Concerning Galois theory, let $A/B$ be a separable extension. Then $$A/B - \text{normal} \Leftrightarrow A \otimes_B A=A \oplus \cdots \oplus A,$$ where the sum has $n$ summands.
Is the same correct in the theory of coverings? Let $A \rightarrow B$ be a connected covering. Is then $$A \rightarrow B - \text{regular} \Leftrightarrow A \times_B A = A \sqcup \cdots \sqcup A?$$
a) For a field extension $k\hookrightarrow K$ of finite dimension $n$ we have the fundamental equivalence: $$ K \text {is Galois over k}\iff K\otimes_k K \cong K\times \cdots K=K^n (\text {isomorphism as} \: k-\text {algebras}) $$
Since "Galois" means "normal + separable", the condition "normal" is not sufficient.
For example if you consider the normal but not separable field extension $\mathbb F_2( t) \hookrightarrow\mathbb F_2(\sqrt t)$ you get $$\mathbb F_2(\sqrt t)\otimes_{F_2(t)}\mathbb F_2(\sqrt t)\cong \mathbb F_2(\sqrt t)[\epsilon ] \;(\text {with}\: \epsilon ^2=0)$$ which is not isomorphic to $\mathbb F_2(\sqrt t)\times \mathbb F_2(\sqrt t)$ since it contains the non-zero nilpotent element $\epsilon$.
b) Yes, a covering is normal (also named Galois) if and only if it trivializes itself.
The amazing similarity between a) and b) is at the heart of Grothendieck's grandiose generalization of Galois theory.