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If $K$ is algebraic over $F$, then the following are equivalent:

(2) If $M$ is an algebraic closure of $K$ and if $\tau: K \rightarrow M$ is an $F$-homomorphism, then $\tau(K)=K$.

(3) If $F \subseteq L \subseteq K \subseteq N$ are fields and if $\sigma: L \rightarrow N$ is an $F$-homomorphism, then $\sigma(L) \subseteq K$, and there is a $\tau \in Gal(K/F)$ with $\tau_L = \sigma$.

--- I have a number of questions regarding the proof (highlighted)---

Proof: Suppose $F \subseteq L \subseteq K \subseteq N$, $\sigma:L \rightarrow N$ is an $F$-homomorphism. Since $L \subseteq K$, the extension $L/F$, the extension $L/F$ is algebraic, and $\sigma(L) \subseteq N$ is algebraic over $F$. Let $M'$ be the algebraic closure of $F$ in $N$ (1) and let $M$ be the algebraic closure of $M'$. By the isomorphism extension theorem(2) there is an extension $\rho:M \rightarrow M$ with $\rho|_L= \sigma$. Let $\tau= \rho|_K$. By 2 we have $\tau(K)=K$, so $\sigma(L) = \tau(L) \subseteq \tau(K)= K$. Thus $\tau \in Gal(K/F)$.


(1) What does this mean? This was not referred anywhere in the text. Do we define $M':=\{ \alpha \in N \,:\, \alpha \text{ is algebraic over } M \}$?

(2) I do not see how the theorem is applied. The theorem is stated as

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  • $\begingroup$ I suppose some books are better than others...and this one doesn't look very good right now. I have that book and there's nothing even similar to that in page 40...though you seem to be referring to theorem 3.28 (page 36 in my book). Imo, the most important and relevant parts in this theorem are $\;(1),\,(2),\,(4)\;$ . Number $\;(3)\;$ looks so unnecessarily messy and cumbersome that I wouldn't care about it much...and it will require a good ammount of work to understand it. $\endgroup$ – DonAntonio Jan 10 '18 at 8:43
  • $\begingroup$ @DonAntonio, yes sorry, it is theorem 3.28, p36. which book is a good reference for the stated equivalence? $\endgroup$ – CL. Jan 10 '18 at 10:16
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    $\begingroup$ @CWK Lang's "Algebra" is excellent in this particular subject. You can also take a peek at people.virginia.edu/~mve2x/7752_Spring2010/lecture17.pdf Also Hungeford's "Algebra" is very good in this. $\endgroup$ – DonAntonio Jan 10 '18 at 10:45
  • $\begingroup$ For the first question, this is Definition 1.25 on page 11 in the book. "Definition 1.25 Let $K$ be a field extension of $F$. The set $\{ a \in K : a \text{ is algebraic over } F \}$ is called the algebraic closure of $F$ in $K$." $\endgroup$ – Brahadeesh Sep 10 '18 at 20:01
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(1) The algebraic closure $M'$ of $F$ in $N$ is $$M':=\{\alpha\in N\mid \alpha\text{ is algebraic over} F\}=M\cap N.$$ Note that $N$ may not be algebraic over $F$, and the $F$-embedding $\sigma$ is actually $\sigma\colon L\to M'\subset M$, into an algebraically closed field containing $L$.

(2) I think the theorem you cited is not sufficient for the deduction (if it is stated for a finite family of polynomials), and it would be much clearer for this deduction if you use the following

Theorem. Let $L/F$ be an algebraic extension and $M$ the algebraic closure of $L$. If $\sigma\colon L\to M$ is an $F$-embedding, and $E/L$ is algebraic, then $\sigma$ can be extended to an $F$-embedding $\tau\colon E\to M$.

Proof of Theorem. Consider the set $S$ consisting of pairs $(k,\varphi)$ where $k$ is an intermediate field $L\subseteq k\subseteq E$ and $\varphi\colon k\to E$ is an $F$-embedding such that $\varphi|_L=\sigma$. Define $(k_1,\varphi_1)\geq (k_2,\varphi_2)$ iff. $k_1\supseteq k_2$ and $\varphi_1|_{k_2}=\varphi_2$. Then $(S,\geq)$ becomes a poset and each chain (totally ordered subset) $\{(k_\lambda,\varphi_\lambda)\}_{\lambda\in\Lambda}$ of $S$ has an upper bound $(\cup_{\lambda\in\Lambda} k_\lambda, \cup_{\lambda\in\Lambda} \varphi_\lambda)$ where $\cup_{\lambda\in\Lambda} \varphi_\lambda$ is defined by $\cup_{\lambda\in\Lambda} \varphi_\lambda|_{k_{\lambda'}}=\varphi_{\lambda'}$. Invoking Zorn's lemma one has a maximal element $(k',\tau)$ in $S$. Then $k'=E$, which can be seen from contraposition: if $k'$ is properly contained in $E$, then one can find an $\alpha\in E\setminus k'$, and the Theorem 3.20 in your post provides a pair $(k'',\varphi'')>(k',\tau)$ (while in fact one only needs to adjoin an element $\alpha\in E\setminus k'$ to $k'$ and extend $\tau$ to some $F$-embedding $\tau'\colon k'(\alpha)\to M$ to get the pair $(k'',\varphi'')$). Q.E.D.

By the Theorem above you can immediately figure out how the questioned deduction in your post is made.

ps. Besides Lang's Algebra, as DonAntonio recommended, Jacobson's Basic Algebra I is also a very good reference for you to learn Galois theory.

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    $\begingroup$ The theorem is stated (and proved) for an arbitrary family of polynomials in Morandi. The index $i$ ranges over an arbitrary indexing set. $\endgroup$ – Brahadeesh Sep 10 '18 at 20:06

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