Extension Fields

For example, let \[\begin{aligned}\end{aligned}\] and let \(E = {\mathbb Q }( \sqrt{2} + \sqrt{3}\,)\) be the smallest field containing both \({\mathbb Q}\) and \(\sqrt{2} + \sqrt{3}\). Both \(E\) and \(F\) are extension fields of the rational numbers. We claim that \(E\) is an extension field of \(F\). To see this, we need only show that \(\sqrt{2}\) is in \(E\). Since \(\sqrt{2} + \sqrt{3}\) is in \(E\), \(1 / (\sqrt{2} + \sqrt{3}\,) = \sqrt{3} - \sqrt{2}\) must also be in \(E\). Taking linear combinations of \(\sqrt{2} + \sqrt{3}\) and \(\sqrt{3} - \sqrt{2}\), we find that \(\sqrt{2}\) and \(\sqrt{3}\) must both be in \(E\).

Let \(p(x) = x^2 + x + 1 \in {\mathbb Z}_2[x]\). Since neither \(0\) nor \(1\) is a root of this polynomial, we know that \(p(x)\) is irreducible over \({\mathbb Z}_2\). We will construct a field extension of \({\mathbb Z}_2\) containing an element \(\alpha\) such that \(p(\alpha) = 0\). By , the ideal \(\langle p(x) \rangle\) generated by \(p(x)\) is maximal; hence, \({\mathbb Z}_2[x] / \langle p(x) \rangle\) is a field. Let \(f(x) + \langle p(x) \rangle\) be an arbitrary element of \({\mathbb Z}_2[x] / \langle p(x) \rangle\). By the division algorithm, \[\begin{aligned}\end{aligned}\], where the degree of \(r(x)\) is less than the degree of \(x^2 + x + 1\). Therefore, \[\begin{aligned}\end{aligned}\]. The only possibilities for \(r(x)\) are then \(0\), \(1\), \(x\), and \(1 + x\). Consequently, \(E = {\mathbb Z}_2[x] / \langle x^2 + x + 1 \rangle\) is a field with four elements and must be a field extension of \({\mathbb Z}_2\), containing a zero \(\alpha\) of \(p(x)\). The field \({\mathbb Z}_2( \alpha)\) consists of elements \[\begin{aligned}0 + 0 \alpha & = 0 \\ 1 + 0 \alpha & = 1 \\ 0 + 1 \alpha & = \alpha \\ 1 + 1 \alpha & = 1 + \alpha\end{aligned}\]. Notice that \({\alpha}^2 + {\alpha} + 1 = 0\); hence, if we compute \((1 + \alpha)^2\), \[\begin{aligned}\end{aligned}\]. Other calculations are accomplished in a similar manner. We summarize these computations in the following tables, which tell us how to add and multiply elements in \(E\).

Let \(p(x) = x^5 + x^4 + 1 \in {\mathbb Z}_2[x]\). Then \(p(x)\) has irreducible factors \(x^2 + x + 1\) and \(x^3 + x + 1\). For a field extension \(E\) of \({\mathbb Z}_2\) such that \(p(x)\) has a root in \(E\), we can let \(E\) be either \({\mathbb Z}_2[x] / \langle x^2 + x + 1 \rangle\) or \({\mathbb Z}_2[x] / \langle x^3 + x + 1 \rangle\). We will leave it as an exercise to show that \({\mathbb Z}_2[x] / \langle x^3 + x + 1 \rangle\) is a field with \(2^3 = 8\) elements.

Both \(\sqrt{2}\) and \(i\) are algebraic over \({\mathbb Q}\) since they are zeros of the polynomials \(x^2 -2\) and \(x^2 + 1\), respectively. Clearly \(\pi\) and \(e\) are algebraic over the real numbers; however, it is a nontrivial fact that they are transcendental over \({\mathbb Q}\). Numbers in \({\mathbb R}\) that are algebraic over \({\mathbb Q}\) are in fact quite rare. Almost all real numbers are transcendental over \({\mathbb Q}\). The probability that a real number chosen at random from the interval \([0, 1]\) will be transcendental over the rational numbers is one. (In many cases we do not know whether or not a particular number is transcendental; for example, it is still not known whether \(\pi + e\) is transcendental or algebraic.)

We will show that \(\sqrt{2 + \sqrt{3} }\) is algebraic over \({\mathbb Q}\). If \(\alpha = \sqrt{2 + \sqrt{3} }\), then \(\alpha^2 = 2 + \sqrt{3}\). Hence, \(\alpha^2 - 2 = \sqrt{3}\) and \(( \alpha^2 - 2)^2 = 3\). Since \(\alpha^4 - 4 \alpha^2 + 1 = 0\), it must be true that \(\alpha\) is a zero of the polynomial \(x^4 - 4 x^2 + 1 \in {\mathbb Q}[x]\).