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Let \(F\) be a field. A vector space over \(F\) is an additive group \(V\), together with scalar multiplication by element of \(F\), satisfying
\[\begin{align*} &a\alpha \in V, \tag{1} \\ &(ab)\alpha = a(b\alpha), \tag{2} \\ &(a+b)\alpha = a\alpha + b\alpha, \tag{3} \\ &a(\alpha+\beta) = a\alpha + a\beta, \tag{4} \\ &1\alpha = \alpha, \tag{5} \end{align*}\]for all \(a, b\in F\) and all \(\alpha,\beta \in V\).
\(\alpha, \beta \in V\) 之間的乘法沒有定義喔！如果有定義，該結構稱為 algebra，如 quaternion algebra。
Let \(E\) be an extension field of a field \(F\). Then \(E\) is a vector space over \(F\).
顯然。
Let \(\alpha\in E\), where \(E\) is a field extension of \(F\).
- If \(\alpha\) is algebraic over \(F\) with \(\deg(\alpha, F) = n\), then \(F(\alpha)\) is a finite-dimensional vector space over \(F\), with basis \(\{1, \alpha, \cdots, \alpha^{n-1} \}\). This means that \(\dim(F(\alpha)) = n\).
- If \(\alpha\) is not algebraic over \(F\), then \(F(\alpha)\) is an infinite-dimensional vector space over \(F\).
Proof
If \(\alpha\) is algebraic over \(F\), then by this theorem, \(\{1, \alpha, \cdots, \alpha^{n-1} \}\) is indeed a basis for \(F(\alpha)\).
If \(\alpha\) is transcendental over \(F\), then the vectors \(1, \alpha, \alpha^2, \cdots\) are linealy independent. Thus, \(F(\alpha)\) is not finite-dimensional. ◼
An extension field \(E\) of a field \(F\) is an algebraic extension of \(F\) if every element in \(E\) is algebraic over \(F\).
Show that \(F(\alpha)\) is an algebraic extension of \(F\), when \(\alpha\) is algebraic over \(F\).
Solution
By this theorem and this corollary, we know that \(F(\alpha)\) is finite and \(F^\times(\alpha)\) is cyclic. Now, for any non-zero \(g(\alpha) \in F(\alpha)\), let \(\text{ord}(g(\alpha)) = d\). Then we have \(f(x) = x - x^{d+1} \in F[x]\) such that
\[f\big(g(\alpha)\big) = g(\alpha) - \big(g(\alpha)\big)^{d+1} = 0.\]Therefore, the statement has been proved. ◼
If an extension field \(E\) of a field \(F\) is of finite dimension \(n\), then \(E\) is a finite extension of degree \(n\) over \(F\). We let \([E:F]\) denote the degree of \(E\) over \(F\).
定義中的 finite 指的是 dimension 有限，也就是 \(\deg(\alpha, E)\) 有限，而非 \(E\) 是 finite field。例如，\(\mathbb{Q}[i]\) is a finite extension of degree \(2\) of \(\mathbb{Q}\)，但 \(\mathbb{Q}[i]\) 有無限多個元素。
A finite extension \(E\) of a field \(F\) is an algebraic extension.
Proof
Let \(\alpha \in E\), and assume \([E:F]=n\). Then the \(n+1\) vectors \(1,\alpha,\cdots,\alpha^n\) are linearly dependent over \(F\). Thus, there are elements \(a_i\), not all zero, such that \(a_0+a_1\alpha+\cdots+a_n\alpha^n = 0\), which implies that \(\alpha\) is algebraic over \(F\). ◼
If \(E\) is a finite extension of a field \(F\), and \(K\) is a finite extension of \(E\), then \(K\) is a finite extension of \(F\) and \([K:F] = [K:E][E:F]\).
Proof
Let \(\{\alpha_1,\cdots,\alpha_m \}\) be a basis for \(E\) over \(F\), and \(\{\beta_1,\cdots,\beta_n \}\) be a basis fo r\(K\) over \(E\). It suffices to prove that
Let \(\gamma \in K\). Then \(\gamma =\sum_{i=1}^m e_i\alpha_i\) for some \(e_i \in E\) since \(\{\alpha_i\}\) is a basis for \(K\) over \(E\). Each \(e_i\) can also be expressed as \(e_i = \sum_{j=1}^n f_{ij}\beta_j\) since \(\{\beta_j\}\) is a basis for \(E\) over \(F\). Thus, \(\gamma = \sum_{i,j}f_{ij}\alpha_i\beta_j\), which implies that every element of \(K\) is a linear combination of \(\{\alpha_i\beta_j\}\) over \(F\).
Now suppose that \(f_{ij}\) are elements in \(F\) such that \(\sum_{i,j}f_{ij}\alpha_i\beta_j = 0\). Write it in the form
\[\sum_{i=1}^m \Big(\sum_{j=1}^n f_{ij}\beta_j \Big)\alpha_i = 0,\]where \(\sum_{j=1}^n f_{ij}\beta_j \in E\). Since \(\{\alpha_i\}\) are linearly independent over \(E\), \(\sum_{j=1}^n f_{ij}\beta_j\) must be zero for all \(i\). Then \(f_{ij}=0\) for all \(j\) since \(\{\beta_j\}\) are linearly independent over \(F\). ◼
If \(F_1\le \cdots\le F_n\) is a series of finite extension of fields, then \([F_n:F_1] = [F_n:F_{n-1}]\cdots[F_2:F_1]\).
Let \(E\) be an extension field of \(F\) and \(\alpha \in E\) is algebraic over \(F\). If \(\beta\in F(\alpha)\), then \(\deg(\beta, F)\) divides \(\deg(\alpha, F)\).
Proof
Since \(\beta\in F(\alpha)\), \(F(\alpha, \beta) = F(\alpha)\). Let \(E=F(\beta)\) and \(K=F(\alpha, \beta)\). Then by the above theorem, this corollary has been proved. ◼
Assume that \(E=F(\alpha, \beta)\) is an algebraic extension of \(F\). Let \(n=\deg(\alpha, F)\) and \(m=\deg(\beta, F)\). Then
- \([E:F]\le mn\),
- \(\text{lcm}(m,n)\big\vert [E:F]\).
Proof
Here goes two nice proofs.
We have \(x^3 - 2 = (x-\sqrt[3]{2})(x-\sqrt[3]{2}\omega)(x - \sqrt[3]{2}\omega^2)\), where \(\omega=e^{2\pi i\over 3}\) is a zero of \(x^2+x+1\). Then apparently, \([\Bbb Q(\sqrt[3]{2}): \Bbb Q] = [\Bbb Q(\sqrt[3]{2}\omega):\Bbb Q] = 3\). How about \([\Bbb Q(\sqrt[3]{2}\omega, \sqrt[3]{2}):\Bbb Q(\sqrt[3]{2})]\)?
First notice that \(\Bbb Q(\sqrt[3]{2}\omega, \sqrt[3]{2}) = \Bbb Q(\sqrt[3]{2}, \omega)\). Since we have already know \(\deg(\omega, \Bbb Q(\sqrt[3]{2})) = 2\), the desired value is therefore \(2\).
因為 \(\omega^2+\omega+1 = 0\)，所以能得到上述 degree 為 \(2\) 的結果。
2.1
Let \(F = \Bbb Q(\sqrt{2}, \sqrt{3})\). We can see that \([\Bbb Q(\sqrt{2}): \Bbb Q] = [\Bbb Q(\sqrt{3}): \Bbb Q] = 2\), but what about \([F:\Bbb Q(\sqrt{2})]\)?
Note that by this corollary, \([F:\Bbb Q]=2\) or \(4\). Suppose \([F:\Bbb Q] = 2\); this means that \([F:\Bbb Q(\sqrt{2})] = 1\), and that there exists \(a_0\) and \(a_1\) in \(\Bbb Q\) such that
\[\sqrt{3} = a_0 + a_1\sqrt{2},\]which is impossible. Therefore, \([F:\Bbb Q]=4\) and \([F:\Bbb Q(\sqrt{2})] = 2\).
2.2
Let \(\alpha =\sqrt{2} + \sqrt{3}\). Show that \(\Bbb Q(\sqrt{2}, \sqrt 3) = \Bbb Q(\alpha)\).
It is clear that \(\Bbb Q(\alpha)\subset \Bbb Q(\sqrt 2, \sqrt 3)\), so it remains to show that \(\Bbb Q(\alpha) \supset \Bbb Q(\sqrt 2, \sqrt 3)\), or equivalently, \([\Bbb Q(\alpha): \Bbb Q]=4\).
Solution \(\rm I\): show that \(\deg(\alpha, \Bbb Q) = 4\).
Note that
\[3 = (\alpha-\sqrt{2})^2 = \alpha^2 - 2\sqrt{2}\alpha + 2, \\ (2\sqrt{2}\alpha)^2 = (\alpha^2-1)^2, \\ \alpha^4 - 10\alpha^2 + 1 = 0.\]Thus we only need to show that \(x^4 - 10x^2 + 1\) is irreducible.
Solution \(\rm II\): show that \(\{1, \alpha, \alpha^2, \alpha^3\}\) are linearly independent, i.e., \([\Bbb Q(\alpha): \Bbb Q]\ge 4\).
Note that a basis of \(\Bbb Q(\sqrt 2, \sqrt 3)\) is
\[\beta = \{1, \sqrt 2\} \cdot \{1, \sqrt 3\} = \{1, \sqrt 2, \sqrt 3, \sqrt 6\}.\]Then
\[\begin{align*} [1]_\beta^T &= (1, 0, 0, 0), \\ [\alpha]^T_\beta &= [\sqrt{2} + \sqrt{3}]_\beta = (0, 1, 1, 0), \\ [\alpha^2]_\beta^T &= [5 + 2\sqrt 6]_\beta = (5, 0, 0, 6), \\ [\alpha^3]_\beta^T &= [11\sqrt 2 + 9\sqrt 3]_\beta = (0, 11, 9, 0). \end{align*}\]We now have a matrix and
\[\det \begin{pmatrix} 1 & 0 & 5 & 0 \\ 0 & 1 & 0 & 11 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 6 & 0 \\ \end{pmatrix} \not = 0.\]Thus \(\{1, \alpha, \alpha^2, \alpha^3\}\) are linearly independent.
以 \(\{1, \sqrt{2}, \sqrt{3}, \sqrt{6}\}\) 為基底，驗證 \(\{1, \alpha, \alpha^2, \alpha^3\}\) 是否線性獨立。因為 \(\Bbb Q(\alpha) \subset \Bbb Q(\sqrt{2}, \sqrt{3})\)，因此我們可以用 \(\Bbb Q(\sqrt{2}, \sqrt{3})\) 的基底來表示 \(\Bbb Q(\alpha)\) 的元素。
知道 \(\{1, \alpha, \alpha^2, \alpha^3\}\) 線性獨立後，我們也就知道 \(\Bbb Q(\alpha)\) over \(\Bbb Q\) 的 degree 至少是 \(4\)！
2.3
Let \(\gamma = \sqrt{2} + \sqrt{3} - \sqrt{6}\). Find \(\text{Irr}(\gamma, \Bbb Q)\).
Solution \(\rm I\)
There is no easy way to directly see a possible \(f(x)\in \Bbb Q\) such that \(f(x) = \text{Irr}(\gamma, \Bbb Q)\), unlike the case when \(\alpha = \sqrt 2 + \sqrt 3\). Thus, we regard \(\gamma\) as a vector of \(\Bbb Q(\sqrt 2, \sqrt 3)\) over \(\Bbb Q\). Then we have
\[a_0 + a_1\gamma + \cdots + a_n\gamma^n \iff a_0[1]_\beta + a_1[\gamma]_\beta + \cdots + a_n[\gamma^n]_\beta = \vec{0},\]and
\[\begin{align*} [1]_\beta^T &= (1, 0, 0, 0), \\ [\gamma]^T_\beta &= (0, 1, 1, -1), \\ [\gamma^2]^T_\beta &= (11, -6, -4, 2), \\ [\gamma^3]_\beta^T &= (-36, 29, 27, -21), \\ [\gamma^4]_\beta^T &= (265, 180, -136, 92). \end{align*}\]Note that \(\det[1, \gamma, \gamma^2, \gamma^3] \not = 0\), and
\[[\gamma^4]_\beta = 23[1]_\beta - 48[\gamma]_\beta + 22[\gamma^2]_\beta.\]Therefore,
\[\text{Irr}(\gamma, \Bbb Q) = x^4 - 22x^2 + 48 - 23.\]找到最大的 \(n\) 使得 \(\{1, \cdots, \gamma^n\}\) 線性獨立，而 \(\text{Irr}\) 就可以從第 \(n+1\) 個向量的線性組合得到！
這種方法得到的 polynomail 一定 irreducible，不須再檢查。
Solution \(\rm II\)
\[\begin{align*} &\gamma + \sqrt{6} = \sqrt{2} + \sqrt{3} \\ \implies &\gamma^2 + 2\sqrt{6}\gamma + 6 = 5 + 2\sqrt 6 \\ \implies &(\gamma^2 + 1)^2 = (2\sqrt{6}(1 - \gamma))^2 \end{align*}\]This way, we can obtain a quartic polynomial, but we still need to check whether it is irreducible.
如果要找 \(a_i\) 使得
\[a_0 + a_1\gamma + \cdots + a_n\gamma^n = 0，\]不如將 \(\gamma\) 視為向量，檢查
\[a_0[1]_\beta + a_1[\gamma]_\beta + \cdots + a_n[\gamma^n]_\beta\]是否線性獨立。如此一來就有好多線性代數工具可以使用！
Q: How to determine if a set of vectors are linearly independent?