这是一份GLA格拉斯哥大学MATHS4105_1 /MATHS5071_1作业代写的成功案例
Now let $\beta \in \mathbf{F}(\alpha)$. If $\beta \in \mathbf{F}$, there is nothing to prove (as then $m_{\beta}(X)=$ $X-\beta$ ) so assume $\beta \notin \mathbf{F}$. Then $\beta=\sum_{i=0}^{m-1} b_{i} \alpha^{i}$ with $b_{i} \in \mathbf{F}$ (and $m=p^{r}$ ), so
$$
\beta^{p^{\prime}}=\left(\sum_{i=0}^{m-1} b_{i} \alpha^{i}\right)^{p^{\prime}}=\sum_{i=0}^{m-1} b_{i}^{p^{r}}\left(\alpha^{p^{\prime}}\right)^{i}=\sum_{i=0}^{m-1} b_{i}^{p^{\prime}} a^{i} \in \mathbf{F}
$$
so $\mathbf{F}=\mathbf{F}\left(\beta^{p^{\prime}}\right)$ and hence $\mathbf{F}\left(\beta^{p^{r}}\right) \subset \mathbf{F}(\beta)$. Then, by Lemma 5.1.4, $\beta$ is inseparable over $\mathbf{F}$. (If $\beta$ were separable over $\mathbf{F}$, we would have $\mathbf{F}(\beta)=\mathbf{F}\left(\beta^{p}\right)=$ $\mathbf{F}\left(\left(\beta^{p}\right)^{p}\right)=\cdots=\mathbf{F}\left(\beta^{p^{r}}\right)$, a contradiction.)
MATHS4105_1 /MATHS5071_1 COURSE NOTES ：
This is an application of Zorn’s Lemma. Let
$$
\mathcal{S}={(\mathbf{D}, \tau)|\mathbf{B} \subseteq \mathbf{D}, \tau: \mathbf{D} \rightarrow \mathbf{D}, \tau| \mathbf{B}=\sigma}
$$
Order $\mathcal{S}$ by $\left(\mathbf{D}{1}, \tau{1}\right) \leq\left(\mathbf{D}{2}, \tau{2}\right)$ if $\mathbf{D}{1} \subseteq \mathbf{D}{2}$ and $\tau_{2} \mid \mathbf{D}{1}=\tau{1}$. Then every chain $\left{\left(\mathbf{D}{i}, \tau{i}\right)\right}$ in $\mathcal{S}$ has a maximal element $(\mathbf{D}, \tau)$ given by
$$
\mathbf{D}=\bigcup \mathbf{D}{i}, \quad \tau(d)=\tau{i}(d) \text { if } d \in \mathbf{D}_{i}
$$