6. Finite Fields（有限域）
Now that we have developed the machinery of Galois theory, we apply it in this chapter to study special classes of field extensions. Sections 9 and 11 are good examples of how we can use group theoretic information to obtain results in field theory. Section 10 has a somewhat different flavor than the other sections. In it, we look into the classical proof of the Hilbert Theorem 90, a result originally used to help describe cyclic extensions, and from that proof we are led to the study of cohomology, a key tool in algebraic topology, algebraic geometry, and the theory of division rings.
In this section, we study finite fields and, more generally, finite extensions of finite fields.
Let \(F\) be a finite field, and say \(\mathrm{char}(F) = p\). We can view \(F\) as an extension field of \(\mathbb{F}_p\). Since \(F\) is finite, \(F\) is a finite dimensional \(\mathbb{F}_p\)-vector space. If \([F : \mathbb{F}_p] = n\), then \(F\) and \(\mathbb{F}_p^n\) are isomorphic as \(\mathbb{F}_p\)-vector spaces, so \(|F| = p^n\). We will first obtain some field theoretic information about \(F\) by investigating the group structure of the multiplicative group \(F^*\). For the next lemma, recall that if \(G\) is an Abelian group, then the exponent \(\exp(G)\) of \(G\) is the least common multiple of elements in \(G\). By a group theory exercise, there is an element of \(G\) whose order is \(\exp(G)\). From this fact, it follows that \(G\) is cyclic if and only if \(|G| = \exp(G)\). These facts are proven in Proposition 1.4 of Appendix C.
Lemma 6.1. If \(K\) is a field and \(G\) is a finite subgroup of \(K^*\), then \(G\) is cyclic.
Proof. Let \(n = |G|\) and \(m = \exp(G)\). Then \(m\) divides \(n\) by Lagrange's theorem. If \(g \in G\), then \(g^m = 1\), so each element of \(G\) is a root of the polynomial \(x^m - 1\). This polynomial has at most \(m\) roots in the field \(K\). However, \(x^m - 1\) has at least the elements of \(G\) as roots, so \(n \leq m\). Therefore, \(\exp(G) = |G|\), so \(G\) is cyclic. □
Corollary 6.2. If \(F\) is a finite field, then \(F^*\) is cyclic.
Example 6.3. Let \(F = \mathbb{F}_p\). A generator for \(F^*\) is often called a primitive root modulo \(p\). For example, \(2\) is a primitive root modulo \(5\). Moreover, \(2\) is not a primitive root modulo \(7\), while \(3\) is a primitive root modulo \(7\). In general, it is not easy to find a primitive root modulo \(p\), and there is no simple way to find a primitive root in terms of \(p\).
In Section 5, the primitive element theorem was stated for arbitrary base fields but was proved only for infinite fields. If \(K/F\) is an extension of finite fields, then there are finitely many intermediate fields. Therefore, the hypotheses of the primitive element theorem hold for \(K/F\). The following corollary finishes the proof of the primitive element theorem.
Corollary 6.4. If \(K/F\) is an extension of finite fields, then \(K\) is a simple extension of \(F\).
Proof. By the previous corollary, the group \(K^*\) is cyclic. Let \(\alpha\) be a generator of the cyclic group \(K^*\). Every nonzero element of \(K\) is a power of \(\alpha\), so \(K = F(\alpha)\). Therefore, \(K\) is a simple extension of \(F\). □
The following theorem exploits group theoretic properties of finite groups to give the main structure theorem of finite fields.
Theorem 6.5. Let \(F\) be a finite field with \(\mathrm{char}(F) = p\), and set \(|F| = p^n\). Then \(F\) is the splitting field of the separable polynomial \(x^{p^n} - x\) over \(\mathbb{F}_p\). Thus, \(F/\mathbb{F}_p\) is Galois. Furthermore, if \(\sigma\) is defined on \(F\) by \(\sigma(a) = a^p\), then \(\sigma\) generates the Galois group \(\mathrm{Gal}(F/\mathbb{F}_p)\), so this Galois group is cyclic.
Proof. Let \(|F| = p^n\), so \(|F^*| = p^n - 1\). By Lagrange's theorem, if \(a \in F^*\), then \(a^{p^n - 1} = 1\). Multiplying by \(a\) gives \(a^{p^n} = a\). This equation also holds for \(a = 0\). Therefore, the elements of \(F\) are roots of the polynomial \(x^{p^n} - x\). However, this polynomial has at most \(p^n\) roots, so the elements of \(F\) are precisely the roots of \(x^{p^n} - x\). This proves that \(F\) is the splitting field over \(\mathbb{F}_p\) of \(x^{p^n} - x\), and so \(F\) is normal over \(\mathbb{F}_p\). Moreover, the derivative test shows that \(x^{p^n} - x\) has no repeated roots, so \(x^{p^n} - x\) is separable over \(\mathbb{F}_p\). Thus, \(F\) is Galois over \(\mathbb{F}_p\).
Define \(\sigma : F \to F\) by \(\sigma(a) = a^p\). An easy computation shows that \(\sigma\) is an \(\mathbb{F}_p\)-homomorphism, and \(\sigma\) is surjective since \(F\) is finite. Hence, \(\sigma\) is an \(\mathbb{F}_p\)-automorphism of \(F\). The fixed field of \(\sigma\) is \(\{a \in F : a^p = a\} \supseteq \mathbb{F}_p\). Each element in \(\mathcal{F}(\sigma)\) is a root of \(x^p - x\), so there are at most \(p\) elements in \(\mathcal{F}(\sigma)\). This proves that \(\mathbb{F}_p = \mathcal{F}(\sigma)\), so \(\mathrm{Gal}(F/\mathbb{F}_p)\) is the cyclic group generated by \(\sigma\). □
The automorphism \(\sigma\) defined above is called the Frobenius automorphism of \(F\).
Corollary 6.6. Any two finite fields of the same size are isomorphic.
Proof. The proof of Theorem 6.5 shows that any two fields of order \(p^n\) are splitting fields over \(\mathbb{F}_p\) of \(x^{p^n} - x\), so the corollary follows from the isomorphic extension theorem. □
We can use Theorem 6.5 to describe any finite extension of finite fields, not only extensions of \(\mathbb{F}_p\).
Corollary 6.7. If \(K/F\) is an extension of finite fields, then \(K/F\) is Galois with a cyclic Galois group. Moreover, if \(\mathrm{char}(F) = p\) and \(|F| = p^n\), then \(\mathrm{Gal}(K/F)\) is generated by the automorphism \(\tau\) defined by \(\tau(a) = a^{p^n}\).
Proof. Say \([K : \mathbb{F}_p] = m\). Then \(\mathrm{Gal}(K/\mathbb{F}_p)\) is a cyclic group of order \(m\) by Theorem 6.5, so the order of the Frobenius automorphism \(\sigma\) of \(K\) is \(m\). The group \(\mathrm{Gal}(K/F)\) is a subgroup of \(\mathrm{Gal}(K/\mathbb{F}_p)\), so it is also cyclic. If \(s = |\mathrm{Gal}(K/F)|\) and \(m = ns\), then a generator of \(\mathrm{Gal}(K/F)\) is \(\sigma^n\). By induction, we see that the function \(\sigma^n\) is given by \(\sigma^n(a) = a^{p^n}\). Also, since \(s = [K : F]\), we have that \(n = [F : \mathbb{F}_p]\), so \(|F| = p^n\). □
We have described finite fields as extensions of \(\mathbb{F}_p\) and have shown that any finite extension of \(\mathbb{F}_p\) has \(p^n\) elements for some \(n\). However, we have not yet determined for which \(n\) there is a field with \(p^n\) elements. Using the fundamental theorem along with the description of finite fields as splitting fields in Theorem 6.5, we now show that for each \(n\) there is a unique up to isomorphism field with \(p^n\) elements.
Theorem 6.8. Let \(N\) be an algebraic closure of \(\mathbb{F}_p\). For any positive integer \(n\), there is a unique subfield of \(N\) of order \(p^n\). If \(K\) and \(L\) are subfields of \(N\) of orders \(p^m\) and \(p^n\), respectively, then \(K \subseteq L\) if and only if \(m\) divides \(n\). When this occurs, \(L\) is Galois over \(K\) with Galois group generated by \(\tau\), where \(\tau(a) = a^{p^m}\).
Proof. Let \(n\) be a positive integer. The set of roots in \(N\) of the polynomial \(x^{p^n} - x\) has \(p^n\) elements and is a field. Thus, there is a subfield of \(N\) of order \(p^n\). Since any two fields of order \(p^n\) in \(N\) are splitting fields of \(x^{p^n} - x\) over \(\mathbb{F}_p\) by Theorem 6.5, any subfield of \(N\) of order \(p^n\) consists exactly of the roots of \(x^{p^n} - x\). Therefore, there is a unique subfield of \(N\) of order \(p^n\).
Let \(K\) and \(L\) be subfields of \(N\), of orders \(p^m\) and \(p^n\), respectively. First, suppose that \(K \subseteq L\). Then
so \(m\) divides \(n\). Conversely, suppose that \(m\) divides \(n\). Each element \(a\) of \(K\) satisfies \(a^{p^m} = a\). Since \(m\) divides \(n\), each \(a\) also satisfies \(a^{p^n} = a\), so \(a \in L\). This proves that \(K \subseteq L\). When this happens \(L\) is Galois over \(K\) by Corollary 6.7. That corollary also shows that \(\mathrm{Gal}(L/K)\) is generated by \(\tau\), where \(\tau\) is defined by \(\tau(a) = a^{|K|}\). □
If \(F\) is a finite field and \(f(x) \in F[x]\), then Theorems 6.5 and 6.8 can be used to determine the splitting field over \(F\) of the polynomial \(f\).
Corollary 6.9. Let \(F\) be a finite field, and let \(f(x)\) be a monic irreducible polynomial over \(F\) of degree \(n\).
- If \(a\) is a root of \(f\) in some extension field of \(F\), then \(F(a)\) is a splitting field for \(f\) over \(F\). Consequently, if \(K\) is a splitting field for \(f\) over \(F\), then \([K : F] = n\).
- If \(|F| = q\), then the set of roots of \(f\) is \(\{a^{q^r} : r \geq 1\}\).
Proof. Let \(K\) be a splitting field of \(f\) over \(F\). If \(a \in K\) is a root of \(f(x)\), then \(F(a)\) is an \(n\)-dimensional extension of \(F\) inside \(K\). By Theorem 6.5, \(F(a)\) is a Galois extension of \(F\); hence, \(f(x) = \min(F, a)\) splits over \(F(a)\). Therefore, \(F(a)\) is a splitting field of \(f\) over \(F\), so \(K = F(a)\). This proves the first statement. For the second, we note that \(\mathrm{Gal}(K/F) = \langle \sigma \rangle\), where \(\sigma(c) = c^q\) for any \(c \in K\), by Theorem 6.8. Each root of \(f\) is then of the form \(\sigma^r(a) = a^{q^r}\) by the isomorphism extension theorem, which shows that the set of roots of \(f\) is \(\{a^{q^r} : r \geq 1\}\). □
Example 6.10. Let \(F = \mathbb{F}_2\) and \(K = F(\alpha)\), where \(\alpha\) is a root of \(f(x) = x^3 + x^2 + 1\). This polynomial has no roots in \(F\), as a quick calculation shows, so it is irreducible over \(F\) and \([K : F] = 3\). The field \(K\) is the splitting field of \(f\) over \(F\), and the roots of \(f\) are \(\alpha\), \(\alpha^2\), and \(\alpha^4\), by Corollary 6.9. Since \(f(\alpha) = 0\), we see that \(\alpha^3 = \alpha^2 + 1\), so \(\alpha^4 = \alpha^3 + \alpha = \alpha^2 + \alpha + 1\). Therefore, in terms of the basis \(\{1, \alpha, \alpha^2\}\) for \(K/F\), the roots of \(f\) are \(\alpha\), \(\alpha^2\), and \(1 + \alpha + \alpha^2\). This shows explicitly that \(F(\alpha)\) is the splitting field of \(f\) over \(F\).
Example 6.11. Let \(F = \mathbb{F}_2\) and \(f(x) = x^4 + x + 1\). By the derivative test, we see that \(f\) has no repeated roots. The polynomial \(f\) is irreducible over \(F\), since \(f\) has no roots in \(F\) and is not divisible by the unique irreducible quadratic \(x^2 + x + 1\) in \(F[x]\). If \(\alpha\) is a root of \(f\), then \(\alpha^4 = \alpha + 1\); hence, the roots of \(f\) are \(\alpha\), \(\alpha + 1\), \(\alpha^2\), and \(\alpha^2 + 1\).
Example 6.12. Let \(f(x) = x^2 + 1\). If \(p\) is an odd prime, then we show that \(f\) is reducible over \(F = \mathbb{F}_p\) if and only if \(p \equiv 1 \pmod{4}\). To prove this, if \(a \in F\) is a root of \(x^2 + 1\), then \(a^2 = -1\), so \(a\) has order \(4\) in \(F^*\). By Lagrange's theorem, \(4\) divides \(|F^*| = p - 1\), so \(p \equiv 1 \pmod{4}\). Conversely, if \(p \equiv 1 \pmod{4}\), then \(4\) divides \(p - 1\), so there is an element \(a \in F^*\) of order \(4\), since \(F^*\) is a cyclic group of order \(p - 1\). Thus, \(a^4 = 1\) and \(a^2 \neq 1\). This forces \(a^2 = -1\), so \(a\) is a root of \(f\).
If \(F\) is a finite field, then we have seen that every finite extension of \(F\) is Galois over \(F\). Hence, every extension of \(F\) is separable over \(F\). Every algebraic extension of \(F\) is then separable over \(F\), so \(F\) is perfect. To note this more prominently, we record this as a corollary. We have already seen this fact in Example 4.14.
Corollary 6.13. Every finite field is perfect.
Given an integer \(n\), Theorem 6.8 shows that there is a finite field with \(p^n\) elements. For a specific \(n\), how do we go about finding this field? To construct finite fields, we can use irreducible polynomials over \(\mathbb{F}_p\). Note that if \(f(x)\) is an irreducible polynomial of degree \(n\) in \(\mathbb{F}_p[x]\), then \(\mathbb{F}_p[x]/(f(x))\) is a field extension of degree \(n\) over \(\mathbb{F}_p\); hence, it has \(p^n\) elements. Conversely, if \(F\) has \(p^n\) elements, and if \(F = \mathbb{F}_p(\alpha)\), then \(\min(\mathbb{F}_p, \alpha)\) is an irreducible polynomial of degree \(n\). Therefore, finding finite fields is equivalent to finding irreducible polynomials in \(\mathbb{F}_p[x]\). For instance, \(\mathbb{Z}_2[x]/(x^2 + x + 1)\) is a field of \(4\) elements, and \(\mathbb{Z}_5[x]/(x^4 - 7)\) is a field of \(5^4 = 625\) elements. The following proposition gives one way of searching for irreducible polynomials over \(\mathbb{F}_p\).
Proposition 6.14. Let \(n\) be a positive integer. Then \(x^{p^n} - x\) factors over \(\mathbb{F}_p\) into the product of all monic irreducible polynomials over \(\mathbb{F}_p\) of degree a divisor of \(n\).
Proof. Let \(F\) be a field of order \(p^n\). Then \(F\) is the splitting field of \(x^{p^n} - x\) over \(\mathbb{F}_p\) by Theorem 6.5. Recall that \(F\) is exactly the set of roots of \(x^{p^n} - x\). Let \(a \in F\), and set \(m = [\mathbb{F}_p(a) : \mathbb{F}_p]\), a divisor of \([F : \mathbb{F}_p]\). The polynomial \(\min(\mathbb{F}_p, a)\) divides \(x^{p^n} - x\), since \(a\) is a root of \(x^{p^n} - x\). Conversely, if \(f(x)\) is a monic irreducible polynomial over \(\mathbb{F}_p\) of degree \(m\), where \(m\) divides \(n\), let \(K\) be the splitting field of \(f\) over \(\mathbb{F}_p\) inside some algebraic closure of \(F\). If \(a\) is a root of \(f\) in \(K\), then \(K = \mathbb{F}_p(a)\) by Corollary 6.9. Therefore, \([K : \mathbb{F}_p] = m\), so \(K \subseteq F\) by Theorem 6.8. Thus, \(a \in F\), so \(a\) is a root of \(x^{p^n} - x\). Since \(f\) is irreducible over \(\mathbb{F}_p\), we have \(f = \min(\mathbb{F}_p, a)\), so \(f\) divides \(x^{p^n} - x\). Since \(x^{p^n} - x\) has no repeated roots, \(x^{p^n} - x\) factors into distinct irreducible factors over \(\mathbb{F}_p\). We have shown that the irreducible factors of \(x^{p^n} - x\) are exactly the irreducible polynomials of degree a divisor of \(n\); hence, the proposition is proven. □
Example 6.15. The monic irreducible polynomials of degree \(5\) over \(\mathbb{F}_2\) can be determined by factoring \(x^{2^5} - x\), which we see factors as
This factorization produces the six monic irreducible polynomials of degree \(5\) over \(\mathbb{F}_2\). Note that we only need one of these polynomials in order to construct a field with \(2^5\) elements. Similarly, the monic irreducible polynomials of degree \(2\), \(3\), or \(6\) over \(\mathbb{F}_2\) can be found by factoring \(x^{2^6} - x\). For example, \(x^6 + x + 1\) is an irreducible factor of \(x^{64} - x\), so \(\mathbb{F}_2[x]/(x^6 + x + 1)\) is a field with \(64\) elements. The factorization of \(x^{32} - x\) and the factor \(x^6 + x + 1\) of \(x^{64} - x\) was found by using the computer algebra program Scientific Workplace.