Let $F,K$ be fields, prove that any element in $K$ that is not in $F$ is irreducible in $F[x]$ of degree $p$.

Question

I am working on the following problem.

Assume $p$ is a prime number and $F,K$ are fields such that $F \subseteq K$ and $[K:F]=p$. Prove that every element of $K$ that is not in $F$ satisfies an irreducible polynomial in $F[x]$ of degree $p$.

I have decided to examine a smaller example to get more of an understanding.

Suppose that $K=\mathbb{Q}(\sqrt{2})$ and $F=\mathbb{Q}$. Now $[K:F]=2$ since the basis for $K$ over $F$ is $\beta=${$1,\sqrt{2}$}. Also, $\mathbb{Q}(\sqrt{2})$={$a+b\sqrt{2}|a,b\in \mathbb{Q}$}. Let $p(x) \in K$ be an irreducible polynomial, with $p(\alpha)=0$.

Now, $F \subset F(\alpha) \subset K$ and $[K:F]=[K:F(\alpha)] \cdot [F(\alpha):F]=[K:F(\alpha)] \cdot deg(p(x))=[K:F(\alpha)] \cdot 2=2$. Thus, $[K:F(\alpha)]=1 \rightarrow K=F(\alpha)$.

I hope that my above argument makes sense, but I am still not sure how to get at showing that "every element of $K$ that is not in $F$ satisfies an irreducible polynomial in $F[x]$ of degree $p$."

Any advice would be very helpful, thank you for looking.

Update

Since $[K:F]=p<\infty$, this means that $K$ is an algebraic extension of $F$, hence every element $\alpha \in K$ is algebraic over $F$ ($\alpha$ is the zero of some nonzero polynomial in $F[x]$).

Let $\alpha \in K$, then there exists a minimal polynomial $p(x) \in F[x]$ such that $p(\alpha)=0$. Then $F \subset F(\alpha) \subset K$. Now, $p=[K:F]=[K:F(\alpha)] \cdot [F(\alpha):F]=[K:F(\alpha)] \cdot deg(p(x))$.

Suppose that $deg(p(x))=1$, then $p(x)=x+c$ for $c \in F$. Since $p(\alpha)=0$, $\alpha + c=0 \rightarrow c=-\alpha$ but $\alpha \notin F$ which is a contradiction, therefore $deg(p(x))=p$.

Moreover, since $\alpha$ is algebraic over $F$ and $p(x)$ is the minimal polynomial in $F[x]$ such that $p(\alpha)=0$, $p(x)$ is irreducible over $F[x]$. $\Box$

Answer 1

Since $[K:F]=p<\infty$, this means that $K$ is an algebraic extension of $F$, hence every element $\alpha \in K$ is algebraic over $F$ ($\alpha$ is the zero of some nonzero polynomial in $F[x]$).

Let $\alpha \in K$, then there exists a minimal polynomial $p(x) \in F[x]$ such that $p(\alpha)=0$. Then $F \subset F(\alpha) \subset K$. Now, $p=[K:F]=[K:F(\alpha)] \cdot [F(\alpha):F]=[K:F(\alpha)] \cdot deg(p(x))$.

Suppose that $deg(p(x))=1$, then $p(x)=x+c$ for $c \in F$. Since $p(\alpha)=0$, $\alpha + c=0 \rightarrow c=-\alpha$ but $\alpha \notin F$ which is a contradiction, therefore $deg(p(x))=p$.

Moreover, since $\alpha$ is algebraic over $F$ and $p(x)$ is the minimal polynomial in $F[x]$ such that $p(\alpha)=0$, $p(x)$ is irreducible over $F[x]$. $\Box$

Answer 2

Your idea works in general.

Take $\alpha \in K \setminus F$ with minimal polynomial $f$ over $F$ and look at the tower of extensions $F \subset F(\alpha) \subset K$. Then $$p = [K:F] = [K : F(\alpha)] [F(\alpha):F] = [K : F(\alpha)] \deg(f).$$ Because $\alpha \not \in F$, $\deg(f) > 1$ and because $p$ is prime it follows that $\deg(f) = p$.

Answer 3

Pick $\alpha \in K$. Notice that the vectors $1,\alpha,\alpha^2,\dots,\alpha^{p}$ are linearly dependent over $K$, this gives a polynomial $P$ with $P(\alpha)=0$ clearly there must be an irreducible polynomial $Q$ that divides $P$ with $Q(\alpha)=0$.

Consider the valuation map $f:\frac{F[x]}{(Q)}\rightarrow K$ given by $f([P])=P(\alpha)$. This map is injective so it gives us an intermediate extension $F\leq f(\frac{F[x]}{(Q)})\leq K$ which has degree $\deg(Q)$, by the dimension theorem we have that $deg(Q)=1$ or $p$, the second case is clearly not true as $\alpha \not \in F$.