This is a proposition in Artin's algebra. But I cannot understand the proof of (b) clearly. Could someone tell me how does he jump from the third step to the fourth step? Thanks so much
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Understanding the inductive proof about field extension in Artin
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$\begingroup$
abstract-algebra
polynomials
field-theory
irreducible-polynomials
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0Maybe insert $\ldots =F(\alpha_1,\ldots,\alpha_{k-1})(\alpha_k)=\ldots$ [and the final $\alpha_{\color{red} n}$ instead of $\alpha_{\color{red} k}$ is clearly a typo] – 2017-02-21
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0@Hagen von Eitzen Yes, it is. But I still don't understead the final step. How does the conclusion come from? – 2017-02-21
1 Answers
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$F(a_1,...,a_{k-1})=F'$ is a field. Thus $F(a_1,...,a_{k-1})[a_k]=F'[a_k]$ and you can reapply a) slightly modified with $F'$ instead of $F$