Let $F$ be a field and $K$ be an extension of $F$ if $ b\in K $ is algebraic of degree $n$ over $F$ then $ [F (b):F]=n $ .
I want to know if I am given that $ [F (b):F] =n $ then can I tell that b is algebraic of degree n over F?
If not can someone give me an example please?
Thank you in advance.
Edit : I have started studying "extension fields" and I have so many doubts , the above question was one of my doubts .
I have this doubt because there is a theorem in my textbook that " The element $ a\in K $ is algebraic over F if and only if F (a) is a finite extension over F"
and the next theorem was " If $ a\in K $ is algebraic of degree n over F, then [F (a):F]=n"
I didn't understand why Mr. Herstein didn't use " if and only if " term in the second case .
In my opinion the "if" part is also true but I thought I should ask the experts.
Yes. In this case $b^n = p(b)$ where $p$ is a polynomial of degree $n-1$, so $b$ is a root of the degree $n$ polynomial $x^n - p(x)$. If $b$ were a root of a lower degree polynomial, then the elements $1,b,\dotsc,b^{n-1}$ would not be linearly independent, so the extension would not be degree $n$.
It's been a while since I studied field theory, so I'm not sure whether the assertion that $1,b,\dotsc,b^{n-1}$ form a basis needs more justification.