3
$\begingroup$

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 $F[x]$. $\Box$

$\endgroup$
  • 1
    $\begingroup$ The degree $[F(\alpha):F]$ is also the degree of the minimal polynomial of $\alpha$ over $F$. $\endgroup$ – Lord Shark the Unknown Jun 15 '17 at 4:25
4
$\begingroup$

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 $\deg(f) = p$.

$\endgroup$
  • $\begingroup$ I am confused, do you mean "take $\alpha \in K$ \ $F$ with minimal polynomial $f$ over $F$ and look at the tower of extensions $F \subset F(\alpha) \subset K$? Since $F \subseteq K$ by assumption? $\endgroup$ – Joe Jun 15 '17 at 15:56
1
$\begingroup$

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 $\alpha \not \in F$.

$\endgroup$
0
$\begingroup$

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 $F[x]$. $\Box$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.