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Under what conditions is the field extension $F(a)=F[a]$?

Where $F[a]$ is the set of all polynomials $\sum c_ia^i$ (finite sum, coefficients in $F$).

I am doing revision of Galois Theory, I remember there are some conditions for this, but I can't seem to find it in the book I am reading.

Offhand, what I can remember is clearly we have $F[a]\subseteq F(a)$.

A trivial sufficient condition is $a\in F$, but that is hardly interesting.

Thanks for any help.

yoyostein
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1 Answers1

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A necessary and sufficient condition is that $a$ should be algebraic over $F$ (Assuming $a$ is an element of a field $K$ which contains $F$). Short argument:

One direction follows from the fact that $a.a^{-1}=1$ and $a^{-1}\in F[a]$ The other direction is a classical theorem in field theory, which follows from the fact that $$F[a] \cong \frac{F[X]}{(P(X))}$$ With $P$ the minimal polynomial of $a$ over $F$. Since $P$ is irreducible, $(P(X))$ is a maximal ideal and $F[a]$ is a field.

Jef
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  • What you call "ther other direction" is the only thing needed, since by definition $;F(a);$ is fractions field of $;F[a];$ , so minimality makes the trick. – DonAntonio Sep 04 '16 at 14:38
  • Thanks, I see the sufficient direction now. Very nice. I am kind of missing out on the necessary condition, is it something obvious? – yoyostein Sep 04 '16 at 14:45
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    I think I get it: something like if $F[a]=F(a)$, so $a^{-1}=\sum c_ia^i$, implies $0=\sum c_ia^{i+1}-1$ so $a$ is algebraic. – yoyostein Sep 04 '16 at 14:48
  • Would love to know how to show that $a$ is algebraic – Ilan Aizelman WS May 22 '19 at 09:00
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    @Ilan If $;F[a]=F(a);$ , then in particular there exists a (obviously non-zero) polynomial $;t(a)\in F[a];$ such that $$\frac1a=t(a)\implies at(a)-1=0$$ and there you have non-zero polynomial over $;F;$ of whom $;a;$ is a root... – DonAntonio May 22 '19 at 20:19