Normality of field extensions via tensor product

Question

Suppose $L/K$ is a finite extension of fields. Some of the properties of the extension can be characterised via properties of the tensor product $L \otimes_{K} L$. For example,

1. $L/K$ is separable iff $L \otimes_{K} L$ is reduced,
2. $L/K$ is purely inseparable iff $L \otimes_{K} L$ is a local ring,
3. $L/K$ is Galois iff $L \otimes_{K} L \cong \sum_{Gal(L/K)} L$.

Is there an analogous statement for $L/K$ normal?

Answer 1

One statement you can make is that $L/K$ is normal iff all the residue fields of $L\otimes_K L$ are isomorphic to $L$ (as extensions of $K$). Indeed, a residue field of $L\otimes_K L$ is just the image of a homomorphism from $L\otimes_K L$ to a field, which is just the compositum of the images of two embeddings of $L$ into an extension of $K$. If $L$ is normal, then the images of any two embeddings are always the same, and so the compositum is equal to that common image which is isomorphic to $L$. Conversely, if $L$ is not normal, then it has two embeddings into an algebraic closure of $K$ with different images. The compositum of these embeddings then has degree larger than $[L:K]$ and thus is not isomorphic to $L$.

You can get another characterization by splitting the extension into its inseparable and separable parts. Suppose $L/K$ is normal, and let $S$ be the separable closure of $K$ in $L$. Then $L/S$ is purely inseparable and $S/K$ is separable, and both are normal. So $S\otimes_K S$ is a product of copies of $S$, one for each automorphism of $S$. Then $L\otimes_K L$ is a product of rings $L\otimes_S L$, where in each copy, the left $L$ is given its usual $S$-algebra structure and the right $L$ is given the $S$-algebra structure coming from a chosen automorphism of $S$. But since $L$ is normal over $K$, this automorphism of $S$ extends to an automorphism of $L$, and so this $S$-algebra structure on $L$ is isomorphic to the usual one. Thus $L\otimes_K L$ is a product of copies of the local ring $L\otimes_S L$ (where both $L$'s have the usual $S$-algebra structure), and in particular $L\otimes_K L$ is a product of isomorphic local rings. Conversely, if $L\otimes_K L$ is a product of isomorphic local rings, then in particular its residue fields are all isomorphic. Since at least one residue field is isomorphic to $L$ (given by the map $L\otimes_K L\to L$ that is the identity on each copy of $L$), all the residue fields are isomorphic to $L$, so $L/K$ is normal as shown above.

To sum up, the following are equivalent for a finite extension $L/K$:

1. $L/K$ is normal.
2. All residue fields of $L\otimes_K L$ are isomorphic to $L$ (as $K$-algebras).
3. All residue fields of $L\otimes_K L$ are isomorphic to each other (as $K$-algebras).
4. $L\otimes_K L$ is isomorphic to a a product of copies of some local $K$-algebra.