Let $F$ be a field and let $\bar{F}$ be an algebraic closure of $F$. $K/F$ is a finite extension of degree $d$. Could you please tell me why the following conditions are equivalent?
Thanks in advance.
$\newcommand\gen[1]{\langle #1\rangle}(1\implies 2)$ Since $K/F$ is a finite separable field extension, by primitive element theorem there exists $a\in K$ such that $K=F[a]$. Let $f\in F[X]$ be the minimum polynomial of $a$ over $F$. Then $$f(X)=\prod_{i=1}^d(x-a_i)$$ where $a_i\in\bar F$ are conjugate to $a$. Consequently, applying chinese remainder theorem: \begin{align} K\otimes_F\bar F &\cong\frac{F[X]}{\gen f}\otimes_F\bar F\\ &\cong\frac{\bar F[X]}{\gen f}\\ &\cong\prod_{i=1}^d\frac{\bar F[X]}{\gen{X-a_i}}\\ &\cong\prod_{i=1}^d\bar F \end{align}
$(2\implies 3)$ Let $(a_i:1\leq i\leq d)\in\bar F^d$ be a nilpotent element so that there exists $n\in\Bbb N$ such that $a_i^n=0$ in $\bar F$. Then $a_i=0$ for every $i$, hence $(a_i:1\leq i\leq d)=0$.
$(3\implies 1)$ Let $b\in K$, $g$ be the minimum polynomial of $b$ over $F$ and $$g(X)=\prod_{i<r}(X-b_i)^{d_i}$$ its factorization in $\bar F$. As before we have \begin{align} F[b]\otimes_F\bar F &\cong\frac{K[X]}{\gen g}\otimes_F\bar F\\ &\cong\frac{\bar F[X]}{\gen g}\\ &\cong\prod_{i<r}\frac{\bar F[X]}{\gen{X-b_i}^{d_i}} \end{align} and since $F[b]\otimes_F\bar F$ is isomorphic to a subring of $K\otimes_F\bar F$, it's a reduced ring as well. Now consider the element $(X-b_i+\gen{X-b_i}^{d_i}:i<r)$ and $m=\max\{d_i:i<r\}$. Then $$(X-b_i+\gen{X-b_i}^{d_i}:i<r)^d=((X-b_i)^d+\gen{X-b_i}^{d_i}:i<r)=0$$ hence $X-b_i\in\gen{X-b_i}^{d_i}$ thus $d_i=1$ for every $i$. Then every element of $K$ is separable over $F$, hence $K/F$ is a separable extension field.
