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$E/F$ is a finite extension. Prove that $E/F$ is a normal extension $\iff$ $E$ is a splitting field for some polynomial $f\in F[X]$.

An extension $E/F$ is called normal if it is algebraic and if $\alpha \in E$ then minimal polynomial of $\alpha$ splits in $E$.

My try:

Let $\{a_1,a_2,...a_n\}$ be a basis of $E/F$. Let $f_i$ be the minimal polynomial of $a_i$ over $F$. Then $f_i$ splits in $E$. Consider $f=f_1f_2\cdots f_n$. Then $f$ splits in $E$. Since $a_i$ are the roots of $f$ then $E$ is the splitting field of $E$ over $F$.

Conversely, let $\alpha \in E$ and $g$ be the minimal polynomial of $\alpha $ over $F$. To show that $g$ splits in $E$

How to show this? Any help

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  • $\begingroup$ I don’t see normality coming into your argument. And while we’re at it, you’d better tell us your definition of normality (there are several). $\endgroup$ – Lubin Jun 25 '15 at 19:51
  • $\begingroup$ @Lubin;I have provided it ;Please help now $\endgroup$ – Learnmore Jun 26 '15 at 2:04
  • $\begingroup$ @learnmore: Apologies for the off-topic comment, but I just wanted to point out that you seem to be in the habit of putting space before, and not after, punctuation. This is incorrect English orthography. For example, We are walking. They are smiling. is correct, whereas We are walking .They are smiling . is incorrect. $\endgroup$ – Zev Chonoles Jun 26 '15 at 2:12
  • $\begingroup$ Will look after that in near future. Well for the time being can you help me out to complete this proof@ZevChonoles $\endgroup$ – Learnmore Jun 26 '15 at 2:20
  • $\begingroup$ Sorry, but the only help I could give passes through another (fairly easily provably equivalent) definition of normality, namely that $E$ is normal over $F$ if every $F$-homomorphism of $E$ into an algebraically closed field containing $E$ sends $E$ into (and thus onto) itself. $\endgroup$ – Lubin Jun 26 '15 at 17:06

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