TITLE: Describing the idempotents in a tensor product $L\otimes_{K} L$ with $L/K$ Galois
QUESTION [5 upvotes]: Let $L/K$ be a finite Galois extension with $\mathrm{Gal}(L/K)=G$. The map $$f\colon L\otimes_{K} L \rightarrow \prod_{\sigma \in G} L$$ can be shown to be an isomorphism using the Chinese remainder theorem. Because of this, there are $n=\#G$ idempotents of $L\otimes_{K}L$ corresponding to the these idempotents on the right hand side which are equal to $1$ in precisely one component and are equal to $0$ elsewhere.
Question: is there a natural description of these idempotents as elements of $L\otimes_{K}L$?
For example, if $x\in L$ gives rise to a normal basis $\{\sigma(x):\sigma \in G\}$ of $L$ over $K$ then $\{\sigma(x)\otimes 1:\sigma \in G\}$ is an $L$-basis of $L\otimes_{K}L$ over $L$, where $L$ is acting on the right. The matrix relating this $L$-basis of $L\otimes_{K} L$ with the basis of idempotents is the matrix $(\sigma_i\sigma_j)(x)$. So there is a way to describe these elements upon choosing a normal basis, and it turns out to be independent of the choice of $x$. I was wondering whether there is a more natural description of these idempotents.
This is related to the fact that I don't really know how to write an explicit inverse for $f$ which is independent of any non canonical choices.
Edit: for example, it will be nice if there is some representation theoretic description (in terms of the $K[G\times G]$-action on $L\otimes_K L$ and the decomposition $L\otimes_K L = \wedge^2L \oplus \mathrm{Sym}^2(L)$) of the $K$-lines in which the idempotents lie.
Thanks!

REPLY [2 votes]: There are not just $n$ idempotents, but $2^n$ idempotents. For each subset $S$ of $G$, let $e_S$ be the $n$-tuple in $\prod_{\sigma \in G} L$ with $1$ in coordinates from $S$ and $0$ in coordinates not in $S$. The specific idempotents you ask about correspond to subsets $S$ of $G$ with size $1$.
What you want to describe is not the individual idempotents in your particular list, but the totality of them.  Idempotents are called orthogonal when their product is $0$: the term is taken from the analogy with orthogonal vectors in $\mathbf R^n$ relative to the dot product. For example, if $e$ is idempotent then $e$ and $1-e$ are orthogonal idempotents. Call an idempotent primitive if it is not a sum of two nonzero orthogonal idempotents.
The specific list of idempotents you are asking about is a set of primitive orthogonal idempotents that sum to $1$, and in fact that description makes them unique. See the last proposition on the page https://mathstrek.blog/2015/03/02/idempotents-and-decomposition/amp/ and keep in mind that rings there might not be commutative (unlike your example), so the words central and centrally are used a few times and that is automatic in your case due to commutativity.
For each of your preferred idempotents $e$, the idempotent $1-e$ generates a maximal ideal in the product of copies of $L$, all maximal ideals arise in this way, and they are in a bijection with the $K$-automorphisms of $L$. A generalization of this to a tensor product of two $K$-isomorphic finite extensions of $K$ (not necessarily Galois over $K$) is in Theorem 4 in https://kconrad.math.uconn.edu/blurbs/galoistheory/splittingisom.pdf.
Note: your description of $L \otimes_K L$ as a direct sum of exterior and symmetric squares is not valid if $K$ (and then also $L$) has characteristic $2$. Exterior and symmetric powers of a vector space $V$ in general should be quotients of a tensor power of $V$, not subspaces of it. Only when certain factorials are nonzero in the scalar field for $V$ can you treat the exterior and symmetric powers as subspaces in the way you did.