$ above.

Irreducible and Primitive polynomials

While any irreducible polynomial can be used to construct the extension field, computation in the field is easier of a primitive polynomial is used. First, we make the following observation:

theorem 4   An irreducible $m$th-degree polynomial $f(x) \in GF(p)[x]$ divides $x^{p^m-1} - 1$

It can be shown that the polynomial $p(x) = x^3+x+1$ used above is primitive in $GF(2)[x]$, since $x^7-1$ is divisible by $p(x)$,

$$x^7-1 = (x^3+x+1)(x^4+ x^2 + x + 1),$$

but no smaller $n$ exists such that $x^n-1$ is divisible by $p(x)$. Not every irreducible polynomial is primitive. The following theorem, provides the motivation for using primitive polynomials.

theorem 5   The roots of an $m$th degree primitive polynomial $p(x) \in GF(p)[x]$ are primitive elements in $GF(p^m)$.

All the nonzero elements of the field can be generated as powers of the roots of the primitive polynomial.