Show that the multiplicative group $K^{\times}=K \setminus \{0\}$ is not cyclic.

Question

Suppose that $p$ is a prime and that K is an infinite field of characteristic $p\gt0$.

(1). Show that there is no algebraic element $\alpha \in K$ with the property that $K=\mathbb Z_p(\alpha)$, where $\mathbb Z_p$ is viewed as the prime field of K.

(2). Show that the multiplicative group $K^{\times}=K \setminus \{0\}$ is not cyclic.

This is an exercise problem from the textbook. I guess the purpose of these two subquestions is to lead us to consider the cases where $\alpha$ is algebraic and transcendental.

So according to subquestion (1), I should be able to prove that suppose $K^{\times}=\langle \alpha \rangle$, $\alpha$ cannot be algebraic over $\mathbb Z_p$. Then to prove (2), I am just to prove $\alpha$ cannot be transcendental either.

But I have no idea how to start. For example, I can't find the connections between $\alpha$ being algebraic, $K=\mathbb Z_p(\alpha)$ and $K^{\times}$ being not cyclic. So any help will be appreciated.

Answer 1

(1) If $\alpha \in K$ is algebraic over $\mathbb Z_p$, then $\mathbb Z_p(\alpha)/\mathbb Z_p$ is a finite extension and so $\mathbb Z_p(\alpha)$ is finite. Thus, $K$ cannot be equal to $\mathbb Z_p(\alpha)$ for $\alpha$ is algebraic, since $K$ is infinite.

(2) follows from (1) because $K^{\times}=\langle \alpha \rangle$ implies $K=\mathbb Z_p(\alpha)$.

Answer 2

For $\alpha$ algebraic over $\Bbb Z_p$, the vector space $\Bbb Z_p(\alpha)$ has $\{\alpha,...,\alpha^n\}$ as basis over the field $\Bbb Z_p$, where $n$ is the degree of minimal polynomial of $\alpha$ over $\Bbb Z_p$. Hence $|\Bbb Z_p(\alpha)|=p^n$.

Now let $\beta$ be a generator of $K^\times$. Then, $\beta\not\in \Bbb Z_p$. Similarly $\beta+1\not\in \Bbb Z_p$. Choose $m\in\Bbb N$ such that, $\beta^m=\beta+1$. Then $\beta$ satisfies the polynomial $x^m-x-1\in \Bbb Z_p[x]$. So $\beta$ is algebraic over $\Bbb Z_p$. Hence we have a contardiction as $K=\Bbb Z_p(\beta)$ and first paragraph suggests that, $|\Bbb Z_p(\beta)|<\infty$ .