Table of Content
A field \(E\) is an extension field of a field \(F\) if \(F\le E\), denoted by \(E/F\).
Let \(F\le E\). If there exists \(\alpha\in E\) such that \(E=F(\alpha)\), then \(E=F(\alpha)\) is a simple extension of \(F\).
Let \(E\) be a field extension of \(F\). For \(\alpha\in E\), let
\[F[\alpha]:=\{f(\alpha): f(x)\in F[x]\},\]which is the minimal subring containing \(F\) and \(\alpha\); let
\[F(\alpha):=\{f(\alpha)/g(\alpha): f(x),g(x)\in F[x], g(\alpha) \not = 0 \},\]which is the minimal subfield containing \(F\) and \(\alpha\).
An element \(\alpha\) of an extension field \(E\) of a field \(F\) is algebraic over \(F\) if \(f(\alpha) = 0\) for some nonzero polynomial \(f(x)\in F[x]\). If such a polynomial does not exist, then \(\alpha\) is transcendental over \(F\).
Example
就看 over 哪個 field！
Show that \(x^2-\pi\) is irreducible over \(\mathbb{Q}(\pi)\).
Solution
Suppose \(f(\pi)/g(\pi)\) is a zero of \(x^2-\pi\) where \(g(\pi)\not = 0\). Then
\[\Big({f(\pi)\over g(\pi)}\Big)^2 - \pi = 0, \\ f(\pi)^2 - \pi g(\pi)^2 = 0,\]which means that \(\pi\) is a zero of \(f(x)^2 - xg(x)^2 \in \mathbb{Q}[x]\). This is a contradiction. Therefore, \(\pi\) is irreducible over \(\mathbb{Q}(\pi)\). ◼
Let \(E\) be an extension field of a field \(F\), and let \(\alpha\in E\). Then \(\alpha\) is transcendental over \(F\) iff the evaluation homomorphism \(\phi_\alpha: F[x]\to E\) is an isomorphism of \(F[x]\) with a subdoamin of \(E\).
Proof
It suffice to show that \(\alpha \in E\) is transcendental over \(F\) iff \(\phi_\alpha\) is one-to-one.
因為只要 isomorphic to a subdomain！
可用 \(\text{ker}(\phi_\alpha) = \{0\}\)。
Remark
這個定理說明了，如果 \(\alpha\) is transcendental over \(F\)，則 \(F[\alpha]\) 和 \(F[x]\) 看起來沒兩樣。
Assume that \(F\le E\) and \(\alpha \in E\) is algebraic over \(F\). Let \(I_\alpha = \{f(x)\in F[x]: f(\alpha)=0 \}\) be a subset of \(F[x]\) and let \(p_\alpha(x)\) be a nonzero polynomial in \(I_\alpha\) of minimal degree.
很像 kernel！
\(p_\alpha(x)\) is irreducible.
Proof
Suppose \(p_\alpha(x) = r(x)s(x)\). Then we have \(r(\alpha)s(\alpha) = 0\). …
If \(p_\alpha(x)\) and \(p_{\alpha}'(x)\) are both nonzero monic polynomials in \(I_\alpha\) of minimal degree, then \(p_\alpha(x) = p_{\alpha}'(x)\).
Proof
\[\deg{(p_\alpha(x) - p_{\alpha}'(x))} < \deg{p_\alpha(x)}...\]For all \(f(x)\in I_\alpha\), \(p_\alpha(x)\mid f(x)\). In other words, \(I_\alpha = p_\alpha(x)F[x]\).
Proof
By the division algorithm, we have
\[f(x) = p_\alpha(x)q(x) + r(x).\]Since \(f(\alpha) = 0\), \(r(\alpha) = 0\). However, \(\deg r(x) < \deg p_\alpha(x)\), but \(p_\alpha(x)\) is the monic polynomial in \(I_\alpha\) with minimal degree. Thus \(r(x) = 0\) and this means that \(p_\alpha(x) \mid f(x)\). ◼
This proposition implies that \(p_\alpha(x)\) is the only monic irreducible polynomial in \(I_\alpha\).
The unique monic irreducible polynomial \(p(x)\) in \(I_\alpha\) is called the irreducible polynomial for \(\alpha\) over \(F\), and is denoted by \(\text{Irr}(\alpha, F)\). The degree of \(\text{Irr}(\alpha, F)\) is the degree of \(\alpha\) over \(F\), and is denoted by \(\deg{(\alpha, F)}\).
\(\text{Irr}(\alpha, F) = \text{Irr}(\alpha, F)(x)\), and that \(\text{Irr}(\alpha, F)(\alpha) = 0\).
如何找出 \(\text{Irr}(\alpha, F)\)？先找到 \(f(x)\) 使得 \(f(\alpha)=0\)，再驗證 \(f(x)\) is irreducible！
如果 \(\deg{(\alpha, F)} = n\) 而且 \(\vert F\vert = m\)，則 \(\vert F(\alpha)\vert = m^n\).
從這裡大小的比較可以看出，\(F(\alpha)\) 的確是 \(F\) 的extension！
Let \(F_1\subset F_2 \subset F_3\) be three fields. Suppose \(\alpha \in F_3\) is algebraic over \(F_1\). Show that \(\text{Irr}(\alpha, F_2)\big\vert \text{Irr}(\alpha, F_1)\).
Solution
Let \(I_{\alpha, i} = \{f(x)\in F_i[x]\mid f(\alpha)=0 \}\). Since \(I_{\alpha, 1} \subseteq I_{\alpha, 2}\), we have \(\deg{(\alpha, F_1)} \ge \deg{(\alpha, F_2)}\). By the division algorithm,
\[\text{Irr}(\alpha, F_1) = \text{Irr}(\alpha, F_2)q(x) + r(x),\]where \(q(x), r(x)\in F_2[x]\). Since \(\text{Irr}(\alpha, F_1)(\alpha) = \text{Irr}(\alpha, F_2)(\alpha) = 0\), \(r(\alpha) = 0\). This means that \(r(x)\) must be the zero polynomial, and therefore the statement is proved. ◼
If \(\alpha\) is algebraic over \(F\), then \(F(\alpha) = F[\alpha]\).
Proof
It is clear that \(F[\alpha] \subset F(\alpha)\). It remains to show that every \(f(\alpha)/g(\alpha)\in F(\alpha)\) is also in \(F[\alpha]\).
Since \(g(\alpha)\not = 0\), \(g(x)\) is not divisible by \(\text{Irr}(\alpha, F)\). Moreover, \(\text{Irr}(\alpha, F)\) is irreducible, so \(1\) is a GCD of \(g(x)\) and \(\text{Irr}(\alpha, F)\). Thus, we can find \(p(x)\) and \(q(x)\) such that
\[p(x)g(x) + q(x)\text{Irr}(\alpha, F) = 1\]Then we have \(p(\alpha)g(\alpha) = 1\), which implies that
\[{f(\alpha)\over g(\alpha)} = {f(\alpha)p(\alpha)\over g(\alpha)p(\alpha)} = f(\alpha)p(\alpha) \in F[\alpha]. \tag*{$\blacksquare$}\]If \(\alpha\) is algebraic over \(F\), then any \(\beta\in F(\alpha)\) can be uniquely expressed in the form \(\beta=b_0+b_1\alpha+\cdots+b_{n-1}\alpha^{n-1}\), where \(n=\deg(\text{Irr}(\alpha, F))\).
By this theorem, we have \(F(\alpha)=F[\alpha]\) since \(\alpha\) is algebraic over \(F\). Thus \(\beta=f(\alpha)\) for some \(f(x)\in F[x]\).
existence
By the division algorithm, there exists \(q(x)\) and \(r(x)\) such that \(f(x) = q(x)\text{Irr}(\alpha, F) + r(x)\) with \(r(x) = 0\) or \(\deg{r(x)} < n\). Then \(r(\alpha)=f(\alpha) = \beta\).
uniqueness
Now suppose \(\beta=f(\alpha)=g(\alpha)\) where \(f(x)\) and \(g(x)\) are two distinct polynomials of degree less than \(n\). However, \(f(x)-g(x)\) then become a non-zero polynomial with the zero \(\alpha\) and of degree less than \(n\), which is a contradiction. ◼
Let \(\beta = 1/(\sqrt[3]{4} - 2\sqrt[3]{2} + 2) \in \mathbb{Q}(\sqrt[3]{2})\). Write \(\beta\) into the form \(a_0 + a_1\sqrt[3]{2} + a_2\sqrt[3]{4}\) where \(a_i\in \mathbb{Q}\).
Solution
Let \(f(x)=\text{Irr}(\sqrt[3]{2}, \mathbb{Q}) = x^3-2\). Let \(g(x) = x^2-2x+2\), and then \(\beta=g^{-1}(\sqrt[3]{2})\).
Now our goal is to find the inverse of \(g(x)\) in \(\mathbb{Q}(\sqrt[3]{2})\Big/f(x)\mathbb{Q}(\sqrt[3]{2})\). This can be achieved by applying the Euclidean Algorithm. After some calculation, we have
\[{1\over 10 }\cdot g(x)(x^2+3x+4) - {1\over 10} \cdot f(x)(x+1) = 1.\]Thus, \(g^{-1}(x) = (x^2+3x+4) / 10\), and \(\beta = (\sqrt[3]{4}^2+3\sqrt[3]{2}+4)/10\). ◼
\(\sqrt[3]{2}\) is algebraic over \(\mathbb{Q}\).
Let \(f(x) = x^2+1 \in \mathbb{Z}_3[x]\). Since \(f(0)=1\) and \(f(1)=f(2)=2\), \(f(x)\) is irreducible over \(\mathbb{Z}_3[x]\). Let \(\alpha\) be a zero of \(f(x)\) in some extension of \(\mathbb{Z}_3\). Then
\[\mathbb{Z}_3(\alpha) = \{a_0+a_1\alpha\mid a_i \in \mathbb{Z}_3\},\]which is a field of \(9\) elements.
Since the multiplicative group of every finite field is cyclic, we can find a generator of \(\mathbb{Z}_3^\times(\alpha)\); let’s first check whether \(\alpha\) is a generator:
\[\alpha^2 = -1 \Rightarrow \alpha \text{ is not a generator.}\]Now let \(\beta=\alpha+1\), and then
\[\beta^2 = (\alpha+1)^2 = 2\alpha, \\ \beta^4 = 4\alpha^2 = -4 = -1 \Rightarrow \beta \text{ is a generotor}.\]Thus, \(\mathbb{Z}_3^\times(\alpha) = \langle \beta\rangle\).
Since \(\mathbb{Z}_3(\alpha) = \mathbb{Z}_3(\alpha+1) = \mathbb{Z}_3(\beta)\) and \(\beta\) is a generator, it is preferred to add \(\beta\) to \(\mathbb{Z}_3\) than \(\alpha\).
\(\beta\) is a zero of \(g(x) = (x-1)^2+1 = x^2-2x+2 = x^2+x+2 = \text{Irr}(\beta, F)\).
If our field extension is \(\mathbb{Z}_3(\beta)\), we have
\[\beta^2 = 2\beta + 1, \\ \beta^3 = 2\beta+2, \\ \cdots \\ \beta^7 = \beta+1, \\ \beta^8 = 1,\]which means that we can express all non-zero elements in \(\mathbb{Z}_3(\beta)\) as some power of \(\beta\)!
用 power 表示，算 inverse 很方便。
Recall the evaluation homomorphism; here we define it as
\[e_\alpha: F[x] \to E,\]such that \(f(x) \mapsto f(\alpha)\). Then
\[\text{ker}(e_\alpha) = \{f(x)\in F[x]\mid f(\alpha)=0 \} = \text{Irr}(\alpha, F)F[x].\]By the first ring isomorphism theorem, we have
\[F[x]\big/\text{Irr}(\alpha, F)F[x] \cong F[\alpha]=F(\alpha).\]Therefore, there are two equivalent ways to construct a algebraic field extension of \(F\):
第一種方法的好處是，我們比較容易同時觀察好幾個 algebraic field extension，因為他們都在同一代數閉包（algebraic closure）之中；第二種方法的好處是，建構過程只牽涉到 \(F\)。
Let \(\alpha\) and \(\beta\) be two distinct zeros of an irreducible polynomials \(f(x)\) over \(F\), then we have
\[F(\alpha) \cong F[x]\Big/f(x)F[x] \cong F(\beta).\]Therefore, the map \(\tau: F(\alpha) \to F(\beta)\) ,given by \(\tau(g(\alpha)) = g(\beta)\) for all \(g(x) \in F[x]\), is a field isomorphism. In other words, for \(F\) itself, it cannot tell the difference between \(\alpha\) and \(\beta\).
對於同一個 irreducible polynomial，不管加入哪個 zero 結果都一樣！
\(\Bbb{Q}(\sqrt[3]{2}) \cong \Bbb{Q}(\sqrt[3]{2}\omega)\), where \(\omega = e^{2\pi i/3}\).