# Degree of minimal polynomial of given degree of field extension

This is a pretty basic question as I'm a beginner but I seem to be confusing myself. Suppose we have some $$\alpha$$ in an extension $$K$$ of $$F$$ such that $$[F(\alpha):F]=2$$. Am I safe to conclude that the minimal polynomial of $$\alpha$$ is of degree two?

To me, it makes sense as $$[F(\alpha):F]=2$$ being finite means that $$\alpha$$ is algebraic over $$F$$. But if its minimal polynomial was not degree two, but instead one, for example, then we would have $$[F(\alpha):F]=1$$, a contradiction.

I am a little hesitant because the textbook I've looked at mentions only the fact that

Hence $$\alpha$$ is algebraic over $$F$$ and satisfies a nonzero polynomial of degree at most $$[F(\alpha):F]$$

which seems to imply that it is possible for the degree of the minimal polynomial to be less than $$[F(\alpha):F]$$. Is that possible?

First, let's prove the statement you quoted from your textbook. Let $$d=[F(\alpha):F]$$. That means, by definition, that $$F(\alpha)$$ has dimension $$d$$ as a vector space over $$F$$. In particular, any $$d+1$$ elements of $$F(\alpha)$$ are linearly dependent; so the set $$\{1,\alpha,\dots,\alpha^d\}$$ is linearly dependent. That means there exists constants $$c_0,\dots,c_d$$, not all zero, such that $$c_0+c_1\alpha+\cdots+c_d\alpha^d=0$$.
It is true that the degree of the extension $$F(\alpha)/F$$ is actually equal to the degree of the minimal polynomial of $$\alpha$$ over $$F$$. I'm guessing that the quote from the textbook is from a place where they are in the middle of the process of proving equality; this one inequality is half the goal, and later the reverse inequality will complete the goal.