# If $[E:F]$ is finite and $\alpha \in E$ then there is an irr. polynomial in $F[x]$ with root $\alpha$

I'm studying for an exam and encountered a confusing proof of the following fact in my notes:

Let $[E:F]$ be finite and $\alpha \in E$ then there is an irreducible polynomial $p(x) \in F[x]$ with $p(\alpha) = 0$

My notes start with that $\{1, \alpha, ..., \alpha^{n-1}\}$ were linearly dependent (shouldn't this say independent?) in $E$ over $F$ and defining a nonempty ideal $I = \{p(x) \in F[x] : p(\alpha) = 0 \}$ that is generated by some irreducible polynomial $p_0(x)$

It then assumes $p_0(x)$ is not irreducible and then the proof was supposed to derive a contradiction, but it seems to taper off at that point, so I probably missed something. More specifically, it says $p_0 = p_1p_2$ with the degrees of each of the polynomials on the RHS being less than the degree of $p_0$ then mentioning extension fields and thus contradiction.

Could anyone clarify?

• Let $n$ be the degree of $E$ over $F$. Then $1,\alpha, \dots, \alpha^{n}$ are linearly dependent. – André Nicolas Mar 12 '14 at 18:55
• @AndréNicolas My notes say $\{1, \alpha, ..., \alpha^{n-1}\}$ so either the professor assumed the degree was $n-1$ which would be out of the ordinary or I wrote the wrong thing. Probably the latter. – Lost Mar 12 '14 at 18:58

1) Let $n$ be the degree of $E$ over $F$. Then $1,\alpha, \dots, \alpha^{n}$ are linearly dependent over $F$.
2) Let $k$ be the smallest integer such that $1,\alpha,\dots,\alpha^k$ are linearly dependent.
• You are welcome. Yes, if we pick the least $k$, it is straightforward to show the polynomial we get is irreducible, for if not, we could come up with a smaller $k$. – André Nicolas Mar 12 '14 at 20:53