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?


1 Answer 1


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.


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