Describing the idempotents in a tensor product $L\otimes_{K} L$ with $L/K$ Galois

Question

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!

Answer 1

What you can do is replace one non-canonical choice (a normal basis) with a different one (a primitive element). This seems to lead to at least a decent-looking expression for the idempotents.

Choose a primitive element $\alpha$ and take its minimal polynomial $f\in K[x]$, so $L\cong K[x]/f$. Then we get $$L\otimes_KL\cong L\otimes_K K[x]/f \cong L[x]/f \cong \prod_{i=1}^n L[x]/(x-\alpha_i)$$ where $n = |G|$ and $\alpha=\alpha_1,\dots,\alpha_n\in L$ are the roots of $f$. (Since the $G$ acts simply transitively on the roots of $f$, we could just as well write the factors on the RHS as $L[x]/(x-\sigma(\alpha))$ as $\sigma$ ranges in $G$.)

Now it's a little clearer how to construct the idempotents; it is a case of Lagrange interpolation. The idempotent $(0,\dots,1,\dots,0)$ with a $1$ in the $i$th slot is given by a polynomial $p_i\in L[x]/f$ where $p_i\equiv \delta_{ij}\mod (x-\alpha_j)$ for all $j$. This is precisely the polynomial $$p_i(x) = \frac{1}{f'(\alpha_i)}\cdot \frac{f(x)}{x-\alpha_1} = \frac{1}{f'(\alpha_i)}\prod_{j\ne i}(x-\alpha_j).$$ Now you can simply read off exactly which element of $L\otimes_K L$ you get, it is $$\sum_{j=0}^{n-1} [x^j]p_i \otimes \alpha^j$$ where $[x^j]p_i$ denotes the coefficient of $x^j$ in $p_i$. This is at least gives a different perspective on how to get a handle on these idempotents.

It seems a little unrealistic to be able to say anything interesting about what the idempotents are without choosing a $K$-basis for $L$ (e.g. a normal basis as you did, or a power basis of $K[x]/f$, as in this answer). But maybe I'm wrong!

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Remark: By actually working out what $[x^j]p_i$ is, we can obtain a pretty nice expression just in terms of $f$ and $\alpha$. For this, we write $f(x) = a_nx^n + \cdots + a_0$ and for each $j\le n$ we write $f_j(x) = a_nx^j + a_{n-1}x^{j-1}+\cdots+a_{n-j}$. Then the expression for the $i$th idempotent becomes $$\sum_{j\ge 0} \frac{f_j(\alpha_i)}{f'(\alpha_i)} \otimes \alpha^j$$ so if instead we index the RHS of the isomorphism as $L\otimes_K L\to \prod_{\sigma \in G} L$ (as in the OP's question) we have that the $\sigma$'th idempotent is $$\sum_{j\ge 0} \sigma\left(\frac{f_j(\alpha)}{f'(\alpha)}\right)\otimes \alpha^j.$$ So this is the concrete realization of the fact that the $\sigma$'th idempotent should be obtained by applying $\sigma$ in the left slot of $L\otimes_K L$ to the "$1$'th" idempotent $\sum_j (f_j(\alpha)/f'(\alpha))\otimes \alpha^j$. This makes computing a lot easier. For fun, if we look at the cyclotomic extension $\mathbb Q(\zeta_{p})/\mathbb Q$ with primitive element $\zeta = e^{2\pi i/p}$ (for a prime $p$), our $f$ is $x^{p-1}+\cdots++1$ and so $f_j(x) = x^j+\cdots+1$ for all $j$. Simplifying all the expressions, one finds the idempotent $$\frac{1}{p}\sum_{j=0}^{p-1} \zeta^j\otimes \zeta^j.$$

Answer 2

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.