Algebraic Closure
An algebraic closure of a field \(F\) is an algebraic extension of \(F\) that is algebraically closed. We will prove:
Theorem. Every field has an algebraic closure, which is unique up to an isomorphism of field extensions.
Its proof invokes the Axiom of Choice when referring to Krull's theorem: if \(R\) is a ring and \(I\subset R\) is a proper ideal, then there exists a maximal ideal of \(R\) containing \(I\). The basic idea of the proof of existence goes back to Emil Artin. (Two excellent notes by Keith Conrad: [https://kconrad.math.uconn.edu/blurbs/galoistheory/algclosureshorter.pdf], [http://math.stanford.edu/~conrad/121Page/handouts/algclosure.pdf].)
Preparation 1 (Splitting Field). Let \(f(X)\) be a monic irreducible polynomial over a field \(K\). Then there exists a field extension \(L/K\) such that \(f(X)\) splits over \(L\), and such a field extension is unique up to an isomorphism. The field \(L\) is called the splitting field of \(f(X)\).
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Preparation 2 (Krull's Theorem).
TODO:
Proof of Existence. Let \(F\) be any field. Denote by \(\{f_j:j\in J\}\) the set of all monic irreducible polynomials over \(F\). Introduce indeterminates \(u_{j,1},\cdots,u_{j,d_j}\) for each \(\lambda\), where \(d_j:=\deg(f_j)\). Let \(R\) be the polynomial ring over \(F\) generated by these indeterminates. For each \(j\), consider the coefficients of the polynomial
\[f_j(x)-\prod_{i=1}^{d_j}(x-u_{j,i})=:\sum_{i=0}^{d_j-1}r_{j,i}x^i\in R[x] \]Denote by \(I\) the ideal in \(R\) generated by these coefficients. Let \(E_j\) be the splitting field of \(f_j\) over \(F\). Subtituting \(u_{j,1},\cdots,u_{j,d_j}\) by the roots of \(f_j\) in \(E_j\), we see that \(r_{j,i}\) vanishes at these roots. Consequently, \(1\notin I\) and so \(I\) is a proper ideal in \(R\). By Krull's theorm, there exists a maximal ideal \(\mathfrak{m}\) in \(R\) such that \(I\subset \mathfrak{m}\). Then the field \(K_1:=R/\mathfrak{m}\) is an algebraic extension of \(F\) such that every polynomial over \(F\) has a root in \(K_1\). By repeating inductively, we may construct a sequence of fields \(\{K_n\}_{n=1}^{\infty}\) such that each \(K_{n+1}\) is an algebraic extension of \(K_n\) and every polynomial over \(K_n\) has a root in \(K_{n+1}\). Define \(K:=\bigcup_{n=1}^{\infty}K_n\). Then \(K\) is an algebraic extension of \(F\), and every polynomial over \(K\) has a root in \(K\) itself, i.e., \(K\) is algebraically closed. Hence, \(K\) is an algebraic closure of \(F\). \(\blacksquare\)
Proof of Uniqueness.
标签：Closure,field,extension,algebraic,over,Algebraic,ideal,polynomial From： https://www.cnblogs.com/chaliceseven/p/17094303.html