Separable extension and tensor product 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?


*

*K/F is separable;

*$K\otimes_F \bar{F}\simeq (\bar{F})^d$;

*The ring $K\otimes_F\bar{F}$ is reduced (i.e. every nilpotent element is zero).


Thanks in advance.
 A: $\newcommand\gen[1]{\langle #1\rangle}(1\implies 2)$ 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.
