4 ms·
That's true but I think it's implicit here we're working with the algebraic extension generated by the number itself (which always contains the Galois conjugate
by bubblyworld 3y ago
That's true but I think it's implicit here we're working with the algebraic extension generated by the number itself (which always contains the Galois conjugates).
- throwaway81523 3y agoI'm not sure if I'm using the jargon properly, but if r is the real cube root of 2, then I thought the extension generated by r is Q(r), which contains only reals. So (x^3 - 2) factors into (x-r) and the irreducible quadratic (x^2 + r*x + r^2). You have to do some extra thing to reach the splitting field, amirite?
- bubblyworld 3y agoPardon me, you're completely right! Thanks for reminding me to refresh my basics before I write =)