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

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!

• I'm sure you have looked at quadratic extensions of $\Bbb{Q}$. Did you get any clues from that effort? May 29, 2021 at 4:59
• @JyrkiLahtonen, if $K=\mathbb{Q}$ and $L=\mathbb{Q}(\sqrt{d})$, I can take $x=(1+\sqrt{d})/2$ which gives the normal basis $(1\pm\sqrt{d})/2$. Using the computation I outlined above, one gets the the idempotents are $e1 = (1+\sqrt{d})/2\otimes(1+\sqrt{d})/2\sqrt{d}+(1-\sqrt{d})/2\otimes(\sqrt{d}-1)/2\sqrt{d}$ and $e2 = (1+\sqrt{d})/2\otimes(\sqrt{d}-1)/2\sqrt{d}+(1-\sqrt{d})/2\otimes(1+\sqrt{d})/2\sqrt{d}$. May 30, 2021 at 13:37
• This suggests that there might be some representation theoretic description of the $K$-lines in $L\otimes_K L$ in which $e_1$ and $e_2$ lie, but I'm not sure what it will be for more general situations. May 30, 2021 at 13:41

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!

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 $$\frac{1}{p}\sum_{j=0}^{p-1} \zeta^j\otimes \zeta^j.$$

• Thanks for the answer, but as I explained in the question, I already knew this sort of thing and it wasn't the answer I was looking for. May 30, 2021 at 13:41

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.

• Thanks for the answer. I know all of these things but I don't really understand how this answers my question. Sorry if there's something I'm missing. May 30, 2021 at 18:36
• For example, for a group algebra $\mathbb{C}[G]$ there is a nice natural description of the idempotents as averages of conjugacy classes. One could ask if there is a natural explicit description of the primitive idempotents in $L\otimes_K L$. I understand they are in correspondence with other things but I am not asking about that. May 30, 2021 at 18:38
• You said in a comment to the other answer that it was also not answering the question you asked. Please edit the question to make it clearer what exactly you want to know. When I read the question "is there a natural description of these idempotents" in the tensor product, I thought a characterization of them (distinguishing them from all other idempotents) would answer your question. Evidently not. So we both misundertood you. Please edit the question to clarify what a "natural" explicit description means and why answers so far don't satisfy you compared to the answer you seek.
– KCd
May 30, 2021 at 19:35