Question

If $\left[K:F\right]$is prime and $u\in K$is algebraic over$F$ , show that either$F\left(u\right)=K$ or$F\left(u\right)=F$ .

Short answer

It is proved$F\left(u\right)=K$ or$F\left(u\right)=F$ .

Step-by-step solution

Definition of algebraic extension

Let $K$be an extension field of$F$ . Then$K$ is said to be an algebraic extension of $F$if every element of$K$ is algebraic over $F$.

Show that Fu=K or Fu=F

Let$\left[K:F\right]=p$ and $p$be prime numbers and$u\in K$ be algebraic over$F$ .

$u\in K$is algebraic over$F$ . Therefore,$F\left(u\right)$ is an algebraic extension over$F$ .

Thus,$F\subseteq F\left(u\right)\subseteq K$ .

$\left[K:F\right]$is finite. That is$\left[K:F\right]=p$ . Therefore, consider the extension

$\left[K:F\right]=\left[K:F\left(u\right)\right]\left[F\left(u\right):F\right]$

.

As $\left[K:F\right]=p$, this implies$\left[K:F\left(u\right)\right]\left[F\left(u\right):F\right]=p$ .

So, the possible factors are

$\left[K:F\left(u\right)\right]\left[F\left(u\right):F\right]=p\times 1$or$\left[K:F\left(u\right)\right]\left[F\left(u\right):F\right]=1\times p$

.

This gives$\left[K:F\left(u\right)\right]=1or\left[F\left(u\right):F\right]=1$ .

So, $\left[K:F\left(u\right)\right]=1or\left[F\left(u\right):F\right]=1$gives

$\begin{array}{rcl}\left[K:F\left(u\right)\right]& =& 1\\ K& =& F\left(u\right)\end{array}$

And

$\begin{array}{rcl}\left[F\left(u\right):F\right]& =& 1\\ F\left(u\right)& =& F\end{array}$

Therefore, $F\left(u\right)=KorF\left(u\right)=F$.