Table of Content
- The characteristic \(\text{char}(F)\) is either a prime or \(0\).
- If \(p = \text{char}(F)\) is a prime, then \(F\) can be regarded as a field extension of \(\Bbb{Z}_p\).
- If \(\text{char}(F)=0\), then \(F\) can be regraded as a field extension of \(\Bbb Q\).
Proof (prop. 1)
Suppose \(\text{char}(F) = n < \infty\). If \(n\) is composite, \(n=ab\) for some \(a,b\in \Bbb{Z}^+\).
Then
\[0 = n \cdot 1 = (a\cdot 1)(b \cdot 1).\]This implies that \(a = 0\) or \(b = 0\), which is a contradiction. Therefore, \(n\) is either prime or \(0\). ◼
Proof (prop. 2)
We know that \(\Bbb{Z}_p\) is of characteristic \(p\). Since \(\Bbb{Z}_p\) is cyclic, it means that \(\Bbb{Z}_p\) is the smallest field of characteristic \(p\). Thus, \(\Bbb{Z}_p \le E\). ◼
Let \(E\) be a finite extension of degree \(n\) over a finite field \(F\). If \(F\) has \(q\) elements, then \(E\) has \(q^n\) elements.
Proof
If \(E\) is a finite field of characteristic \(p\), then the number of elements in \(E\) is \(p^n\) for some positive integer \(n\).
Check proposition 2..
If \(E\) is a vector sapce over \(F\) whose dimension is \(n\), then \(E \cong F^n\).
若要建構一有 \(p^n\) 個元素的 field，就找一 irreducible polynomial \(f(x)\) over \(\Bbb{Z}_p\) of degree \(n\)。令 \(f(\alpha) = 0, \alpha \not \in \Bbb{Z}_p\)，所求的 field 就是 \(\Bbb{Z}_p(\alpha)\)！
A finite extension \(E\) of a finite field \(F\) is a simple extension of \(F\).
Proof
Let \(\alpha\) be a generator of the multiplicative group \(E^\times\). Then \(E=F(\alpha)\). ◼
No finite field is algebraically closed.
Proof
Suppose \(F\) and \(E\) are two finite fields, and \(F\le E\). We have
\[[E:\Bbb Z_p] = [E:F][F:\Bbb Z_p].\]Since \(F, E\) are both finite, \([E:\Bbb Z_p]\) and \([F:\Bbb Z_p]\) are finite, which implies that \([E:F]\) is finite as well. By this lemma, there exists \(\alpha \in E\) such that \(\alpha\) is algebraic over \(F\). Hence, \(F\) is not algebraically closed. ◼
Let \(E\) be a field of \(p^n\) elements contained in \(\overline{\Bbb{Z}}_p\). Then the elements of \(E\) are precisely the zeros of \(x^{p^n}-x\) in \(\overline{\Bbb{Z}}_p\).
Proof
There are \(p^n-1\) elements in \(E^\times\). By Lagrange’s theorem, every nonzero element of \(E\) is a zero of \(x^{p^n-1}-1\). Thus, every elements if \(E\) is a zero of \(x^{p^n}-x = x(x^{p^n-1}-1)\).
On the other hand, since \(\overline{\Bbb Z}_p\) is a field, a polynomial of degree \(p^n\) has at most \(p^n\) zeros.
Therefore the theorem is proved. ◼
A finite field of \(p^n\) elements exists for every prime power \(p^n\).
Here, we shall show that the converse of this theorem is true, i.e., the set \(S\) containing zeros of \(x^{p^n}-x\) in \(\overline{\Bbb Z}_p\) forms a field of \(p^n\) elements. Note that since \(\overline{\Bbb Z}_p\) is algebraically closed, \(x^{p^n}-x\) factors completely into linear factors over \(\overline{\Bbb Z}_p\). We will show
Before our proof begins, let’s define an useful property.
Let \(F\) be a field. The map \(D: F[x]\to F[x]\) defined by \(D(a_0+a_1x\cdots+a_nx^n) = a_1+2a_2x + \cdots+na_nx^{n-1}\) is called the derivation, and \(D(f(x))\) is called the derivative of \(f(x)\).
形式上的微分！
Let \(F\) be a field. The derivation \(D\) on \(F[x]\) satisfies
- \(D(f+g) = D(f) + D(g)\).
- \(D(fg) = gD(f) + fD(g)\).
Here we go.
1. zeros are distinct
Suppose that \(\alpha\) is a repeated zero of \(x^{p^n}-x\), say, \(x^{p^n}-x = (x-\alpha)^2g(x)\) for some \(g(x)\in \overline{\Bbb Z}_p[x]\). We have
\[D((x-\alpha)^2g(x)) = 2(x-\alpha)g(x) + (x-\alpha)^2D(g(x)),\]and thus \(\alpha\) is also a zero of
\[D((x-\alpha)^2g(x)) = D(x^{p^n}-x) = p^nx^{p^n-1}-1.\]However, since \(\alpha\) is a zero of \(x^{p^n}-x = x(x^{p^n-1}-1)\), \(\alpha^{p^n-1} = 1\). And since \(\text{char}(\overline{\Bbb Z}_p) = p\), i.e. \(p\cdot 1 = 0\), we have \(p^n\alpha^{p^n-1} = 0\). That is,
\[D((x-\alpha)^2g(x)) = -1,\]which is a contradiction. Therefore, \(x^{p^n}-x\) has no repeated root.
2. \(S\) is a sundomain
Recall that any subring of a field is a subdomain. Thus it suffices to show that \(S\) is a subring; we know that \(a\in S\) if \(a^{p^n} = a\).
additive identity
\(0^{p^n} = 0\). Thus \(0 \in S\).
closed
Let \(\alpha\) and \(\beta\) be two elements in \(S\).
Recall that \(p \mid {p \choose k}\) for all \(1\le k \le p-1\). Since \(p\cdot 1 = p = 0\), for \(k\) in that range we have
\[{p\choose k}u^kv^{p-k} = 0,\]for all \(u, v \in \overline{\Bbb Z}_p\). This means that \((u+v)^p = u^p + v^p\) for all \(u, v \in \overline{\Bbb Z}_p\), and thus
\[(\alpha+\beta)^{p^n} = ((\alpha+\beta)^p)^{p^{n-1}} = (\alpha^p + \beta^p)^{p^{n-1}} = \cdots = \alpha^{p^n} + \beta^{p^n} = \alpha + \beta.\]Therefore, \(\alpha+\beta \in S\).
On the other hand, closedness of multiplication is easily proved, which is omitted here.
additive inverse
Suppose \(\alpha \in S\). If \(p=2\), then \(-\alpha=\alpha \in S\). Otherwise, \(p\) is odd and
\[(-\alpha)^{p^n} = -\alpha^{p^n} = -\alpha,\]which means \(-\alpha\in S\). ◼
Let \(\alpha\) be a zero of \(f(x)\in \Bbb Z_p[x]\) in \(\overline{\Bbb Z}_p\). Then \(\alpha^p\) is also a zero of \(f(x)\).
Proof
Let \(f(x) = \sum_{i=0}^n a_ix^i\), where \(a_i\in \Bbb{Z}_p\). This means that \(a_i^p = a_i\) for all \(i \in \{0, 1, \cdots, (n-1)\}\). Then
\[0 = \big(f(\alpha) \big)^p = \Big(\sum_{i=0}^n a_i\alpha^i\Big)^p.\]By the same argument of the closed part in the previous proof, we have
\[\Big(\sum_{i=0}^n a_i\alpha^i\Big)^p = \sum_{i=0}^n a_i^p\alpha^{pi} = \sum_{i=0}^n a_i\alpha^{pi} = f(\alpha^p) = 0. \tag*{$\blacksquare$}\]綜合以上兩 theorem，對於任意 prime power \(p^n\)，存在唯一的 finite field，其包含 \(p^n\) 個 \(\overline{\Bbb Z}_p\) 中的元素，稱為 the Galois field，以 \(\Bbb{F}_{p^n}\) 或 \(GF(p^n)\) 表示。
Let \(F\) be a field of \(p^n\) elements. Then \(F \cong \Bbb{F}_{p^n}\).
Proof
Let \(\alpha\) be a generator of \(F^\times\) and \(\alpha'\) be a zero of \(f(x) = \text{Irr}(\alpha, \Bbb{Z}_p)\) in \(\overline{\Bbb Z}_p\). Then
\[F = \Bbb{Z}_p(\alpha) \cong \Bbb{Z}_p[x]\Big/f(x)\Bbb{Z} _p[x] \cong \Bbb{Z}_p(\alpha') = GF(p^n). ◼\]參考 Field extension。
Recall that \(n\)-th cyclotomic polynomial is defined to be
\[\Phi_p(x) = \prod_{\substack{a\in\mathbb{C}^\times \\ \text{ord}(a) = n}} (x-a),\]which can be computed inductively using the formula
\[x^n - 1 = \prod_{d\mid n}\Phi_d(x).\]If we replace \(\Bbb C\) by \(\overline{\Bbb Z}_p\) in the definition of \(\Phi_n(x)\) (and denote the resulted function as \(\Phi_{n,p}(x)\)), then the above formula still holds if \(p\not \mid n\). Especially, the above formula shows that when \(\text{char}(F)=p\), then \(\Phi_{n, p}(x) \in \Bbb{Z}_p[x]\). Moreover, we also have \(\Phi_n(x) \equiv \Phi_{n,p}(x) \pmod p\), given \(p\not \mid n\).
What if \(p \mid n\)?
Let us denote \(\Phi_{n, p}(x)\) still by \(\Phi_n(x)\) for short. For \(GF(16)^\times\), its elements are zeros of \(x^{15}-1\). By the above formula, we have
\[x^{15}-1 = \Phi_1(x)\Phi_3(x)\Phi_5(x)\Phi_{15}(x).\]Moreover, over \(\Bbb{Z}_2\), we have
\[\begin{align*} \Phi_1(x) &= x - 1 \\ \Phi_3(x) &= x^2 + x + 1\\ \Phi_5(x) &= x^4 + x^3 + x^2 + x + 1 \\ \Phi_{15}(x) &= (x^4+x+1)(x^4+x^3+1) . \end{align*}\]Because of this corollary, \(\Phi_{15}(x)\) must be factored into irreducible polynomials of degree \(1, 2\) or \(4\).
Therefore, there are exactly three irreducible monic polynomials of degree \(4\) over \(\Bbb Z_2\). Besides, zeros of \(x^4 + x^3 + x^2 + x + 1\) are of multiplicative order \(5\), and zeros of \(x^4+x+1\) and \(x^4+x^3+1\) are of multiplicative order \(15\).
Q: How to find the order of those zeros?
A: This is by definition. (See the substack of \(\prod\).) To it state explicitly: All the zeros of \(\Phi_n(x)\) are of degree \(n\). (any exceptions?)
Now if \(\alpha\) is a zero of \(x^4+x+1\), then
\[\Bbb{Z}_2(\alpha)^\times = \{1, \alpha, \cdots, \alpha^{14}\} \cong \Bbb{Z}_{15}.\]因為 \(\alpha\) 的 order 是 \(15\)，所以他是 \(\Bbb Z_2(\alpha)^\times\) 的 generator！
The elements of order \(3\) are \(\alpha^5\) and \(\alpha^{10}\), which implies that
\[x^2 + x + 1 = (x-\alpha^5)(x-\alpha^{10}).\]Similarly, the elements of order \(5\) are \(\alpha^3, \alpha^6, \alpha^9\) and \(\alpha^{12}\), which implies that
\[x^4+x^3+x^2+x+1 = (x-\alpha^3) (x-\alpha^6) (x-\alpha^9 )(x-\alpha^{12}).\]For elements of order \(15\), they are \(\{1,2,4,7,8,11,13,14 \}\), and they can be divided into two subsets: \(\{1,2,4,8 \}\) and \(\{7, 14, 13, 11\}\). We can see that
\[1 \xrightarrow[]{\times 2}2 \xrightarrow[]{\times 2}4 \xrightarrow[]{\times 2}8,\]and
\[7 \xrightarrow[]{\times 2}14 \xrightarrow[]{\times 2}13 \xrightarrow[]{\times 2}11,\]modulo \(15\). Then
\[x^4+x+1 = (x-\alpha)(x-\alpha^2)(x-\alpha^4)(x-\alpha^8), \\ x^4+x^3+1 = (x-\alpha^7)(x-\alpha^{14})(x-\alpha^{13})(x-\alpha^{11}).\]But how are they factored?
為什麼要 \(\times 2\)？見 Frobenius Automorphism！
綜合以上定義與觀察，回答
並試圖歸納以上理論！
In \(\overline{\Bbb Z}_p\), \(\Bbb{F}_{p^m}\) is contained in \(\Bbb{F}_{p^n}\) iff \(m\mid n\).
Proof
Suppose \(\Bbb{F}_{p^m} \le \Bbb{F}_{p^n}\). Then
\[n = [\Bbb{F}_{p^n}:\Bbb{Z}_p]=[\Bbb{F}_{p^n}:\Bbb{F}_{p^m}][\Bbb{F}_{p^m}:\Bbb{Z}_p] = m[\Bbb{F}_{p^n}:\Bbb{F}_{p^m}].\]Thus \(m\mid n\).
Conversely, assume that \(m \mid n\). It suffice to prove that if \(\alpha\in \overline{\Bbb Z}_p\) is a zero of \(x^{p^m}-x\), then it is also a zero of \(x^{p^n}-x\).
For all \(\alpha\in \Bbb{F}_{p^m}\), we have \(\alpha^{p^m} = \alpha\). Then
\[\alpha^{p^n} = (\alpha^{p^m})^{p^{n-m}} = \alpha^{p^{n-m}} = \cdots.\]Since \(m\mid n\), we eventually arrive at \(\alpha^{p^n}=\alpha\). ◼
The polynomial \(x^{p^n}-x\) is equal to the product of all monic irreducible polynomials over \(\Bbb{Z}_p\) with degree \(d\) dividing \(n\).
Proof
Let \(f(x)\) be a monic irreducible polynomial of degree \(d\) over \(\Bbb{Z}_p\), with \(d\mid n\) and \(\beta\) be its zero in \(\overline{Z}_p\). Since \(d\mid n\), we have
\[\Bbb{Z}_p(\beta) = \Bbb{F}_{p^d} \le \Bbb{F}_{p^n}.\]Therefore, \(\beta\) is a zero of \(x^{p^n}-x\), which implies that \(f(x)\mid x^{p^n}-x\).
Conversely, let \(f(x)\) be a monic irreducible factor of \(x^{p^n}-x\) of degree \(d\), with a zero \(\beta\) in \(\overline{\Bbb Z}_p\). Since \(\beta\) is also a zero of \(x^{p^n}-x\), we have \(\beta \in \Bbb{F}_{p^n}\). Thus,
\[\Bbb{F}_{p^d} = \Bbb{Z}_p(\beta) \le \Bbb{F}_{p^n},\]and then we conclude that \(d\mid n\). ◼
先證所有 \(d\mid n\) 都是 factor，再證所有 factor 都 \(d\mid n\)！
Find the product of monic irreducible polynomials of degree four over \(\Bbb Z_3\).
We know that \(x^{3^4}-x\) is equal to the product of all monic irreducible polynomials over \(\Bbb Z_3\) with degree \(d\) dividing \(4\), i.e. \(1, 2\) and \(4\). However we only need the product of those of degree \(4\), which is
\[{x^{3^4}-x\over x^{3^2}-x} = {x^{81}-x\over x^{9}-x} = x^{72} + x^{64} + \cdots + 1. \tag*{$\blacksquare$}\]Show that \(\Phi_{11}(x)\) is irreducible over \(\Bbb Z_3\).
Suppose \(\alpha\) is a zero of \(\Phi_{11}(x)\). By definition, the order of \(\alpha\) is \(11\).
Then, suppose \([\Bbb Z_3(\alpha):\Bbb Z_3] = d\), which can only be one of \(1, 2, 5\) and \(10\), since all the irreducible factors of a cyclotomic polynomial have the same degree. By Lagrange’s Theorem, \(\vert \langle \alpha \rangle\vert\) divides \(\vert \Bbb{Z}_3(\alpha)^\times\vert\), which is to say
\[11 \mid 3^d-1.\]After some calculation, we can discover that only when \(d=10\), \(11\) is a factor of \(3^d-1\). Thus \([\Bbb Z_3(\alpha):\Bbb Z_3] = 10\) and \(\Phi_{11}(x)\) is irreducible over \(\Bbb Z_3\). ◼
In general, \(\Phi_{n}(x)\) is irreducible over \(\Bbb Z_p\) iff \(p\) is a generator of \(\Bbb Z_n^\times\).
Check whether \(\Phi_9(x)\) is irreducible over \(\Bbb Z_3\).
First note that \(\deg\Phi_9(x) = 6\). Suppose \(\alpha\) is a zero of \(\Phi_9(x)\), and suppose \([\Bbb Z_3(\alpha): \Bbb Z_3] = d\). Then by Lagrange’s Theorem, we have \(9 \mid 3^d - 1\), where \(d=1, 2, 3\) or \(6\). Unfortunately, none of the values fit. How? Let’s then try to factorize \(\Phi_9(x)\) over \(\Bbb Z_3\).
\(\Phi_9(x) = x^6 + x^3 + 1 = x^6 -2x^3 + 1\). After some more calculations, we have
\[\Phi_9(x) = (x-1)^6,\]which yields that \(\alpha = 1\in \Bbb Z_3\)! It turns out that our belief in “\(\text{ord}(\alpha) = 9\)” is incorrect. This is because \(3\mid 9\), which violates our assumption that \(p\not \mid n\) here.
In this case, \(\Phi_9(x)\) is indeed reducible over \(\Bbb Z_3\). ◼
在這裡，我們說明了對於 field extension，加入 generator 是比較好的。但是如何找到 a zero \(\alpha\) of an irreducible polynomial \(f(x)\)，使得 \(\alpha\) 是 generator？
如果 \(\alpha\) 是 generator，則 \(\alpha\) 的 order 必須是 \(p^n-1\)，等同於，\(\alpha\) 是 \(\Phi_{p^n-1}(x)\) 的 zero 之一；也就是說，\(f(x)\) 必須是 \(\Phi_{p^n-1}(x)\) 的不可約因式之一。
於是，做體擴張的時候都選 \(\Phi_{p^n-1}(x)\) 的 zero 加入，就保證是 generator 了！
Every finite field \(F\) is a finite extension of \(\Bbb Z_p\) in \(\overline{\Bbb Z}_p\), and \(\vert F\vert = p^n\) where \(n=[F:\Bbb Z_p]\).
To construct \(\Bbb F_{p^n}\), we have to find an irreducible polynomial \(f(x)\) of degree \(n\) over \(\Bbb Z_p\). Then, \(\Bbb Z_p(\alpha)\) is the desired field where \(\alpha\) is a zero of \(f(x)\) in \(\overline{\Bbb Z}_p\).