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.
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.