Tensor product $L \otimes_K L$ has no nilpotent elements iff $I/I^2=0$ Let $L \supset K$ be  a finite extension of fields.
The diagonal $L \otimes_K L$ we can endow
with structure of $L$ algebra via
$L \to L \otimes_K L,\ l \mapsto l \otimes 1_L$.
Especially $L \otimes_K L$ carries structure
of a $L$-module via $L$-multiplication
in first factor $l \cdot (a \otimes b) :=
la \otimes b$.
Let $d: L \otimes_K L \to L, a \otimes b \mapsto ab $
be the diagonal map and $I$ its kernel.

I want to show that $L \otimes_K L$ is reduced (has no nilpotent elements)
iff $I = I^2$, i.e. $I/I^2=0$.

Ideas: We can simplify the problem if we
choose an more accessible system of generators
for $L$-module $I/I^2$. I claim that
$a \otimes 1_L -1_L \otimes a$ for $a \in L$
generate $I/I^2$ as $L$-module.
That's because $a \otimes b =
ab \otimes 1 - a \cdot (b \otimes 1 - 1 \otimes b)$
(recall $I$ carries $L$-module structure by multiplication in first factor) and therefore for $\sum a_i \otimes b_i \in I$
we obtain by linearity
$$\sum a_i \otimes b_i =
(\sum a_i b_i) \otimes 1  - \sum a_i \cdot
 (b_i \otimes 1 - 1 \otimes b_i)= \sum a_i \cdot
 (b_i \otimes 1 - 1 \otimes b_i)$$
Therefore it suffice to show that
$L \otimes_K L$ is reduced iff
every $a \otimes 1 - 1 \otimes a, a \in L$ is
modulo $I^2$ a product of two other such
$x \otimes 1 - 1 \otimes x, y \otimes 1 - 1 \otimes y,
x, y \in L$.
How can I do it?
Note: If we realize that $I/I^2$ can be interpreted as module of Kahler differentials $\Omega_L$ and we use another more usual construction of $\Omega_L$ as certain quotient of some free $\bigoplus_iK[x_1,..., x_n] dx_i$ use some standard exact sequences between Kahler differentials then the claim
follows immediately. But I want to know if it possible to show the claim directly(!) working only with definition of $I/I^2$ instead of making a detour over quoting results known for Kahler differntials.
 A: Question: "How can we show the opposite direction."
Answer: Assume $k \subseteq l:=k\{e_1,..,e_n\}$ is finite and let $K$ be the algebraic closure of $k$ with $L:=K\otimes_k l$. There is an inclusion
$$l\otimes_k l \subseteq L \otimes_k L$$
and isomorphisms $l\otimes_k l:=A \cong k\{e_i\otimes e_j\}$ and $L\otimes_K L:=B \cong K\{e_i\otimes e_j\}$. Hence $l\otimes_k l$ is a finite dimensional $k$-algebra and $L\otimes_k L$ is a finite dimensional $K$-algebra. If $I/I^2\cong \Omega^1_{l/k}=0$ it follows $\Omega^1_{B/K}=0$. If $nil(A) \neq (0)$ there is a maximal ideal $(0)\neq m \subseteq B$ with $B/m \cong K$. There is an isomorphism
$$\Omega^1_{B/K}\otimes_B B/m \cong m/m^2=0$$
hence $m^n=m$ for all $n \geq 2$. Since $dim_K(B)<\infty$ it follows $krdim(B)=0$ hence there is a direct sum decomposition
$$B\cong B_1\oplus \cdots \oplus B_d$$
with $B_i$ artinian $K$-algebra with maximal ideal $m_i \subseteq B_i$. Since $\Omega^1_{B/K}=0$ it follows $\Omega^1_{B_i/K}=0$ hence
$$\Omega^1_{B_i/K}\otimes_{B_i} B_i/m_i \cong m_i/m_i^{2}=0$$
hence $m_i^n=m_i$ for all $n\geq 2$. Since $dim_K(B_i)< \infty$ there is an $N_i$ with $(0)=m_i^{N_i}=m_i=0$. Hence $B_i$ is a field and hence $L\otimes_k L$ is a direct sum of fields, hence $L\otimes_k L$ is reduced. There is an inclusion
$$l\otimes_k l \subseteq L\otimes_K L$$
and it follows $l\otimes_k l$ is reduced - a contradiction, hence $nil(A)=(0)$.
Hint: $K\otimes_k \Omega^1_{l/k} \cong \Omega^1_{L/K}$ and $\Omega^1_{B/K} \cong L\otimes_K \Omega^1_{L/K}\oplus \Omega^1_{L/K} \otimes_K L$. Since $B:=L\otimes_K L$ is a finite dimensional non-reduced algebra, $K$-algebra there is a maximal ideal $m \subseteq B$ and since $K$ is algebraically closed it follows $B/m \cong K$.
Example: If $k:=\mathbb{Q}$ and $K:=k(\sqrt{2})$ it follows $k \subseteq K$ is separable and $\Omega^1_{K/k}=0$. You may check explicitly that
$$K\otimes_k K \cong K \oplus K$$
is a direct sum of fields, hence $K\otimes_k K$ is reduced.
Conversely: If $A$ is reduced we may argue as follows: There is a direct sum decomposition
$$A \cong A_1 \oplus \cdots \oplus A_d$$
with $A_i$ artinian, and since $A$ is reduced it follows $A_i$ is a field for all $i$. Since $I\subseteq A$ is a maximal ideal it follows $I:=A_1\oplus \cdots \oplus (0) \oplus \cdots \oplus A_d$ and hence for any element $a\in I$ it follows the element $xa=a \in I^2$ where
$$x:=(1,..,1,0,1,..,1)\in I.$$
Hence $I^2=I$. We have proved:
Lemma: Let $k \subseteq l$ be a finite extension of fields. It follows $\Omega^1_{l/k}=0$ iff $l\otimes_k l$ is reduced.
