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Let \(E\) be an extension field of a field \(F\). If \(\alpha, \beta \not = 0 \in E\) are algebraic over \(F\), then so are \(\alpha+\beta\) and \(\alpha/\beta\).
Proof
It suffices to prove that \(F(\alpha, \beta)\) is a finite extension of \(F\).
Regarding \(f(x) = \text{Irr}(\beta, F)\) as a polynomial in \(F(\alpha)[x]\), we see that \(\beta\) is algebraic over \(F(\alpha)\). This implies that \(F(\alpha, \beta) = F(\alpha)(\beta)\) and we have
\[[F(\alpha, \beta):F] = [F(\alpha)(\beta):F(\alpha)][F(\alpha):F] < \infty.\]Therefore, \(F(\alpha, \beta)\) is a finite extension of \(F\). ◼
For all \(\alpha \in E\), \(\alpha\) is algebraic over \(F\) iff \([F(\alpha):F] < \infty\).
\(F(\alpha)(\beta)\) 就是 \(\big(F(\alpha) \big)(\beta)\)：先放 \(\alpha\) 再放 \(\beta\)。
Let \(E\) be an extension field of \(F\). Then the set
\[\overline{F}_E = \{\alpha\in E\mid \alpha \text{ is algebraic over } F \}\]is a subfield of \(E\), the algebraic closure of \(F\) in \(E\).
THe set \(\overline{\Bbb Q}\) of all algebraic numbers forms a field.
A field \(F\) is algebraically closed if every nonconstant polynomial in \(F[x]\) has a zero in \(F\).
An algebraically closed field \(F\) can be characterized by the property that every polynomail \(f(x)\) in \(F[x]\) factors into a product if linear factors over \(F\).
用 division algorithm 驗證 linear factor 的存在性！
This means that if \(F\) is algebraically closed, then we will not get anything new by joining zeros of polynomials in \(F[x]\) to \(F\).
所以才說 closed！
An algebraic closure \(\overline F\) of a field \(F\) is an algebraic extension of \(F\) that is algebraic closed.
An algebraic closure of a field \(F\) can be thought of as the largest field one may obtain by algebraic means.
The set \(\overline{\Bbb Q}\) of algebraic numbers is an algebraic closure of \(\Bbb Q\).
Proof
By definition, \(\overline{\Bbb Q}\) is an algebraic extension over \(\Bbb Q\). It remains to prove that \(\overline{\Bbb Q}\) is algebraically closed, i.e., if \(\alpha\) is a zero of \(f(x) \in \overline{\Bbb Q}[x]\), then \(\alpha\) is in \(\overline{\Bbb Q}\).
Suppose that \(\alpha\) is a zero of \(a_nx^n + \cdots + a_0\), where \(a_i \in \overline{\Bbb Q}\). We shall show that \([\Bbb{Q}(\alpha): \Bbb Q] < \infty\), by this lemma.
We have
\[\begin{align*} [\Bbb{Q}(\alpha): \Bbb Q] &\le [\Bbb Q(\alpha, a_0, \cdots, a_n): \Bbb{Q}] \\ &= [\Bbb Q(\alpha, a_0, \cdots, a_n): \Bbb{Q}(a_0, \cdots, a_n)][\Bbb{Q}(a_0, \cdots, a_n):\Bbb Q]. \end{align*}\]Since \(a_n(\alpha)^n + \cdots + a_0 = 0\), \(\deg(\alpha, \Bbb{Q}(a_0, \cdots, a_n)) \le n\), which implies that
\[[\Bbb Q(\alpha, a_0, \cdots, a_n): \Bbb{Q}(a_0, \cdots, a_n)] \le n.\]By using the same argument as in this proof, we see that \([\Bbb{Q}(a_0, \cdots, a_n):\Bbb Q] < \infty\).
Therefore, \([\overline{\Bbb Q}(\alpha): \Bbb Q] < \infty\) and \(\alpha\) is algebraic over \(\Bbb Q\). ◼
Every field has an algebraic closure.
A field may have several algebraic closures, but they are all isomorphic.