Exercise 1.13

Question

Let $K$ be a field and let $\Sigma$ be the set of all irreducible monic polynomials $f$ in one indeterminate with coefficients in $K$. Let $A$ be the polynomial ring over $K$ generated by indeterminates $x_f$, one for each $f \in \Sigma$. Let $\mathfrak{a}$ be the ideal of $A$ generated by the polynomials $f(x_f)$ for all $f \in \Sigma$. Show that $\mathfrak{a} 
e (1)$.
  \par Let $\mathfrak{m}$ be a maximal ideal of $A$ containing $\mathfrak{a}$, and let $K_1 = A/\mathfrak{m}$. Then $K_1$ is an extension field of $K$ in which each $f \in \Sigma$ has a root. Repeat the construction with $K_1$ in place of $K$, obtaining a field $K_2$, and so on. Let $L = \bigcup_{n=1}^\infty K_n$. Then $L$ is a field in which each $f \in \Sigma$ splits completely into linear factors. Let $\overline{K}$ be the set of all elements of $L$ which are algebraic over $K$. Then $\overline{K}$ is an algebraic closure of $K$.

Amit Awasthi · Accepted Answer

\begin{proof}
  Suppose $\mathfrak{a} = (1)$, and so there exist $g_i \in A$ such that $\sum g_if_i(x_{f_i}) = 1$. The residue of this sum in $A/(f_1(x_{f_1}),\ldots,f_n(x_{f_n}))$ equals zero, which is a contradiction since $A/(f_1(x_{f_1}),\ldots,f_n(x_{f_n}))$ is isomorphic to the polynomial ring over the field $K[x_{f_1},\ldots,x_{f_n}]/(f_1(x_{f_1}),\ldots,f_n(x_{f_n}))$ with indeterminates $\{x_f \mid f \in \Sigma \setminus \{f_1,\ldots,f_n\}\}$.
  \par Every $f$ has a root $\alpha_1 \coloneqq x_f + \mathfrak{m} \in K_1$ since $f(\alpha_1) = f(x_f) + \mathfrak{m} = \mathfrak{m}$ in $A/\mathfrak{m}$.
  \par We claim if $f \in K[x]$ is of degree $m>0$, then it splits completely in $K_{m-1}$. We proceed by induction on $m$. This is clear if $m=1$ where $K_0 \coloneqq K$. If $m > 1$, by the above $f$ has a root $\alpha_1 \in K_{1}$, hence $f/(x-\alpha_1)$ has degree $m-1$ in $K_{1}$. By inductive hypothesis, $f/(x-\alpha_1)$ splits completely in $K_{m-1}$, hence $f$ splits completely in $K_{m-1}$.
  \par Now let $\overline{K}$ be the set of all elements in $L$ which are algebraic over $K$. It is a ring by Cor.~5.3, and moreover a field since if $\alpha$ is a root of $f$, then $\alpha^{-1}$ is a root of $x^{\deg f}f(x^{-1})$. It is algebraically closed since every polynomial $g \in K[x]$ has an irreducible factor $f$ which has a root in $\overline{K}$ by construction.
\end{proof}

Exercise 1.13

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Exercise 1.13

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