I hadn't seen that but I just looked at it and it reminded me of stuff I used to do in my previous life on rings, algebras, semigroups, semigroup algebras, etc. In fact if I am not mistaken I believe I have a short proof of your theorem, namely, that one cannot have a basis S (which is closed under multiplication) for a field F relative to a proper subfield K: Pf: Suppose that the field F has a basis S closed under multiplication over the proper subfield K of F. Note that since F is a field the only ideals it has are the whole ring F and the trivial ideal {0}. Let J denote the set of all sums of the form a_1*s_1 + ... + a_n*s_n where the a_i are in K and the s_i are in S and a_1 + ... + a_n = 0. Then with a little effort one can show that J is an ideal of F. Since J contains s_1 - s_2 for distinct s_1,s_2 in S, J cannot be {0}. But J does not contain s where s is in S. So J cannot be F. This is a contradiction. // By the way, an algebraist would say that you have proved that a semigroup algebra over a nontrivial semigroup cannot be a field. In fact the proof above shows that a semigroup algebra cannot be a simple ring. The ideal in the above proof is called the "augmentation ideal". You can find it in the book at amazon: "Semigroup Algebras" by Okninski. Do a search in the book for augmentation ideal and you will find it at the bottom of page 35.