Exercise 10.2.6 (Extension over finite field is simple)

Question

In the proof of Corollary 10.2.5, once we know that the group $M^	imes$ is generated by $\alpha$,
how does it follow that $M=K(\alpha)$ ?

Hassaan Naeem · Accepted Answer

extbf{Solution:}
$M^	imes = M\backslash\{0\}$ under multiplication.
We then have that  $K(\alpha) = \{ a + b \alpha :  a,b \in K\} = M$.

Richard Ganaye · Answer

\begin{proof} 
Let $L$ be a subfield of $M$ such that $K \subseteq L \subseteq M$, and $\alpha \in L$. We must show that $L = M$.

Since the group $M^	imes = \langle \alpha angle$ is generated by $\alpha \in L$, every $x \in M^	imes$ is of the form $x = \alpha^k$ for some integer $k$, so $x \in L$. Morever $0 \in L$, since $L$ is a subfield. Therefore $M \subseteq L$, where $L$ is a subfield of $M$, so $L = M$. This shows that the smallest subfield of $M$ which contains $K$ and $\alpha$ is $M$, so  $M = K(\alpha)$.
\end{proof}

Exercise 10.2.6 (Extension over finite field is simple)

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Exercise 10.2.6 (Extension over finite field is simple)

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