Applied Discrete Structures

By the last condition $\sqrt{2}$ must be an element of $S\text{,}$ and, if $S$ is to be a field, the sum, product, difference, and quotient of elements in $S$ must be in $S\text{.}$ So operations involving this number, such as $\sqrt{2}\text{,}$ ${\left(\sqrt{2}\right)}^2\text{,}$ ${\left(\sqrt{2}\right)}^3\text{,}$ $\sqrt{2}+\sqrt{2}\text{,}$ $\sqrt{2}-\sqrt{2}\text{,}$ and $\frac{1}{\sqrt{2}}$ must all be elements of $S\text{.}$ Further, since $S$ contains $\mathbb{Q}$ as a subset, any element of $\mathbb{Q}$ combined with $\sqrt{2}$ under any field operation must be an element of $S\text{.}$ Hence, every element of the form $a+b\sqrt{2}\text{,}$ where $a$ and $b$ can be any elements in $\mathbb{Q}\text{,}$ is an element of $S\text{.}$ We leave to the reader to show that $S=\{a+b\sqrt{2}\mid a,b\in \mathbb{Q}\}$ is a field (see Exercise 1 of this section). We note that the second zero of $x^2-2\text{,}$ namely $-\sqrt{2}\text{,}$ is an element of this set. To see this, simply take $a=0$ and $b=-1\text{.}$ The field $S$ is frequently denoted as $\mathbb{Q}\left(\sqrt{2}\right)\text{,}$ and it is referred to as an extension field of $\mathbb{Q}\text{.}$ Note that the polynomial $x^2-2=(x-\sqrt{2})(x+\sqrt{2})$ factors into linear factors, or splits, in $\mathbb{Q}\left(\sqrt{2}\right)[x]\text{;}$ that is, all coefficients of both factors are elements of the field $\mathbb{Q}\left(\sqrt{2}\right)\text{.}$

Therefore, $a+1$ is also a zero of $x^2+x+1\text{.}$ Hence, $S=\{0,1,a,a+1\}$ is the smallest field that contains ${\mathbb{Z}}_2=\{0,1\}$ as a subfield and contains all zeros of $x^2+x+1\text{.}$ This extension field is denoted by ${\mathbb{Z}}_2(a)\text{.}$ Note that $x^2+x+1$ splits in ${\mathbb{Z}}_2(a)\text{;}$ that is, it factors into linear factors in ${\mathbb{Z}}_2(a)\text{.}$ We also observe that ${\mathbb{Z}}_2(a)$ is a field containing exactly four elements. By Theorem 16.2.10, we expected that ${\mathbb{Z}}_2(a)$ would be of order $p^2$ for some prime $p$ and positive integer $n\text{.}$ Also recall that all fields of order $p^n$ are isomorphic. Hence, we have described all fields of order $2^2=4$ by finding the extension field of a polynomial that is irreducible over ${\mathbb{Z}}_2\text{.}$

Next, we briefly describe the field $GF(8)$ and how an error correcting code can be build on a the same observation about that field.

First, we start with the irreducible polynomial $p(x)=x^3+x+1$ over ${\mathbb{Z}}_2\text{.}$ There is another such cubic polynomial, but its choice produces essentially the same result. Just as we did in the previous example, we assume we have a zero of $p(x)$ and call it $\beta \text{.}$ Since we have assumed that $p(\beta )={\beta }^3+\beta +1=0\text{,}$ we get the recurrence relation for powers ${\beta }^3=\beta +1$ that lets us reduce the seven powers ${\beta }^k\text{,}$ $0\le k\le 6\text{,}$ to linear combinations of 1, $\beta \text{,}$ and ${\beta }^2\text{.}$ Higher powers will reduce to these seven, which make up the elements of a field with $2^3=8$ elements when we add zero to the set. We leave as an exercise for you to set up a table relating powers of $\beta$ with the linear combinations.