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Let \(K\) be a cyclic number field over \(\mathbb{Q}\) of degree \(q=7\) with Galois group \(G\) and consider the \(\mathbb{F}_2\) algebra over \(G\), also known as the group ring. This is isomorphic to \(\mathbb{F}_2[x]/(x^q+1)\). In the case \(q=7\), the ideal \((x^q+1)\) splits mod \(2\) as \((x-1)(x^4+x^2+x+1)(x^4+x^2+x)\). So even though the Galois group is cyclic, the \(\mathbb{F}_2\) algebra over it is not.

... calculating idempotents. One merely CRT's the thing and they emerge. The only problem is, they are \(x\) and \(x+x^2\), so I'm then not sure how to construct something out of those, because I missed the clue as to what scalars the 'linear combination' was using. I need to decompose \(x+x^3+x^4\). This should be simple but... see missing clue noted above.

Kim Reece@kimreece@mathstodon.xyzThen any Galois action in this \(\mathbb{F}_2\) algebra can be taken to have a component in each of these split parts. We ignore \((x-1)\) for reasons I need to articulate better, it's somehow 'already handled' by cyclotomic context in my case, but... I am told the means to represent the Galois action in this way is to develop idempotents corresponding to the particular components and then take it as a linear combination of these idempotents. That is feasible and not a problem so far as ...